Question 15 Marks
Find a particular solution of the differential equation (x - y) (dx + dy) = dx - dy, given that y = -1, when x = 0. (Hint: put x - y = t)
AnswerIt is given that (x - y)(dx + dy) = dx - dy
⇒ (x - y + 1)dy = (1 - x + y)dx
$\Rightarrow \frac{d y}{d x}=\frac{1-x+y}{x-y+1}$ ......(i)
Let x – y = t
$\Rightarrow \frac{d}{d x} (x-y)=\frac{d t}{d x}$
$\Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$
Now, let us substitute the value of x-y and $\frac{d y}{d x}$ in equation (i), we get,
$1-\frac{d t}{d x}=\frac{1-t}{1+t}$
$\Rightarrow \frac{d t}{d x}=1-\left(\frac{1-t}{1+t}\right)$
$\Rightarrow \frac{d t}{d x}=\frac{(1+t)-(1-t)}{1+t}$
$\Rightarrow \frac{d t}{d x}=\frac{2 t}{1+t}$
$\Rightarrow\left(\frac{1+t}{t}\right) d t=2 d x$
$\Rightarrow\left(1+\frac{1}{t}\right) d t=2 d x$ .....(ii)
On integrating both side, we get,
t + log|t| = 2x + C
$\Rightarrow$ ( x - y) + log |x - y| = 2x + C
$\Rightarrow$ log|x – y| = x + y + C .......(iii)
Now, y = -1 at x = 0
Then, equation (iii), we get,
log 1 = 0 - 1 + C
$\Rightarrow$ C = 1
Substituting C = 1 in equation (iii), we get,
log|x - y| = x + y + 1
Therefore, a particular solution of the given differential equation is log|x - y| = x + y + 1.
View full question & answer→Question 25 Marks
Prove that $x^2 - y^2 = c(x^2 + y^2)^2$ is the general solution of differential equation $(x^3-3xy^2)dx=(y^3-3x^2y)dy$, where c is a parameter.
AnswerGiven differential equation can be rewritten as
$ \frac { d y } { d x } = \frac { x ^ { 3 } - 3 x y ^ { 2 } } { y ^ { 3 } - 3 x ^ { 2 }y }$ ...(i)
This is a homogeneous differential equation, so, put y = vx
$\Rightarrow \quad \frac { d y } { d x } = v + x \frac { d v } { d x }$
Then, Eq. (i) becomes
$ v + x \frac { d v } { d x } = \frac { x ^ { 3 } - 3 x ( v x ) ^ { 2 } } { ( v x ) ^ { 3 } - 3 x ^ { 2 } ( v x ) }$
$ \Rightarrow \quad v + x \frac { d v } { d x } = \frac { 1 - 3 v ^ { 2 } } { v ^ { 3 } - 3 v }$
$ \Rightarrow \quad x \frac { d v } { d x } = \frac { 1 - 3 v ^ { 2 } } { v ^ { 3 } - 3 v } - v$
$ \Rightarrow \quad x \frac { d v } { d x } = \frac { 1 - 3 v ^ { 2 } - v ^ { 4 } + 3 v ^ { 2 } } { v ^ { 3 } - 3 v }$
$\Rightarrow \quad x \frac { d v } { d x } = \frac { 1 - v ^ { 4 } } { v ^ { 3 } - 3 v }$
$ \Rightarrow \left( \frac { v ^ { 3 } - 3 v } { 1 - v ^ { 4 } } \right) d v = \frac { d x } { x }$
On integrating both sides, we get
$ \int \left( \frac { v ^ { 3 } - 3 v } { 1 - v ^ { 4 } } \right) d v = \int \frac { d x } { x }$
$ \Rightarrow \int \frac { v ^ { 3 } } { 1 - v ^ { 4 } } d v - 3 \int \frac { v } { 1 - v ^ { 4 } } d v = \log x + \log C$ ...(ii)
$ \Rightarrow - \frac{1}{4}\log \left( {1 - {v^4}} \right) - \frac{3}{4}\log \left| {\frac{{1 + {v^2}}}{{1 - {v^2}}}} \right| = \log x + \log C$
$\Rightarrow - \frac { 1 } { 4 } \log \left[ \left( 1 - v ^ { 4 } \right) \left( \frac { 1 + v ^ { 2 } } { 1 - v ^ { 2 } } \right) ^ { 3 } \right] = \log ( C x )$
$\Rightarrow - \frac { 1 } { 4 } \log \left[ \left( 1 - v ^ { 2 } \right) \left( 1 + v ^ { 2 } \right) \times \frac { \left( 1 + v ^ { 2 } \right) ^ { 3 } } { \left( 1 - v ^ { 2 } \right) ^ { 3 } } \right] = \log ( C x )$
$\Rightarrow \quad \log \left[ \frac { \left( 1 + v ^ { 2 } \right) ^ { 4 } } { \left( 1 - v ^ { 2 } \right) ^ { 2 } } \right] ^ { - 1 / 4 } = \log Cx$
$\Rightarrow \quad \frac { \left( 1 + v ^ { 2 } \right) ^ { 4 } } { \left( 1 - v ^ { 2 } \right) ^ { 2 } } = ( C x ) ^ { - 4 }$
$\Rightarrow \frac { \left( 1 + y ^ { 2 } / x ^ { 2 } \right) ^ { 4 } } { \left( 1 - y ^ { 2 } / x ^ { 2 } \right) ^ { 2 } } = \frac { 1 } { C ^ { 4 } x ^ { 4 } }$ $[ \because y = v x ]$
$\Rightarrow \quad \frac { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 4 } } { x ^ { 4 } \left( x ^ { 2 } - y ^ { 2 } \right) ^ { 2 } } = \frac { 1 } { C ^ { 4 } x ^ { 4 } }$
$ \Rightarrow \quad \left( x ^ { 2 } - y ^ { 2 } \right) = C ^ { 2 } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 }$ [taking square root]
$ \Rightarrow \quad \left( x ^ { 2 } - y ^ { 2 } \right) = C _ { 1 } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 }$(where $C_1$ = C$^2$).
View full question & answer→Question 35 Marks
Find a particular solution of the differential equation $ \frac { d y } { d x } + y \cot x = 4 x \; cosec \; x$, x $\neq$ 0 given that y = 0, when $ x = \frac { \pi } { 2 }$.
AnswerGiven differential equation is
$\frac { d y } { d x } + y \cot x = 4 x \; cosec \; x$
which is a linear differential equation of the form
$ \frac { d y } { d x } + P y = Q$, here P = cot x and Q = 4x cosec x
$\therefore \quad \mathrm { IF } = e ^ { \int \mathrm { Pd } \mathrm { x } } = e ^ { \int \cot x d x }$
$= e ^ { \log | \sin x | } = \sin x \quad \left[ \because e ^ { \log | x | } = x \right]$
The solution of linear differential equation is given by
$y \times \mathrm { IF } = \int ( Q \times \mathrm { IF } ) d x + C$
$\Rightarrow \quad y \times \sin x = \int 4 x\ cosec x \cdot \sin x d x + C$
$\Rightarrow \quad y \sin x = \int 4 x \cdot \frac { 1 } { \sin x } \cdot \sin x d x + c$
$\Rightarrow \quad y \sin x = \int 4 x d x + C$
$ \Rightarrow \quad y \sin x = 2 x ^ { 2 } + C$ ...(i)
Also, given that y = 0, when $ x = \frac { \pi } { 2 }$.
On putting y = 0 and $ x = \frac { \pi } { 2 }$ in Eq. (i), we get
$ 0 = 2 \times \frac { \pi ^ { 2 } } { 4 } + C \Rightarrow C = \frac { - \pi ^ { 2 } } { 2 }$
On putting $ C = - \frac { \pi ^ { 2 } } { 2 }$ in Eq. (i), we get
$ y \sin x = 2 x ^ { 2 } - \frac { \pi ^ { 2 } } { 2 }$
$ \therefore \quad y = 2 x ^ { 2 } \ cosec x - \frac { \pi ^ { 2 } } { 2 } \; cosec \; x$ [dividing both sides by sin x]
which is the required solution.
View full question & answer→Question 45 Marks
Solve the differential equation $\left( {\frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}} \right)\frac{{dx}}{{dy}}$ = 1 (x $\neq$ 0)
Answer$\left( {\frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}} \right)\frac{{dx}}{{dy}}$ = 1
$\frac{{dy}}{{dx}} = \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}$
$\frac{{dy}}{{dx}} + \frac{y}{{\sqrt x }} = \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }}$
Given differential equation is of the form:
$\frac{{dy}}{{dx}} + Py = Q$
$I.F = {e^{\int {\frac{1}{{\sqrt x }}dx} }} = {e^{2\sqrt x }}$
Solution is,
$y \times {e^{2\sqrt x }} = \int {{e^{2\sqrt x }} \times \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }}dx + c}$
$y{e^{2\sqrt x }} = \int {\frac{1}{{\sqrt x }}dx + C} $
$y{e^{2\sqrt x }} = 2\sqrt x + C$
View full question & answer→Question 55 Marks
Find the general solution $ x \frac { d y } { d x } + y - x + x y \cot x = 0 \ (x \neq 0)$.
AnswerGiven differential equation is,
$ x \frac { d y } { d x } + y - x + x y \cot x = 0$
Above equation can be written as
$ x \frac { d y } { d x } + y (1 + x \cot x ) = x$
On dividing both sides with x, we get
$ \frac { d y } { d x } + y \left( \frac { 1 + x \cot x } { x } \right) = 1$
$ \Rightarrow \frac { d y } { d x } + y \left( \frac { 1 } { x } + \cot x \right) = 1$
which is a linear differential equation of the form $ \frac { d y } { d x } + P y = Q$,
where $ P = \frac { 1 } { x } + \cot x$ and $ Q = 1.$
we know that ,
$\mathrm { IF } = e ^ { \int { Pdx } } = e ^ { \int \left( \frac { 1 } { x } + \cot x \right) d x } = e ^ { \log | x | + \log \sin x }$
$= e ^ { \log | x \sin x | } [ \because \log m + \log n = \log m n ]$
$ \Rightarrow$ IF = x sin x
$y \times { IF } = \int ( Q \times {I F } ) d x + C$
$\therefore \quad y \times x \sin x = \int 1 \times x \sin x d x + C$
$ \Rightarrow yx\sin x = \int {\mathop x\limits_I } \mathop {\sin }\limits_{II} xdx + C$
$\Rightarrow y \cdot x \sin x = x \int \sin x d x$ $- \int \left( \frac { d } { d x } ( x ) \int \sin x d x \right) d x + C$ [using integration by parts]
$\Rightarrow y x \sin x = - x \cos x - \int 1 ( - \cos x ) d x + C$
$\Rightarrow y x \sin x = - x \cos x + \int \cos x d x + C$
$ \Rightarrow y x \sin x = - x \cos x + \sin x + C$
On dividing both sides by x sin x, we get
$y = \frac { - x \cos x + \sin x + C } { x \sin x }$
$ \therefore y = - \cot x + \frac { 1 } { x } + \frac { C } { x \sin x }$
which is the required solution.
View full question & answer→Question 65 Marks
Find the general solution of $\left(1+x^{2}\right) d y+2 x y d x=\cot x d x(x \neq 0)$
AnswerIt is given that $\left(1+x^{2}\right) d y+2 x y d x=c o t x d x$
$\Rightarrow \frac{d y}{d x}+\frac{2 x y}{\left(1+x^{2}\right)}=\frac{\cot x}{1+x^{2}}$
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where $p=\frac{2 x}{\left(1+x^{2}\right)}$ and $Q=\frac{\cot x}{1+x^{2}}$)
Now, I.F. = $e^{\int p d x}=e^{\int \frac{2 x}{\left(1+x^{2}\right)} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} )=\int(\mathrm{Q} \times \mathrm{I.F}) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot\left(1+\mathrm{x}^{2}\right)=\int\left[\frac{\cot x}{1+\mathrm{x}^{2}} \cdot\left(1+\mathrm{x}^{2}\right)\right] \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot\left(1+\mathrm{x}^{2}\right)=\int \cot \mathrm{x} \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{y}\left(1+\mathrm{x}^{2}\right)=\log |\sin \mathrm{x}|+\mathrm{C}$
Therefore, the required general solution of the given differential equation is
$y\left(1+x^{2}\right)=\log |\sin x|+C$
View full question & answer→Question 75 Marks
Find the general solution of $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$
AnswerIt is given that $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$
$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}$,
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where, p = $\frac{1}{x \log x}$ and $Q=\frac{2}{x^{2}}$ )
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{xlog} \mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log |\log \mathrm{x}|}=\log \mathrm{x}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} )=\int(\mathrm{Q} \times \mathrm{I.F}) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{y} . \log \mathrm{x}=\int\left[\frac{2}{\mathrm{x}^{2}} \cdot \log \mathrm{x}\right] \mathrm{d} \mathrm{x}+\mathrm{C}$ ....(i)
Now, $\int\left[\frac{2}{x^{2}} \cdot \log x\right] d x=2 \int\left(\log x \cdot \frac{1}{x^{2}}\right) d x$
= $2\left[\log x . \int \frac{1}{x^{2}} d x-\int\left\{\frac{d}{d x}(\log x) \cdot \int \frac{1}{x^{2}} d x\right\} d x\right]$
= $2\left[\log _{\mathrm{X}}\left(-\frac{1}{\mathrm{x}}\right)-\int\left(\frac{1}{\mathrm{x}} \cdot\left(-\frac{1}{\mathrm{x}}\right)\right) \mathrm{d} \mathrm{x}\right]$
= $2\left[-\frac{\log x}{x}+\int \frac{1}{x^{2}} d x\right]$
= $2\left[-\frac{\log x}{x}-\frac{1}{x}\right]$
= $-\frac{2}{x}(1+\log x)$
Now, substituting the value in (i), we get,
$\Rightarrow \mathrm{y} \cdot \log \mathrm{x}=-\frac{2}{\mathrm{x}}(1+\log \mathrm{x})+\mathrm{C}$
Therefore, the required general solution of the given differential equation is
$\text { y. } \log x=-\frac{2}{x}(1+\log x)+C$
View full question & answer→Question 85 Marks
Find the general solution of $x \frac{d y}{d x}+2 y=x^{2} \log x$
AnswerIt is given that $x \frac{d y}{d x}+2 y=x^{2} \log x$
$\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x$
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where, p = $\frac{2}{x}$ and Q = x log x)
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{2}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{2(\log \mathrm{x})}=\mathrm{e}^{\log \mathrm{x}^{2}}=\mathrm{x}^{2}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} )=\int(\mathrm{Q} \times \mathrm{I.F}) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{x}^{2}=\int\left(\mathrm{x} \log \mathrm{x} \cdot \mathrm{x}^{2}\right) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow x^{2} y=\int\left(x^{3} \log x\right) d x+c$
$\Rightarrow x^{2} y=\log x . \int x^{3} d x-\int\left[\frac{d}{d x}(\log x) \cdot \int x^{3} d x\right] d x+c$
$\Rightarrow x^{2} y=\log x \cdot \frac{x^{4}}{4}-\int\left(\frac{1}{x} \cdot \frac{x^{4}}{4}\right) d x+c$
$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \int x^{3} d x+C$
$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \cdot \frac{x^{4}}{4}+c$
$\Rightarrow x^{2} y=\frac{1}{16} x^{4}(4 \log x-1)+c$
$\Rightarrow \mathrm{y}=\frac{1}{16} \mathrm{x}^{2}(4 \log \mathrm{x}-1)+\mathrm{Cx}^{-2}$
Therefore, the required general solution of the given differential equation
$y=\frac{1}{16} x^{2}(4 \log x-1)+C x^{-2}$
View full question & answer→Question 95 Marks
Find the general solution of $\cos ^{2} x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)$
AnswerIt is given that $\cos ^{2} \frac{d y}{d x}+y=\tan x$
$\Rightarrow \frac{d y}{d x}+\sec ^{2} x \cdot y=\sec ^{2} x \tan x$
This is equation in the form of $\frac{d y}{d x}+p y=Q ($where, $p = \sec^2 x$ and $Q = \sec^2 x \tan x)$
Now, I.F. = $\mathrm{e}^({^\int}^ \mathrm{(pdx)}=\mathrm{e}^{\int \sec ^{2} \mathrm{x} \mathrm{dx}}=\mathrm{e}^{\tan \mathrm{x}}$
Thus, the solution of the given differential equation is given by the relation:
$y(I. F )=\int(Q \times I . F ) d x+C$
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\tan \mathrm{x}}=\int \mathrm{e}^{\tan \mathrm{x}} \mathrm{dx}+\mathrm{C} ......(i)$
Now, Let t = tanx
$\Rightarrow \frac{d}{d x}(\tan x)=\frac{d t}{d x}$
$\Rightarrow \sec ^{2} x=\frac{d t}{d x}$
$\Rightarrow \sec^2x\ dx = dt$
Thus, the equation (i) becomes,
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\tan \mathrm{x}}=\int\left(\mathrm{e}^{\mathrm{t}} \cdot \mathrm{t}\right) \mathrm{d} \mathrm{t}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\tan \mathrm{x}}=\int\left(\mathrm{t} \cdot \mathrm{e}^{\mathrm{t}}\right) \mathrm{dt}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\mathrm{tanx}}=\mathrm{t} \cdot \int \mathrm{e}^{\mathrm{t}} \mathrm{dt}-\int\left(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t}) \cdot \int \mathrm{e}^{\mathrm{t}} \mathrm{dt}\right) \mathrm{dt}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\mathrm{tanx}}=\mathrm{t} \cdot \mathrm{e}^{\mathrm{t}}-\int \mathrm{e}^{\mathrm{t}} \mathrm{dt}+\mathrm{c}$
$\Rightarrow te^{\tan x} = (t – 1)e^t + C$
$\Rightarrow te^{\tan x} = (\tan x – 1)e^{\tan x} + C$
$\Rightarrow y = (\tan x -1) + C e^{-\tan x}$
Therefore, the required general solution of the given differential equation is
$y = (\tan x -1) + C e^{-\tan x}.$
View full question & answer→Question 105 Marks
Find the equation of a curve passing through the point $(0, 2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by $5.$
AnswerLet $F(x, y)$ be the curve and let $(x, y)$ be a point on the curve.
We know the slope of the tangent to the curve at $(x, y)$ is $\frac{\mathrm{d} y}{\mathrm{d} x}$
According to the given conditions, we get,
$\frac{d y}{d x}+5=x+y$
$\Rightarrow \frac{d y}{d x}-y=x-5$
This is equation in the form of $\frac{d y}{d x}+p y=Q ($where, $p = -1$ and $Q = x - 5)$
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int(-1) \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I.F.}) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{ye}^{-\mathrm{x}}=\int(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}} \mathrm{dx}+\mathrm{C}$ ...(i)
Now, $\int(\mathrm{x}-5) \mathrm{e}^{-\mathrm{x}} \mathrm{dx}=(\mathrm{x}-5) \int \mathrm{e}^{-\mathrm{x}} \mathrm{dx}-\int\left[\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-5) \cdot \int \mathrm{e}^{-\mathrm{x}} \mathrm{dx}\right] \mathrm{dx}$
$= (x-5)\left(-e^{-x}\right)-\int\left(-e^{-x}\right) d x$
$= (x-5)\left(-e^{-x}\right)+\left(-e^{-x}\right)$
$= (4-x) e^{-x}$
Thus, from equation (i), we get,
$\Rightarrow \mathrm{ye}^{-\mathrm{x}}=(4-\mathrm{x}) \mathrm{e}^{-\mathrm{x}}+\mathrm{C}$
$\Rightarrow y = 4 - x + Ce^x$
$\Rightarrow x + y - 4 = Ce^x$
Now, it is given that curve passes through $(0, 2).$
Thus, equation (ii) becomes:
$0 + 2 - 4 = C e^0$
$\Rightarrow - 2 = C$
$\Rightarrow C = -2$
Substituting $C = -2$ in equation (ii), we get,
$x + y - 4 =-2e^x$
$\Rightarrow y = 4 - x - 2e^x$
Therefore, the required general solution of the given differential equation is
$y = 4 - x - 2e^x$
View full question & answer→Question 115 Marks
Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.
AnswerLet $F(x,y)$ be the curve passing through origin and let $(x,y)$ be a point on the curve.
We know the slope of the tangent to the curve at $(x,y)$ is $\frac{d y}{d x}$.
According to the given conditions, we get,
$\frac{d y}{d x}=x+y$
$\Rightarrow \frac{d y}{d x}-y=x$
This is equation in the form of $\frac{d y}{d x}+p y=Q ($where, $p = -1$ and $Q = x)$
Now, $e^{\int p d x}=e^{\int(-1) d x}=e^{-x}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{ye}^{-\mathrm{x}}=\int \mathrm{xe}^{-\mathrm{x}} \mathrm{dx}+\mathrm{C}$ ......(i)
Now, $\int \mathrm{xe}^{-\mathrm{x}} \mathrm{dx}=\mathrm{x} \int \mathrm{e}^{-\mathrm{x}} \mathrm{dx}-\int\left[\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}) \cdot \int \mathrm{e}^{-\mathrm{x}} \mathrm{dx}\right] \mathrm{d} \mathrm{x}$
$= -x\left(e^{-x}\right)-\int\left(-e^{-x}\right) d x$
$= -\mathrm{x}\left(\mathrm{e}^{-\mathrm{x}}\right)+\left(-\mathrm{e}^{-\mathrm{x}}\right)$
$= -e^{-x}(x+1)$
Thus, from equation $(i),$ we get,
$\Rightarrow \mathrm{ye}^{-\mathrm{x}}=-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)+\mathrm{C}$
$\Rightarrow y = -(x+1) + Ce^x$
$\Rightarrow x + y + 1 = Ce^x .......(ii)$
Now, it is given that curve passes through origin.
Thus, equation (ii) becomes:
$1 = C$
$\Rightarrow C = 1$
Substituting $C = 1$ in equation $(ii),$ we get,
$x + y + 1 = e^x$
Therefore, the required general solution of the given differential equation is
$x + y + 1 = e^x$
View full question & answer→Question 125 Marks
Find the general solution of $(x+y) \frac{d y}{d x}=1$
AnswerIt is given that $(x+y) \frac{d y}{d x}=1$
$\Rightarrow \frac{d y}{d x}=\frac{1}{x+y}$
$\Rightarrow \frac{d x}{d y}=x+y$
$\Rightarrow \frac{d x}{d y}-x=y$
This is equation in the form of $\frac{d x}{d y}+p x=Q$(where, p = -1 and Q = y)
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int-\mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}$
Thus, the solution of the given differential equation is given by the relation:
$x(I . F .)=\int(Q \times I . F .) d y+C$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=\int\left[\mathrm{y} \cdot \mathrm{e}^{-\mathrm{y}}\right] \mathrm{d} \mathrm{y}+\mathrm{C}$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=\mathrm{y} \int \mathrm{e}^{-\mathrm{y}} \mathrm{dy}-\int\left[\frac{\mathrm{d}}{\mathrm{dy}}(\mathrm{y}) \int \mathrm{e}^{-\mathrm{y}} \mathrm{dy}\right] \mathrm{dy}+\mathrm{C}$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=\mathrm{y}\left(-\mathrm{e}^{-\mathrm{y}}\right)-\int\left(-\mathrm{e}^{-\mathrm{y}}\right) \mathrm{d} \mathrm{y}+\mathrm{C}$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=-\mathrm{ye}^{-\mathrm{y}}+\int \mathrm{e}^{-\mathrm{y}} \mathrm{dy}+\mathrm{C}$
$\Rightarrow \mathrm{xe}^{-\mathrm{y}}=-\mathrm{ye}^{-\mathrm{y}}-\mathrm{e}^{-\mathrm{y}}+\mathrm{C}$
$\Rightarrow x = - y – 1 + Ce^y$
$\Rightarrow x + y + 1 = Ce^y$
Therefore, the required general solution of the given differential equation is
$x + y + 1 = Ce^y$
View full question & answer→Question 135 Marks
Find the general solution of $\frac{d y}{d x}+2 y=\sin x$
AnswerIt is given that $\frac{d y}{d x}+2 y=\sin x$
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where, p = 2 and Q = sin x)
Now, I.F = $e^{\int p d x}=e^{\int 2 d x}=e^{2 x}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I.F.}) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{ye}^{2 \mathrm{x}}=\int \sin \mathrm{x} \cdot \mathrm{e}^{2 \mathrm{x}} \mathrm{d} \mathrm{x}+\mathrm{C}$ .....(i)
Let $I=\int \sin x \cdot e^{2 x} d x$
$\Rightarrow \mathrm{I}=\sin \mathrm{x} \int \mathrm{e}^{2 \mathrm{x}} \mathrm{d} \mathrm{x}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x}) \cdot \int \mathrm{e}^{ 2x }\mathrm{d} \mathrm{x}\right) \mathrm{d} \mathrm{x}$
= $\sin x \cdot \frac{e^{2 x}}{2}-\int\left(\cos x \cdot \frac{e^{2 x}}{2}\right) d x$
= $\frac{\mathrm{e}^{2 \mathrm{x}} \sin \mathrm{x}}{2}-\frac{1}{2}\left[\cos \mathrm{x} \int \mathrm{e}^{2 \mathrm{x}}-\int\left(\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\cos \mathrm{x}) \cdot \int \mathrm{e}^{2 \mathrm{x}} \mathrm{d} \mathrm{x}\right) \mathrm{d} \mathrm{x}\right]$
= $\frac{\mathrm{e}^{2 \mathrm{x}} \sin \mathrm{x}}{2}-\frac{1}{2}\left[\cos \mathrm{x} \frac{\mathrm{e}^{2 \mathrm{x}}}{2}-\int\left[(-\sin \mathrm{x}) \cdot \frac{\mathrm{e}^{2 \mathrm{x}}}{2}\right] \mathrm{d} \mathrm{x}\right]$
= $\frac{\mathrm{e}^{2 \mathrm{x}} \sin \mathrm{x}}{2}-\frac{\mathrm{e}^{2 \mathrm{x}} \cos \mathrm{x}}{4}-\frac{1}{4} \int\left(\sin \mathrm{x} \cdot \mathrm{e}^{2 \mathrm{x}}\right) \mathrm{d} \mathrm{x}$
= $\frac{e^{2 x}}{4}(2 \sin x-\cos x)-\frac{1}{4} I$
$\Rightarrow \frac{5}{4} \mathrm{I}=\frac{\mathrm{e}^{2 \mathrm{x}}}{4}(2 \sin \mathrm{x}-\cos \mathrm{x})$
$\Rightarrow \mathrm{I}=\frac{\mathrm{e}^{2 \mathrm{x}}}{5}(2 \sin \mathrm{x}-\cos \mathrm{x})$
Now, putting the value of I in (i), we get,
$\Rightarrow \mathrm{ye}^{2 \mathrm{x}}=\frac{\mathrm{e}^{2 \mathrm{x}}}{5}(2 \sin \mathrm{x}-\cos \mathrm{x})+\mathrm{C}$
$\Rightarrow \mathrm{y}=\frac{1}{5}(2 \sin \mathrm{x}-\cos \mathrm{x})+\mathrm{Ce}^{-2 \mathrm{x}}$
Therefore, the required general solution of the given differential equation is
$y=\frac{1}{5}(2 \sin x-\cos x)+C e^{-2 x}$
View full question & answer→Question 145 Marks
Show that the differential equation $y d x + x \log \left| \frac { y } { x } \right| d y - 2 x d y = 0$ is homogeneous and solve it.
AnswerAccording to the question ,
Given differential equation is
$y d x + x \log \left| \frac { y } { x } \right| d y - 2 x d y = 0$
$\Rightarrow \quad y d x = \left[ 2 x - x \log \left| \frac { y } { x } \right| \right] d y$
$\Rightarrow \quad \frac { d y } { d x } = \frac { y } { 2 x - x \log \left| \frac { y } { x } \right| }$
Now, let$F ( x , y ) = \frac { y } { 2 x - x \log \left| \frac { y } { x } \right| }$
On replace x by $\lambda x$ and y by $\lambda y$ both sides, we get
$F ( \lambda x , \lambda y ) = \frac { \lambda y } { 2 \lambda x - \lambda x \log \left| \frac { \lambda y } { \lambda x } \right| }$
$= \frac { \lambda y } { \lambda \left[ 2 x - x \log \left| \frac { y } { x } \right| \right] }$
$\Rightarrow F ( \lambda x , \lambda y ) = \lambda ^ { 0 } \frac { y } { 2 x - x \log \left| \frac { y } { x } \right| } = \lambda ^ { 0 } F ( x , y )$
So, the given differential equation is homogeneous.
On putting y = vx $\Rightarrow \frac { d y } { d x } = v + x \frac { d v } { d x }$ in Eq. (i).
we get $v + x \frac { d v } { d x } = \frac { v x } { 2 x - x \log \left| \frac { v x } { x } \right| } = \frac { v } { 2 - \log | v | }$
$\Rightarrow \quad x \frac { d v } { d x } = \frac { v } { 2 - \log | v | } - v = \frac { v - 2 v + v \log | v | } { 2 - \log | v | }$
$\Rightarrow \quad x \frac { d v } { d x } = \frac { - v + v \log | v | } { 2 - \log | v | }$
$\Rightarrow \quad \frac { 2 - \log | v | } { v \log | v | - v } d v = \frac { d x } { x }$
On integrating both sides, we get
$\int \frac { 2 - \log | v | } { v ( \log | y | - 1 ) } d v = \int \frac { d x } { x }$
On putting log |v| = t $\Rightarrow \frac { 1 } { v } d v = d t$
Then, $\int \frac { 2 - t } { t - 1 } d t = \log | x | + C$
$\Rightarrow \int \left( \frac { 1 } { t - 1 } - 1 \right) d t = \log | x | + C$
$\Rightarrow \quad \log | t - 1 | - t = \log | x | + C$
$\Rightarrow \quad \log | \log v - 1 | - \log v = \log | x | + C$ [put t = log |v|]
$\Rightarrow \quad \log \left| \frac { \log v - 1 } { v } \right| = \log | x | + C$ $\left[ \because \log m - \log n = \log \left( \frac { m } { n } \right) \right]$
$\Rightarrow \log \left| \frac { \log v - 1 } { v } \right| - \log | x | = C \Rightarrow \log \left| \frac { \log v - 1 } { v x } \right| = C$
$\therefore \quad \log \left| \frac { \log \frac { y } { x } - 1 } { y } \right| = c \left[ \because y = v x \Rightarrow v = \frac { y } { x } \right]$
which is the required solution.
View full question & answer→Question 155 Marks
Show that the differential equation $x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$ is homogeneous and solve it.
AnswerWe have
$x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right)$
$\frac{d y}{d x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}$
Let $f(x, y)=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}$
Here, putting x = kx and y = ky
$f(\mathrm{kx}, \mathrm{ky})=\frac{\mathrm{ky}-\mathrm{kx} \sin \left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)}{\mathrm{kx}}$
$f(\mathrm{kx}, \mathrm{ky})=\frac{\mathrm{k}}{\mathrm{k}} \cdot \frac{\mathrm{y}-\mathrm{x} \sin \left(\frac{\mathrm{y}}{\mathrm{x}}\right)}{\mathrm{x}}$
$= k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
$\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}-\mathrm{xsin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$
$\frac{d y}{d x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}$
To solve it we make the substitution.
y = vx
Differentiating eq. with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{vx}-\mathrm{xsin}\left(\frac{\mathrm{vx}}{\mathrm{x}}\right)}{\mathrm{x}} $
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}-\sin \mathrm{v}$
$\mathrm{x} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=-\sin \mathrm{v}$
$\frac{1}{\sin v} d v=-\frac{1}{x} d x$
$cosec \ v dv =-\frac{1}{x} d x$
Integrating both side, we get
$\int cosec v d v=-\int \frac{1}{x} d x$
log(cosecv – cotv) = -logx + logC
$\log \left(cosec \frac{y}{x}-\cot \frac{y}{x}\right)=\log \frac{c}{x}$
$cosec \frac{y}{x}-\cot \frac{y}{x}=\frac{C}{x}$
$\frac{1}{\sin \frac{y}{x}}-\frac{\cos \frac{y}{x}}{\sin \frac{y}{x}}=\frac{c}{x}$
$1-\cos \frac{y}{x}=\frac{C}{x} \cdot \sin \frac{y}{x}$
$x\left(1-\cos \frac{y}{x}\right)=\operatorname{csin} \frac{y}{x}$
The required solution of the differential equation.
View full question & answer→Question 165 Marks
Show that the differential equation of $\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y$ is homogeneous and solve it.
Answer$\frac{d y}{d x}=\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}$
Let $f(x, y)=\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}$
Here, putting x = kx and y = ky
$f(\mathrm{kx}, \mathrm{ky})=\frac{\left\{\mathrm{kx} \cos \left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)+\mathrm{kysin}\left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)\right\} \mathrm{ky}}{\left\{\mathrm{kysin}\left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)-\mathrm{kx} \cos \left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)\right\} \mathrm{kx}}$
$f(\mathrm{kx}, \mathrm{ky})=\frac{\mathrm{k}^{2}}{\mathrm{k}^{2}} \cdot \frac{\left\{\mathrm{x} \cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{y} \sin \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\} \mathrm{y}}{\left\{\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{x} \cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\} \mathrm{x}}$
$= k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
$\frac{d y}{d x}=\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}$
To solve it we make the substitution.
y = vx
Differentiating eq. with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\{\mathrm{x} \cos (\mathrm{v})+\mathrm{vxsin}(\mathrm{v})\} \mathrm{vx}}{\{\mathrm{vxsin}(\mathrm{v})-\mathrm{xcos}(\mathrm{v})\} \mathrm{x}}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\{\cos (\mathrm{v})+\mathrm{vsin}(\mathrm{v})\} \mathrm{v}}{\{\mathrm{vsin}(\mathrm{v})-\cos (\mathrm{v})\}}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\{\cos (\mathrm{v})+\operatorname{vsin}(\mathrm{v})\} \mathrm{v}}{\{\mathrm{vsin}(\mathrm{v})-\cos (\mathrm{v})\}}-\mathrm{v}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v} \cos (\mathrm{v})+\mathrm{v}^{2} \sin (\mathrm{v})-\mathrm{v}^{2} \sin (\mathrm{v})+\mathrm{v} \cos (\mathrm{v})}{\operatorname{vsin}(\mathrm{v})-\cos (\mathrm{v})}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{2 \mathrm{v} \cos (\mathrm{v})}{\mathrm{vsin}(\mathrm{v})-\cos (\mathrm{v})}$
$\frac{v \sin (v)-\cos v}{2 v \cos v} d v=\frac{1}{x} d x$
$\frac{\operatorname{vsin} \mathrm{v}}{2 \mathrm{v} \cos \mathrm{v}} \mathrm{dv}-\frac{\cos \mathrm{v}}{2 \mathrm{v} \cos \mathrm{v}} \mathrm{dv}=\frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}$
$\frac{1}{2} \operatorname{tan vd} v-\frac{1}{2} \cdot \frac{1}{v} d v=\frac{1}{x} d x$
Integrating both sides, we get
$\frac{1}{2} \int \tan v d v-\frac{1}{2} \cdot \int \frac{1}{v} d v=\int \frac{1}{x} d x$
$\frac{1}{2} \log \sec v-\frac{1}{2} \log v=\log x+\log k$
log sec v – log v = 2 log kx
$\log \sec \left(\frac{y}{x}\right)-\log \left(\frac{y}{x}\right)=2 \log k x$
$\log \left(\frac{\mathrm{x}}{\mathrm{y}} \sec \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right)=\log (\mathrm{kx})^{2}$
$\frac{x}{y} \sec \left(\frac{y}{x}\right)=k^{2} x^{2}$
$\frac{1}{x y \cos \left(\frac{y}{x}\right)}=k^{2}$
$\operatorname{xycos}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\frac{1}{\mathrm{k}^{2}}$
$C=\frac{1}{k^{2}}$
$\operatorname{xycos}\left(\frac{y}{x}\right)=c$
The required solution of the differential equation.
View full question & answer→Question 175 Marks
Show that the differential equation of xdy - ydx $= \sqrt { x ^ { 2 } + y ^ { 2 } } d x$, is homogeneous and solve it.
AnswerGiven differential equation is,
$ x d y - y d x = \sqrt { x ^ { 2 } + y ^ { 2 } } d x$
$ \Rightarrow \quad \left( y + \sqrt { x ^ { 2 } + y ^ { 2 } } \right) d x = x d y$
$ \Rightarrow \quad \frac { d y } { d x } = \frac { y } { x } + \sqrt { 1 + \frac { y ^ { 2 } } { x ^ { 2 } } }$
which is a homogeneous differential equation as $ \frac { d y } { d x } = F \left( \frac { y } { x } \right)$
put, y = vx
$ \Rightarrow \frac { d y } { d x } = v + x \frac { d v } { d x }$
$ v + x \frac { d v } { d x } = v + \sqrt { 1 + v ^ { 2 } }$
$ \Rightarrow \quad x \frac { d v } { d x } = \sqrt { 1 + v ^ { 2 } } \Rightarrow \frac { d v } { \sqrt { 1 + v ^ { 2 } } } = \frac { d x } { x }$
On integrating both sides, we get
$ \int \frac { d v } { \sqrt { 1 + v ^ { 2 } } } = \int \frac { d x } { x }$
$ \Rightarrow \quad \log \left| v + \sqrt { 1 + v ^ { 2 } } \right| = \log | x | + C$ $ \left[ {\because \int {\frac{{dx}}{{\sqrt {{a^2} + {x^2}} }}} = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right|and\int {\frac{{dx}}{x}} = \log |x|} \right]$
$ \Rightarrow \log \left| \frac { y } { x } + \sqrt { 1 + \frac { y ^ { 2 } } { x ^ { 2 } } } \right| = \log | x | + C \left[ \text { put } v = \frac { y } { x } \right]$
$ \Rightarrow \quad \log \left| \frac { y + \sqrt { x ^ { 2 } + y ^ { 2 } } } { x } \right| - \log | x | = C$
$\Rightarrow \quad \log \frac{{\left| {\frac{{y + \sqrt {{x^2} + {y^2}} }}{x}} \right|}}{x} = c$ $ \left[ \because \log m - \log n = \log \left( \frac { m } { n } \right) \right]$
$ \Rightarrow \frac { y + \sqrt { x ^ { 2 } + y ^ { 2 } } } { x ^ { 2 } } = e ^ { C } \left[ \begin{array} { l } { \text { if } \log y = x } \ { \text { then } y = e ^ { x } } \end{array} \right]$
$ \Rightarrow \quad y + \sqrt { x ^ { 2 } + y ^ { 2 } } = x ^ { 2 } \cdot e ^ { C }$
$ \therefore y + \sqrt { x ^ { 2 } + y ^ { 2 } } = A x ^ { 2 }$, where $A = e^C$
which is the required solution.
View full question & answer→Question 185 Marks
Show that the differential equation of $x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$ is homogeneous and solve it.
AnswerClearly, $\frac{d y}{d x}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$
Let, $f(x, y)=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$
Here, putting x = kx and y = ky
$f(k x, k y)=\frac{k^{2} x^{2}-2 k^{2} y^{2}+k x k y}{k^{2} x^{2}}$
$f(k x, k y)= \frac{x^{2}-2 y^{2}+x y}{x^{2}}$
$= k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
Now, $x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$
$\frac{d y}{d x}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$
To solve it we make the substitution.
y = vx
Differentiating with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{x^{2}-2 v^{2} x^{2}+x \cdot v x}{x^{2}}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}={1-2 \mathrm{v}^{2}+\mathrm{v}}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=1-2 \mathrm{v}^{2}+\mathrm{v}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=1-2 \mathrm{v}^{2}$
$\frac{1}{1-2 v^{2}} d v=\frac{1}{x} d x$
Integrating both sides, we get
$\int \frac{1}{1-2 v^{2}} d v=\int \frac{1}{x} d x$
$or ~\int \frac{1}{1^{2}-(\sqrt{2} v)^{2}} d v=\int \frac{1}{x} d x$
$\Rightarrow \frac{1}{2.\sqrt{2}} \cdot \log \left|\frac{1+\sqrt{2} v}{1-\sqrt{2 v}}\right|=\log |x|+C$
$\Rightarrow \frac{1}{2 \sqrt{2}} \log \left|\frac{1+\sqrt{2} \frac{y}{x}}{1-\sqrt{2} \frac{y}{x}}\right|=\log |x|+C$
$\Rightarrow \frac{1}{2 \sqrt{2}} \log \left|\frac{1+\sqrt{2} \frac{y}{x}}{1-\sqrt{2} \frac{y}{x}}\right|=\log |x|+C$
$\Rightarrow \frac{1}{2 \sqrt{2}} \log \left|\frac{x+\sqrt{2 y}}{x-\sqrt{2 y}}\right|=\log |x|+C$
Which is the required solution of the differential equation.
View full question & answer→Question 195 Marks
Show that the differential equation of $\left(x^{2}-y^{2}\right) d x+2 x y d y=0$ is homogeneous and solve it.
AnswerWe have
$2 x y d y=-\left(x^{2}-y^{2}\right) d x$
$\frac{d y}{d x}=-\frac{x^{2}-y^{2}}{2 x y}$
Let $f(x, y)=-\frac{x^{2}-y^{2}}{2 x y}$
Here, putting x = kx and y = ky
$f(k x, k y)=-\frac{k^{2} x^{2}-k^{2} y^{2}}{2 k^{2} x y}$
$f(k x, k y)=-\frac{k^{2}}{k^{2}} \cdot \frac{x^{2}-y^{2}}{2 x y}$
$= k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
$\left(x^{2}-y^{2}\right) d x+2 x y d y=0$
$2 x y d y=-\left(x^{2}-y^{2}\right) d x$
$\frac{d y}{d x}=-\frac{x^{2}-y^{2}}{2 x y}$
To solve it we make the substitution.
y = vx
Differentiating eq. with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{\mathrm{x}^{2}-\mathrm{v}^{2} \mathrm{x}^{2}}{2 \mathrm{x} \cdot \mathrm{vx}}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{\mathrm{x}^{2}\left(1-\mathrm{v}^{2}\right)}{2 \mathrm{vx}^{2}}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}^{2}-1}{2 \mathrm{v}}-\mathrm{v}$
$x \frac{d v}{d x}=\frac{-1-v^{2}}{2 v}$
$-\frac{2 \mathrm{v}}{1+\mathrm{v}^{2}} \mathrm{dv}=\frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}$
$\frac{2 v}{1+v^{2}} d v=-\frac{1}{x} d x$
Integrating both sides, we get
$\int \frac{2 v}{1+v^{2}} d v=-\int \frac{1}{x} d x$ .......(i)
Let $I_{1}=\int \frac{2 v}{1+v^{2}} d v$
Put $1 + v^2 = t$
2vdv = dt
$\mathrm{vdv}=\frac{1}{2} \mathrm{dt}$
$\Rightarrow \int \frac{2 v}{1+v^{2}} d v=\int \frac{1}{t} d t$ = log(t)
$\therefore log(1 + v^2)$ = -logx + logC ($\therefore$ From (i) eq.)
$\log \left(1+\left(\frac{y}{x}\right)^{2}\right)=-\log x+\log c$
$\Rightarrow x^{2}+y^{2}=C x$ is the required solution of the differential equation.
View full question & answer→Question 205 Marks
Show that the differential equation of $(x-y) d y-(x+y) d x=0$ is homogeneous and solve it.
AnswerWe have (x - y)dy = (x + y)dx
$\frac{d y}{d x}=\frac{x+y}{x-y}$
Let $f(x, y)=\frac{x+y}{x-y}$
Here, putting x = kx and y = ky
$f(k x, k y)=\frac{k x+k y}{k x-k y}$
$f(k x, k y)=\frac{x+y}{x-y}$
$= k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
(x - y) dy - (x + y) dx = 0
$\frac{d y}{d x}=\frac{x+y}{x-y}$
To solve it we make the substitution as follows ,
y = vx
Differentiating above equation with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$v+x \frac{d v}{d x}=\frac{x+v x}{x-v x}$
$v+x \frac{d v}{d x}=\frac{1+v}{1-v}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}}{1-\mathrm{v}}-\mathrm{v}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}-\mathrm{v}+\mathrm{v}^{2}}{1-\mathrm{v}}$
$x \frac{d v}{d x}=\frac{1+v^{2}}{1-v}$
$\frac{1-v}{1+v^{2}} d v=\frac{1}{x} d x$
Integrating both sides we get,
$\int \frac{1-v}{1+v^{2}} d v=\int \frac{1}{x} d x$
$\int \frac{1}{1+v^{2}} d v-\int \frac{v}{1+v^{2}} d v=\int \frac{1}{x} d x$ ......(i)
Let $I_{1}=\int \frac{v}{1+v^{2}} d v$
Put $1+v^{2}=t$
2vdv = dt
$\mathrm{vdv}=\frac{1}{2} \mathrm{dt}$
$\Rightarrow \frac{1}{2} \int \frac{1}{t} d t$
$= \frac{1}{2} \log t$
$=\frac{1}{2} \log \left|1+\mathrm{v}^{2}\right|$
$\therefore \tan ^{-1} \mathrm{v}-\frac{1}{2} \log \left(1+\mathrm{v}^{2}\right)=\log \mathrm{x}+\mathrm{C}$ ($\because$ from eq. (i))
$\Rightarrow \tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left(1+\left(\frac{y}{x}\right)^{2}\right)=\log x+C$
$\Rightarrow\tan ^{-1} \frac{y}{x}=\log x+\frac{1}{2} \log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)+C$
$\Rightarrow\tan ^{-1} \frac{y}{x}=\frac{1}{2}\left(2 \log x+\log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)\right)+C$
$\Rightarrow\tan ^{-1} \frac{y}{x}=\frac{1}{2}\left(\log \left(\frac{x^{2}+y^{2}}{x^{2}} \times x^{2}\right)\right)+c$
$\Rightarrow\tan ^{-1} \frac{y}{x}=\frac{1}{2}\log (x^{2}+y^{2})+c$
The required solution of the differential equation.
View full question & answer→Question 215 Marks
Show that the differential equation of $y' = \frac{{x + y}}{x}$, is homogeneous and solve it.
AnswerGiven: Differential equation $y' = \frac{{x + y}}{x}$
$\frac{{dy}}{{dx}} = \frac{x}{x} + \frac{y}{x}$
$ \Rightarrow \frac{{dy}}{{dx}} = 1 + \frac{y}{x} = f\left( {\frac{y}{x}} \right)$ …(i)
Therefore, eq. (i) is homogeneous.
Putting $\frac{y}{x} = v$
$\Rightarrow y = vx$
$\Rightarrow \frac{{dy}}{{dx}} = v.1 + x\frac{{dv}}{{dx}} = v + x\frac{{dv}}{{dx}}$
Putting value of y and $\frac{{dy}}{{dx}}$ in eq. (i)
$ \Rightarrow v + x\frac{{dv}}{{dx}} = 1 + v$
$ \Rightarrow x\frac{{dv}}{{dx}} = 1$
$ \Rightarrow xdv = dx$
$\Rightarrow dv = \frac{{dx}}{x}$ [Separating variables]
Integrating both sides,
$ \Rightarrow \int {1dv} = \int {\frac{{dx}}{x}} $
$\Rightarrow v = \log \left| x \right| + c$
Putting $\frac{y}{x} = v$,
$\Rightarrow \frac{y}{x} = \log \left| x \right| + c$
$\Rightarrow y = x\log \left| x \right| + xc$
View full question & answer→Question 225 Marks
Show that the differential equation $2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0$, is homogenous and find the particular solution, given that $y = 2$ when $x = 1$.
AnswerWe have
$2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$
Let $f(x, y)=\frac{2 x y+y^{2}}{2 x^{2}}$
Here, putting x = kx and y = ky
$f(k x, k y)=\frac{2 k x k y+(k y)^{2}}{2(k x)^{2}}$
= $\frac{\mathrm{k}^{2}}{\mathrm{k}^{2}} \cdot \frac{2 \mathrm{xy}+\mathrm{y}^{2}}{2 \mathrm{x}^{2}}$
= $k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
$2 \mathrm{xy}+\mathrm{y}^{2}-2 \mathrm{x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}=0$
$\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$
To solve it we make the substitution.
y = vx
Differentiating eq. with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{2 \mathrm{x} \cdot \mathrm{vx}+(\mathrm{vx})^{2}}{2 \mathrm{x}^{2}}$
$v+x \frac{d v}{d x}=\frac{2 v x^{2}+v^{2} x^{2}}{2 x^{2}}$
$v+x \frac{d v}{d x}=\frac{2 v+v^{2}}{2}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}+\frac{1}{2} \mathrm{v}^{2}$
$x \frac{d v}{d x}=\frac{1}{2} v^{2}$
$2 \frac{1}{\mathrm{v}^{2}} \mathrm{dv}=\frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}$
Integrating both sides, we get
$\int 2 \frac{1}{\mathrm{v}^{2}} \mathrm{d} \mathrm{v}=\int \frac{1}{\mathrm{x}} \mathrm{d} \mathrm{x}$
$-\frac{2}{\mathrm{v}}=\log \mathrm{x}+\mathrm{C}$
$-\frac{2}{\mathrm{y} / \mathrm{x}}=\log \mathrm{x}+\mathrm{C}$
$-\frac{2 x}{y}=\log x+c$
y = 2 when x = 1
$-\frac{2 \times 1}{2}=\log 1+c$
- 1 = C
$\therefore-\frac{2 \mathrm{x}}{\mathrm{y}}=\log \mathrm{x}-1$
$\frac{2 x}{y}=1-\log x$
$y=\frac{2 x}{1-\log |x|}: x \neq e, x \neq 0$
The required solution of the differential equation.
View full question & answer→Question 235 Marks
Show that the differential equation $\frac{d y}{d x}-\frac{y}{x}+cosec \left(\frac{y}{x}\right)=0$ is homogenous and find the particular solution, given that $y = 0$ when $x = 1$.
AnswerWe have
$\frac{d y}{d x}-\frac{y}{x}+cosec \left(\frac{y}{x}\right)=0$
$\frac{d y}{d x}=\frac{y}{x}-cosec \left(\frac{y}{x}\right)$
Let $f(x, y)=\frac{y}{x}-cosec \left(\frac{y}{x}\right)$
Here, putting x= kx and y = ky
$f(k x, k y)=\frac{k y}{k x}-cosec \left(\frac{k y}{k x}\right)$
= $\frac{y}{x}-cosec \left(\frac{y}{x}\right)$
= $k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
$\frac{d y}{d x}-\frac{y}{x}+cosec \left(\frac{y}{x}\right)=0$
$\frac{d y}{d x}=\frac{y}{x}-cosec \left(\frac{y}{x}\right)$
To solve it we make the substitution.
y = vx
Differentiating eq. with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{vx}}{\mathrm{x}}-cosec \left(\frac{\mathrm{vx}}{\mathrm{x}}\right)$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}-\operatorname{cosecv}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-cosec$
$\frac{1}{cosec\ v} d v=-\frac{1}{x} d x$
Integrating both sides, we get
$\int \sin v d v=-\int \frac{1}{x} d x$
- cosv = - logx + C
$-\cos \frac{y}{x}=-\log x+C$
y = 0 when x = 1
$-\cos \frac{0}{1}=-\log 1+C$
- 1 = C
$\therefore-\cos \frac{y}{x}=-\log x-1$
$\cos \frac{y}{x}=\log x+\log e$
$\cos \frac{y}{x}=\log |e x|$
The required solution of the differential equation.
View full question & answer→Question 245 Marks
Show that the differential equation $\left[ x \sin ^ { 2 } \left( \frac { y } { x } \right) - y \right] d x + x d y = 0$ is homogenous and find the particular solution, given that $y = \frac { \pi } { 4 }$, when $x = 1.$
Answer$[x \sin^2(\frac yx) − y]dx + xdy = 0$
$xdy = -[x \sin^2(\frac{y}{x}) − y]dx$
$\Rightarrow \frac{dy}{dx} = - \frac{[x \sin ^2(\frac{y}{x})-y]}{x} ....(1)$
$F(x, y) = -\frac{[x \sin ^2(\frac{y}{x})-y]}{x}$
$F(tx, ty) = \frac{-\left(t x \sin ^{2}(\frac{t y} {t x})-t y\right)}{t x}$
$= t^0F(x, y)$
Hence $F(x, y)$ is a homogenous function of zero degree.
Now put $y = vx$ in $(1),$
$v + x\frac{dv}{dx}$ = $\frac{-(x \sin ^2v-vx)}{x}$
Therefore $v + x\frac{dv}{dx} = -\sin^2v + v$
$\Rightarrow x \frac{dv}{dx} = -\sin^2v + v - v$
$\Rightarrow x \frac{dv}{dx} = -\sin^2v$
$\Rightarrow \large\frac{dv}{\sin ^2 v}=-\frac{dx}{x}$
$\Rightarrow \operatorname {cosec}^2v\ d=\large\frac{-dx}{x}$
Integrating on both sides,
$\int cosec ^2 v\; dv=-\int \large\frac{dx}{x}$
$-\cot v = -\log x - \log c$
Now substituting for v,
$-\cot\left( {\frac{y}{x}} \right) = -(\log cx)$
$\cot\left( {\frac{y}{x}} \right) = \log cx$
$e^{\cot (y/x)}=cx$
When $y = \frac{\pi}{4}$ and $x = 1,$
$\cot \frac{\pi}{4} = 1$, so $e^1 = c$
Hence $e^{\cot( \frac{y}{x})}= ex$
$\Rightarrow$ $e^{\cot (x / y)-1} = x$
View full question & answer→Question 255 Marks
Show that the differential equation $x^{2} d y+\left(x y+y^{2}\right) d x=0$ is homogenous and find the particular solution, given that $y = 1$ when $x = 1$.
AnswerWe have
$x^2dy + (xy + y^2)dx = 0$
$\frac{d y}{d x}=-\frac{\left(x y+y^{2}\right)}{x^{2}}$
Let $f(x, y)=-\frac{\left(x y+y^{2}\right)}{x^{2}}$
Here, putting x = kx and y = ky
$f(k x, k y)=-\frac{\left(k x k y+k^{2} y^{2}\right)}{k^{2} x^{2}}$
= $\frac{\mathrm{k}^{2}}{\mathrm{k}^{2}} \cdot-\frac{\left(\mathrm{xy}+\mathrm{y}^{2}\right)}{\mathrm{x}^{2}}$
= $k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
$x^2dy + (xy + y^2)dx = 0$
$\frac{d y}{d x}=-\frac{\left(x y+y^{2}\right)}{x^{2}}$
To solve it we make the substitution.
y = vx
Differentiating above equation with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{\left(\mathrm{x} \cdot \mathrm{vx}+\mathrm{v}^{2} \mathrm{x}^{2}\right)}{\mathrm{x}^{2}}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{\left(\mathrm{vx}^{2}+\mathrm{v}^{2} \mathrm{x}^{2}\right)}{\mathrm{x}^{2}}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\mathrm{v}-\mathrm{v}^{2}$
$\mathrm{x} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=-\mathrm{v}(\mathrm{v}+2)$
$\frac{1}{v(v+2)} d v=-\frac{1}{x} d x$
Integrating both sides, we get
$\int \frac{1}{v(v+2)} d v=-\int \frac{1}{x} d x$
$\frac{1}{2} \int \frac{2}{v(v+2)} d v=-\int \frac{1}{x} d x$
$\frac{1}{2} \int \frac{2+v-v}{v(v+2)} d v=-\int \frac{1}{x} d x$
$\frac{1}{2} \int\left(\frac{2+v}{v(v+2)}-\frac{v}{v(v+2)}\right) d v=-\int \frac{1}{x} d x$
$\frac{1}{2} \int\left(\frac{1}{v}-\frac{1}{v+2}\right) d v=-\int \frac{1}{x} d x$
$\frac{1}{2}(\log v-\log (v+2))=-\log x+\log C$
$\frac{1}{2}\left(\log \frac{v}{v+2}\right)=\log \frac{C}{x}$
$\log \left(\frac{\frac{y}{x}}{\frac{y}{x}+2}\right)=2 \log \frac{c}{x}$
$\log \left(\frac{\mathrm{y}}{\mathrm{y}+2 \mathrm{x}}\right)=\log \left(\frac{\mathrm{c}}{\mathrm{x}}\right)^{2}$
$\frac{\mathrm{y}}{\mathrm{y}+2 \mathrm{x}}=\left(\frac{\mathrm{c}}{\mathrm{x}}\right)^{2}$
$\frac{x^{2} y}{y+2 x}=c^{2}$
y = 1 when x = 1
$C^{2}=\frac{1}{1+2}=\frac{1}{3}$
$\therefore \frac{x^{2} y}{y+2 x}=\frac{1}{3}$
$3x^2y = y + 2x$
$y + 2x = 3x^2y$
The required solution of the differential equation.
View full question & answer→Question 265 Marks
Show that the differential equation $(x + y) dy + (x - y) dx = 0$, is homogenous and find the particular solution, given that $y = 1$ when $x = 1$.
Answer(x + y)dy + (x - y)dx = 0
$\frac{d y}{d x}=-\frac{(x-y)}{(x+y)}$
Let $f(x, y)=-\frac{(x-y)}{(x+y)}$
Here, putting x= kx and y = ky
$f(\mathrm{kx}, \mathrm{ky})=-\frac{(\mathrm{kx}-\mathrm{ky})}{(\mathrm{kx}+\mathrm{ky})} = k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
Now,
$\frac{d y}{d x}=-\frac{(x-y)}{(x+y)}$
To solve it we make the substitution.
y = vx ...(i)
Differentiating equation (i), with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{(\mathrm{x}-\mathrm{vx})}{(\mathrm{x}+\mathrm{vx})}$
$\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{(1-\mathrm{v})}{(1+\mathrm{v})}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{(1-\mathrm{v})}{(1+\mathrm{v})}-\mathrm{v}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-1+\mathrm{v}-\mathrm{v}-\mathrm{v}^{2}}{(1+\mathrm{v})}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-1-\mathrm{v}^{2}}{(1+\mathrm{v})}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-\left(1+\mathrm{v}^{2}\right)}{(1+\mathrm{v})}$
$\frac{1+v}{1+v^{2}} d v=-\frac{1}{x} d x$
Integrating both sides, we get
$\int \frac{1+v}{1+v^{2}} d v=-\int \frac{1}{x} d x$
$\int \frac{1}{1+v^{2}} d v+\int \frac{v}{1+v^{2}} d v=-\int \frac{1}{x} d x$
$\tan ^{-1} v+\frac{1}{2} \log \left(1+v^{2}\right)=-\log x+C$
$\tan ^{-1} \frac{y}{x}+\frac{1}{2} \log \left(1+\left(\frac{y}{x}\right)^{2}\right)=-\log x+C$
y = 1 when x = 1
$\tan ^{-1} \frac{1}{1}+\frac{1}{2} \log \left(1+\left(\frac{1}{1}\right)^{2}\right)=-\log 1+C$
$\frac{\pi}{4}+\frac{1}{2} \log 2=0+c$
$C=\frac{\pi}{4}+\frac{1}{2} \log 2$
$\therefore \tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}+\frac{1}{2} \log \left(1+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}\right)=-\log \mathrm{x}+\mathrm{C}$
where, $C=\frac{\pi}{4}+\frac{1}{2} \log 2$
$\therefore \tan ^{-1} \frac{y}{x}+\frac{1}{2} \log \left(1+\left(\frac{y}{x}\right)^{2}\right)$
= $-\log x+\frac{\pi}{4}+\frac{1}{2} \log 2$
$2 \tan ^{-1} \frac{y}{x}+\log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)$
= $-2 \log x+\frac{\pi}{2}+\log 2$
$2 \tan ^{-1} \frac{y}{x}+\log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)+\log x^{2}=\frac{\pi}{2}+\log 2$
$2 \tan ^{-1} \frac{y}{x}+\log \left(x^{2}+y^{2}\right)=\frac{\pi}{2}+\log 2$
The required solution of the differential equation.
View full question & answer→Question 275 Marks
Show that the differential equation $\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$ is homogeneous and solve it.
AnswerWe have
$\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right) \mathrm{dx}+\mathrm{e^x} ^/\mathrm{^y}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}=0$
$\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right) \mathrm{dx}=-\mathrm{e^x} \mathrm{^{/y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}$
$\frac{d y}{d x}=\frac{-e^{x / y}\left(1-\frac{x}{y}\right)}{\left(1+e^{x / y}\right)}$
Let $f(x, y)=\frac{-e^{x / y}\left(1-\frac{x}{y}\right)}{\left(1+e^{x / y}\right)}$
Here, putting x = kx and y = ky
$\mathrm{f}(\mathrm{kx}, \mathrm{ky})=\frac{-\mathrm{e}^{\mathrm{kx} / \mathrm{ky}}\left(1-\frac{\mathrm{kx}}{\mathrm{ky}}\right)}{\left(1+\mathrm{e}^{\mathrm{kx} / \mathrm{ky}}\right)}$
$=\frac{-\mathrm{e}^{\mathrm{x} / \mathrm{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right)}{\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right)}$
$= k^0f(x,y)$
Therefore, the given differential equation is homogeneous.
$\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right) \mathrm{dx}+\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}=0$
$\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right) \mathrm{dx}=-\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}$
$\frac{d x}{d y}=\frac{-e^{x / y}\left(1-\frac{x}{y}\right)}{\left(1+e^{x / y}\right)}$
To solve it we make the substitution.
x = vy
Differentiation above equation with respect to x, we get
$\frac{d x}{d y}=v+y \frac{d v}{d y}$
$\mathrm{v}+\mathrm{y} \frac{\mathrm{dv}}{\mathrm{dy}}=\frac{-\mathrm{e}^{\mathrm{vy} / \mathrm{y}}\left(1-\frac{\mathrm{vy}}{\mathrm{y}}\right)}{(1+\mathrm{ev} / \mathrm{y})}$
$\mathrm{y} \frac{\mathrm{dv}}{\mathrm{dy}}=\frac{-\mathrm{e}^{\mathrm{v}}(1-\mathrm{v})}{1+\mathrm{e}^{\mathrm{v}}}$
$\frac{1+\mathrm{e}^{\mathrm{v}}}{\mathrm{e}^{\mathrm{v}}+\mathrm{v}} \mathrm{dv}=-\frac{1}{\mathrm{y}} \mathrm{dy}$
Integrating both sides, we get
$\int \frac{1+\mathrm{e}^{\mathrm{v}}}{\mathrm{e}^{\mathrm{v}}+\mathrm{v}} \mathrm{dv}=\int \frac{1}{\mathrm{y}} \mathrm{d} \mathrm{y}$ ......(i)
Let $I_{1}=\int \frac{1+e^{v}}{e^{v}+v} d v$
Put $e^v + v = t$
$(e^v + 1)dv = dt$
$\mathrm{e}^{\mathrm{v}}+1=\frac{\mathrm{dt}}{\mathrm{dv}}$
$\mathrm{dv}=\frac{\mathrm{dt}}{\mathrm{e}^{\mathrm{v}}+1}$
$\int \frac{1}{t} d t$
log t
$\log(e^v + v)$
$\therefore \log(e^v + v) = - logy + logC$ ($\therefore$ From (i) eq.)
$\log \left(e^{x / y}+\frac{x}{y}\right)=-\log y+\log C$
$\log \left(\mathrm{e}^{\mathrm{x} / \mathrm{y}}+\frac{\mathrm{x}}{\mathrm{y}}\right)=\log \frac{\mathrm{C}}{\mathrm{y}}$
$\mathrm{e}^{\mathrm{x} / \mathrm{y}}+\frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{C}}{\mathrm{y}}$
Multiply by y on both side, we get
$ye^{x/y} + x = C$
$x + ye^{x/y} = C$
The required solution of the differential equation.
View full question & answer→Question 285 Marks
Show that the differential equation of $\left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x$ is homogeneous and solve it.
AnswerWe have $\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}$
Let $f(x, y)=\frac{x^{2}+y^{2}}{x^{2}+x y}$
Here, putting x = kx and y = ky
$f(k x, k y)=\frac{(k x)^{2}+(k y)^{2}}{(k x)^{2}+k x . k y}$ = $k^0.f(x,y)$
Therefore, the given differential equation is homogeneous.
$(x^2 + xy)dy = (x^2 + y^2)dx$
$\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}$
To solve it we make the substitution.
y = vx
Differentiating above eq. with respect to x, we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$v+x \frac{d v}{d x}=\frac{x^{2}+(v x)^{2}}{x^{2}+x \cdot v x}$
$v+x \frac{d v}{d x}=\frac{x^{2}\left(1+v^{2}\right)}{x^{2}(1+v)}$
$v+x \frac{d v}{d x}=\frac{1+v^{2}}{1+v}$
$\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{v}^{2}}{1+\mathrm{v}}-\mathrm{v}=\frac{1+\mathrm{v}^{2}-\mathrm{v}-\mathrm{v}^{2}}{1+\mathrm{v}}$
$x \frac{d v}{d x}=\frac{1-v}{1+v}$
$\frac{1+v}{1-v} d v=\frac{1}{x} d x$
Integrating on both side,
$\int \frac{1+v}{1-v} d v=\int \frac{1}{x} d x$
$\int\left(-1+\frac{2}{1-v}\right) d v=\int \frac{1}{x} d x$
- v - 2log|1 - v| = log|x| + log c
$-\frac{y}{x}-2 \log \left|1-\frac{y}{x}\right|=\log |x|+\log C$
$-\frac{y}{x}=2 \log \left|1-\frac{y}{x}\right|+\log |x|+\log C$
$-\frac{y}{x}=\log \frac{(x-y)^{2}}{x^{2}}+\log |x|+\log C$
$-\frac{y}{x}=\log \frac{(x-y)^{2}}{x^{2}} \cdot C x$
$-\frac{y}{x}=\log \frac{(x-y)^{2}}{x} c$
$-\frac{y}{x}=\log \frac{(x-y)^{2}}{x} c$
$\frac{\mathrm{C}(\mathrm{x}-\mathrm{y})^{2}}{\mathrm{x}}=\mathrm{e}^{-\mathrm{y} / \mathrm{x}}$
$\mathrm{C}(\mathrm{x}-\mathrm{y})^{2}=\mathrm{xe}^{-\mathrm{y} / \mathrm{x}}$
$(x-y)^{2}=k x e^{-y / x}$
Which is the required solution of the given differential equation.
View full question & answer→Question 295 Marks
In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
AnswerLet y be the number of bacteria at any instant t.
Given that the rate of growth of bacteria is proportional to the number present
$\therefore \frac{d y}{d t} \propto y$
$\Rightarrow \frac{d y}{d t}=k y$ (k is a constant)
Separating variables,
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=\mathrm{kdt}$
Integrating both sides,
$\Rightarrow \int \frac{d y}{y}=k \int d t$
$\Rightarrow$ log y = kt + c.....(i)
Let y’ be the number of bacteria at t = 0.
$\Rightarrow$ log y’ = c
Substituting the value of c in (i)
$\Rightarrow$ log y = kt + log y’
$\Rightarrow$ log y - log y’ = kt
$\Rightarrow \log \frac{y}{y^{\prime}}=k t$ .....(ii)
Also, given that number of bacteria increases by 10% in 2 hours.
$\Rightarrow \mathrm{y}=\frac{110}{100} \mathrm{y}^{\prime}$
$\Rightarrow \frac{\mathrm{y}}{\mathrm{y}^{\prime}}=\frac{11}{10}$ ......(iii)
Substituting this value in (ii)
$\Rightarrow \mathrm{k} \times 2=\log \frac{11}{10}$
$\Rightarrow \mathrm{k}=\frac{1}{2} \log \frac{11}{10}$
So, (ii) becomes
$\Rightarrow \frac{1}{2} \log \frac{11}{10} \times \mathrm{t}=\log \frac{\mathrm{y}}{\mathrm{y}^{\prime}}$
$\Rightarrow \mathrm{t}=\frac{2 \log \frac{\mathrm{y}}{\mathrm{y}^{\prime}}}{\log \frac{11}{10}}$ ......(iv)
Now, let 't' be the time when the number of bacteria increases from 100000 to 200000.
$\Rightarrow \mathrm{y}=2 \mathrm{y}^{\prime} \text { at } \mathrm{t}=\mathrm{t}^{\prime}$
So from (iv)
$\Rightarrow \mathrm{t}^{\prime}=\frac{2 \log \frac{\mathrm{y}}{\mathrm{y}^{\prime}}}{\log \frac{11}{10}}=\frac{2 \log 2}{\log \frac{11}{10}}$
So bacteria increases from 100000 to 200000 in $\frac{2 \log 2}{\log _{\frac{11}{10}}}$ hours.
View full question & answer→Question 305 Marks
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
AnswerLet the rate of change of the volume of the balloon be k. (k is a constant)
$\therefore \frac{d y}{d t}=k$
Or,
$\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=k$ {Volume of sphere = $\frac{4}{3} \pi r^{3}$ }
$\Rightarrow \frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}=k$
$\Rightarrow 4 \pi r^{2} d r=k d t$
Integrating both sides,
$\Rightarrow 4 \pi \int r^{2} d r=k \int d t$
$\Rightarrow \frac{4 \pi r^{3}}{3}=k t+c$ .....(i)
Now, given that
At t = 0, r = 3 units,
$\Rightarrow \frac{4 \pi(3)^{3}}{3}=k(0)+C$
$\Rightarrow 4 \pi(3)^{2}=C$
$\Rightarrow C=36 \pi$
Now, at t = 3, r = 6 units:
$\Rightarrow 4 \pi \times (6)^{3}=3(\mathrm{k} \times 3+\mathrm{c})$
$\Rightarrow$ k = $84 \pi$
Substituting the values of k and c in (i)
$\Rightarrow 4 \pi r^{3}=3(84 \pi t+36 \pi)$
$\Rightarrow 4 \pi r^{3}=4 \pi(63 t+27)$
$\Rightarrow r^{3}=63 t+27$
$\Rightarrow r=\sqrt[3]{63 t+27}$
$\therefore$ Radius of balloon after t seconds is $\sqrt[3]{63 t+27}$ units.
View full question & answer→Question 315 Marks
Find a solution of $ x \left( x ^ { 2 } - 1 \right) \frac { d y } { d x } = 1 , $ which satisfy the condition $y = 0$ when $x = 2$.
AnswerWe have differential equation,
$ x \left( x ^ { 2 } - 1 \right) \frac { d y } { d x } = 1$
$ \Rightarrow \quad \frac { d y } { d x } = \frac { 1 } { x \left( x ^ { 2 } - 1 \right) }$
$\Rightarrow \quad \frac { d y } { d x } = \frac { 1 } { x ( x - 1 ) ( x + 1 ) }$ $\left[ \because a ^ { 2 } - b ^ { 2 } = ( a - b ) ( a + b ) \right]$
$\Rightarrow \quad d y = \frac { d x } { x ( x - 1 ) ( x + 1 ) }$
Therefore, on integrating both sides, we get
$\int d y = \int \frac { d x } { x ( x - 1 ) ( x + 1 ) }$
$ \Rightarrow \quad y = I + K$ ...(i)
where, $ I = \int \frac { d x } { x ( x - 1 ) ( x + 1 ) }$
By using partial fraction method,
let $ \frac { 1 } { x ( x - 1 ) ( x + 1 ) } = \frac { A } { x } + \frac { B } { x - 1 } + \frac { C } { x + 1 }$
$\Rightarrow 1 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)$
$\Rightarrow 1 = A(x^2 - 1) + B(x^2 + x) + C(x^2 - x)$
Therefore, on comparing the coefficients of $x^2, x$ and constant terms from both sides, we get
A + B + C = 0
B - C = 0
and -A = 1
$\Rightarrow$ A = -1
Therefore, on solving above equations, we get
A = -1, $B = \frac { 1 } { 2 }$ and $C = \frac { 1 } { 2 }$,
then $\frac { 1 } { x ( x - 1 ) ( x + 1 ) } = \frac { - 1 } { x } + \frac { 1 / 2 } { x - 1 } + \frac { 1 / 2 } { x + 1 }$
Therefore,on integrating both sides w.r.t. x, we get
$I = \int \frac { 1 } { x ( x - 1 ) ( x + 1 ) } d x$ $= \int \frac { - 1 } { x } d x + \frac { 1 } { 2 } \int \frac { d x } { x - 1 } + \frac { 1 } { 2 } \int \frac { d x } { x + 1 }$
$\Rightarrow \quad I = - \log | x | + \frac { 1 } { 2 } \log | x - 1 | + \frac { 1 } { 2 } \log | x + 1 |$
Therefore,on putting the value of I in Eq. (i), we get
$y = - \log | x | + \frac { 1 } { 2 } \log | x - 1 | + \frac { 1 } { 2 } \log | x + 1 | + K$ ....(ii)
Also, we have, y = 0, when x = 2
Therefore,on putting y = 0 and x = 2 in Eq. (ii), we get
$0 = - \log 2 + \frac { 1 } { 2 } \log 1 + \frac { 1 } { 2 } \log 3 + K$
$\Rightarrow \quad K = \log 2 - \frac { 1 } { 2 } \log 1 - \frac { 1 } { 2 } \log 3$
$\Rightarrow \quad K = \log 2 - \log \sqrt { 3 } \quad \quad [ \because \log 1 = 0 ]$
$\Rightarrow \quad K = \log \frac { 2 } { \sqrt { 3 } }$
Therefore,on putting the value of K in Eq. (i), we get
$y = - \log | x | + \frac { 1 } { 2 } \log | x - 1 | + \frac { 1 } { 2 } \log | x + 1 | + \log \frac { 2 } { \sqrt { 3 } }$
which is the required solution.
View full question & answer→Question 325 Marks
In a bank, principal increases continuously at the rate of $5\%$ per year. In how many years $₹\ 1000$ double itself?
AnswerLet $P$ be the principal at any time t. According to the given problem, $\frac{d p}{d t}=\left(\frac{5}{100}\right) \times \mathrm{P}$
or $\frac{d p}{d t}=\frac{\mathrm{P}}{20} ...(i)$
separating the variables in equation (i), we get
$\frac{d p}{\mathrm{P}}=\frac{d t}{20} ...(ii)$
Integrating both sides of equation (ii), we get
$\log P = \frac{t}{20} + C_1$
or $P = e^{\frac{t}{20}} \cdot e^{\mathrm{C}_{1}}$
or $P = \mathrm{C} e^{\frac{t}{20}}$ (where $e^{C_1}= C) ...(iii)$
Now, $P = 1000$, when $t = 0$
Substituting the values of $P$ and t in (iii), we get $C = 1000.$ Therefore, equation $(iii)$, gives
$P = 1000 e^{t/20}$
Let t years be the time required to double the principal. Then $2000 = 1000 e^{t/20} \Rightarrow t = 20 \log_e2.$
View full question & answer→Question 335 Marks
Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point.
AnswerWe know that the slope of the tangent to the curve is $\frac{d y}{d x}$
According to question, $\frac{d y}{d x}=x+x y$
or $\frac{d y}{d x}-x y=x$ ......(i)
This is a linear differential equation of the type $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ where P = -x and Q = x.
Therefore, I.F = $e^{\int-x d x}=e^{\frac{-x^{2}}{2}}$
Hence, the solution of equation is given by
$y \cdot e^{\frac{-x^{2}}{2}}=\int(x)\left(e^{\frac{-x^{2}}{2}}\right) d x+C$ ......(ii)
Let $I=\int(x) e^{\frac{-x^{2}}{2}} d x$
Now, let $\frac{-x^{2}}{2}=t$, then -x dx = dt or x dx = -dt
Therefore, $\mathrm{I}=-\int e^{t} d t=-e^{t}=-e^{\frac{-x^{2}}{2}}$
Substituting the value of I in equation (ii), we get
$y e^{\frac{-x^{2}}{2}}=-e^{\frac{-x^{2}}{2}}+C$ .....(iii)
Now (iii) represents the equation of family of curves. But we are interested in finding a particular member of the family passing through (0, 1). Substituting x = 0 and y = 1 in equation (iii) we get
1 = -1 + C or C = 2
Substituting the value of C in equation (iii), we get
$y=-1+2 e^{\frac{x^{2}}{2}}$
which is the equation of the required curve.
View full question & answer→Question 345 Marks
Find the particular solution of the differential equation $ \frac { d y } { d x } + y \cot x = 2 x + x ^ { 2 } \cot x $ ($ x \neq 0$) given that y = 0, when $ x = \frac { \pi } { 2 }$.
AnswerWe have, $ \frac { d y } { d x } + y \cot x = 2 x + x ^ { 2 } \cot x , ( x \neq 0 )$
This is a linear differential equation of the form $ \frac { d y } { d x } + P y = Q$.
Here, P = cot x and $Q = 2x + x^2 \cot x$.
$ \therefore \quad \mathrm { IF } = e ^ { \int \mathrm { Pdx } } = e ^ { \int \cot x d x }$ $ = e ^ { \log | \sin x | } = \sin x$
The general solution is given by
$y \cdot \mathrm { IF } = \int ( \mathrm { IF } \times Q ) d x + C$
$\Rightarrow y \cdot \sin x = \int \left( 2 x + x ^ { 2 } \cot x \right) \sin x d x + C$
$= 2 \int x \sin x d x + \int x ^ { 2 } \cos x d x + c$
$= 2 \int x \sin x d x + x ^ { 2 } \sin x - \int 2 x \sin x d x + C$
$\Rightarrow \quad y \cdot \sin x = x ^ { 2 } \sin x + C$ ...(i)
On putting $x = \frac { \pi } { 2 }$ and y = 0 in Eq. (i), we get
$0 \cdot \sin \frac { \pi } { 2 } = \left( \frac { \pi } { 2 } \right) ^ { 2 } \cdot \sin \frac { \pi } { 2 } + C \Rightarrow C = - \frac { \pi ^ { 2 } } { 4 }$
On putting $ C = \frac { - \pi ^ { 2 } } { 4 }$ in Eq. (i), we get
$y \cdot \sin x = x ^ { 2 } \sin x - \frac { \pi ^ { 2 } } { 4 }$
$\therefore \quad y = x ^ { 2 } - \frac { \pi ^ { 2 } } { 4 } cosec x$
[divding both sides by sin x]
View full question & answer→Question 355 Marks
Find the general solution of the differential equation $\frac{d y}{d x}-y=\cos x$
AnswerGiven differential equation is of the form
$\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$, where P = -1 and Q = cos x
Therefore, $\mathrm{I} \cdot \mathrm{F}=e^{\int-1 d x}=e^{-x}$
Multiplying both sides of equation by I.F, we get
$e^{-x} \frac{d y}{d x}-e^{-x} y=e^{-x} \cos x$
or $\frac{d }{d x}\left(y e^{-x}\right)=e^{-x} \cos x$
On integrating both sides with respect to x, we get
$y e^{-x}=\int e^{-x} \cos x d x+C$ ......(i)
Let $I=\int e^{-x} \cos x d x$
= $\cos x\left(\frac{e^{-x}}{-1}\right)-\int(-\sin x)\left(-e^{-x}\right) d x$
= $-\cos x e^{-x}-\int \sin x e^{-x} d x$
= $-\cos x e^{-x}-\left[\sin x\left(-e^{-x}\right)-\int \cos x\left(-e^{-x}\right) d x\right]$
= $-\cos x e^{-x}+\sin x e^{-x}-\int \cos x e^{-x} d x$
or $I=-e^{-x} \cos x+\sin x e^{-x}-I$
or $2 \mathrm{I}=(\sin x-\cos x) e^{-x}$
or $I=\frac{(\sin x-\cos x) e^{-x}}{2}$
Substituting the value of I in equation (i), we get
$y e^{-x}=\left(\frac{\sin x-\cos x}{2}\right) e^{-x}+C$
or $y=\left(\frac{\sin x-\cos x}{2}\right)+\mathrm{C} e^{x}$
which is the general solution of the given differential equation.
View full question & answer→Question 365 Marks
Show that the family of curves for which the slope of the tangent at any point (x, y) on it is $ \frac { x ^ { 2 } + y ^ { 2 } } { 2 x y }$, is given by $x^2 - y^2 = cx$.
AnswerWe have, $ \frac { d y } { d x } = \frac { x ^ { 2 } + y ^ { 2 } } { 2 x y }$
Clearly, each of the function $x^2 + y^2$ and 2xy is a homogeneous function of degree 2, so the given equation is homogeneous.
Put y = vx and $ \frac { d y } { d x } = v + x \frac { d v } { d x }$
The given equation becomes
$ v + x \frac { d v } { d x } = \frac { x ^ { 2 } + v ^ { 2 } x ^ { 2 } } { 2 v x ^ { 2 } }$
$ \Rightarrow v + x \frac { d v } { d x } = \frac { v ^ { 2 } + 1 } { 2 v }$
$ \Rightarrow \quad x \frac { d v } { d x } = \left( \frac { v ^ { 2 } + 1 } { 2 v } - v \right)$
$\Rightarrow \quad x \frac { d v } { d x } = \frac { v ^ { 2 } + 1 - 2 v ^ { 2 } } { 2 v } = \frac { 1 - v ^ { 2 } } { 2 v }$
$\Rightarrow \quad x \frac { d v } { d x } = \frac { - \left( v ^ { 2 } - 1 \right) } { 2 v }$ $\Rightarrow - \frac { 2 v } { v ^ { 2 } - 1 } d v = \frac { d x } { x }$ [using variable separable form]
On integrating both sides, we get
$-\log|v^2 - 1| = \log x - \log C_1$
$\Rightarrow - \log \left| v ^ { 2 } - 1 \right| - \log x = - \log C _ { 1 }$
$\Rightarrow \quad \log \left| x \left( v ^ { 2 } - 1 \right) \right| = \log C _ { 1 } \Rightarrow x \left( v ^ { 2 } - 1 \right) = C _ { 1 }$
$\Rightarrow \quad x \left( \frac { y ^ { 2 } } { x ^ { 2 } } - 1 \right) = C _ { 1 }$ $\Rightarrow x \left( \frac { y ^ { 2 } - x ^ { 2 } } { x ^ { 2 } } \right) = C _ { 1 }$
$\Rightarrow \quad \frac { y ^ { 2 } - x ^ { 2 } } { x } = C _ { 1 }$ $\Rightarrow x ^ { 2 } - y ^ { 2 } = - C _ { 1 } x$
$ \Rightarrow \quad x ^ { 2 } - y ^ { 2 } = C x$ $ \left[ \because C = - C _ { 1 } \right]$
View full question & answer→Question 375 Marks
Show that the differential equation $2 y e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0$ is homogeneous and find its particular solution, given that, x = 0 when y = 1.
AnswerThe given differential equation can be written as
$\frac{d x}{d y}=\frac{2 x e^{\frac{x}{y}}-y}{2 y e^{\frac{x}{y}}}$ ......(i)
Let $F(x, y)=\frac{2 x e^{\frac{x}{y}}-y}{2 y e^{\frac{x}{y}}}$
Then, $\mathrm{F}(\lambda x, \lambda y)=\frac{\lambda\left(2 x e^{\frac{x}{y}}-y\right)}{\lambda\left(2 y e^{\frac{x}{y}}\right)}=\lambda^{0}[\mathrm{F}(x, y)]$
Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. To solve it, we make the substitution
x = vy ... (ii)
Differentiating equation (ii) with respect to y, we get
$\frac{d x}{d y}=v+y \frac{d v}{d y}$
Substituting the value of x and $\frac{d x}{d y}$ in equation (i) we get
$v+y \frac{d v}{d y}=\frac{2 v e^{v}-1}{2 e^{v}}$
or $y \frac{d v}{d y}=\frac{2 v e^{v}-1}{2 e^{v}}-v$
or $y \frac{d v}{d y}=\frac{2 v e^{v}-1}{2 e^{v}}-v$
or $2 e^{v} d v=\frac{-d y}{y}$
or $\int 2 e^{v} \cdot d v=-\int \frac{d y}{y}$
or $2 e^{v}=-\log |y|+\mathrm{C}$
and replacing v by $\frac{x}{y}$, we get
$2 e^{\frac{x}{y}}+\log |y|=C$ .......(iii)
Substituting x = 0 and y = 1 in equation (iii), we get
$2 e^{0}+\log |1|=\mathrm{C} \Rightarrow \mathrm{C}=2$
Substituting the value of C in equation (iii), we get
$2 e^{\frac{x}{y}}+\log |y|=2$
which is the particular solution of the given differential equation.
View full question & answer→Question 385 Marks
Show that the differential equation $x \cos \left( \frac { y } { x } \right) \frac { d y } { d x } = y \cos \left( \frac { y } { x } \right) + x$ is homogeneous and solve it.
AnswerGiven differential equation is
$x \cos \left( \frac { y } { x } \right) \frac { d y } { d x } = y \cos \left( \frac { y } { x } \right) + x$
which is a homogeneous differential equation because $\frac { d y } { d x } = F \left( \frac { y } { x } \right)$.
put y = vx
$\Rightarrow \frac { d y } { d x } = v + x \frac { d v } { d x }$
$\Rightarrow x \cos v \left[ v + x \frac { d v } { d x } \right] = v x \cos v + x$
$v x \cos v + x ^ { 2 } \cos v \frac { d v } { d x } = v x \cos v + x$
$\Rightarrow \quad x ^ { 2 } \cos v \frac { d v } { d x } = x$
$\Rightarrow \quad \cos v d v = \frac { d x } { x }$
On integrating both sides, we get
$\int \cos v d v = \int \frac { d x } { x }$
$\Rightarrow \quad \sin v = \log | x | + C$
$\Rightarrow \quad \sin \left( \frac { y } { x } \right) = \log | x | + c \left[ \text { put } v = \frac { y } { x } \right]$
This is the required solution of given differential equation.
View full question & answer→Question 395 Marks
Show that the differential equation $(x-y) \frac{d y}{d x}=x+2 y$ is homogeneous and solve it.
AnswerThe given differential equation can be expressed as
$\frac{d y}{d x}=\frac{x+2 y}{x-y}$ .....(i)
Let $\mathrm{F}(x, y)=\frac{x+2 y}{x-y}$
Now $\mathrm{F}(\lambda x, \lambda y)=\frac{\lambda(x+2 y)}{\lambda(x-y)}=\lambda^{0} \cdot f(x, y)$
Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.
$\frac{d y}{d x}=\left(\frac{1+\frac{2 y}{x}}{1-\frac{y}{x}}\right)=g\left(\frac{y}{x}\right)$ ......(ii)
R.H.S. of differential equation (ii) is of the form $g\left(\frac{y}{x}\right)$ and so it is a homogeneous function of degree zero. Therefore, equation (i) is a homogeneous differential equation. To solve it we make the substitution
y = vx ......(iii)
Differentiating equation (iii) with respect to, x we get
$\frac{d y}{d x}=v+x \frac{d v}{d x}$ ......(iv)
Substituting the value of y and $\frac{d y}{d x}$ in equation (i) we get
$v+x \frac{d v}{d x}=\frac{1+2 v}{1-v}$
or $x \frac{d v}{d x}=\frac{1+2 v}{1-v}-v$
or $x \frac{d v}{d x}=\frac{v^{2}+v+1}{1-v}$
or $\frac{v-1}{v^{2}+v+1} d v=\frac{-d x}{x}$ .....(v)
Integrating both sides of equation (v), we get
$\int \frac{v-1}{v^{2}+v+1} d v=-\int \frac{d x}{x}$
or $\frac{1}{2} \int \frac{2 v+1-3}{v^{2}+v+1} d v=-\log |x|+C_{1}$
or $\frac{1}{2} \int \frac{2 v+1}{v^{2}+v+1} d v-\frac{3}{2} \int \frac{1}{v^{2}+v+1} d v=-\log |x|+\mathrm{C}_{1}$
or $\frac{1}{2} \log \left|v^{2}+v+1\right|-\frac{3}{2} \int \frac{1}{\left(v+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d v=-\log |x|+\mathrm{C}_{1}$
or $\frac{1}{2} \log \left|v^{2}+v+1\right|-\frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)=-\log |x|+\mathrm{C}_{1}$
or $\frac{1}{2} \log \left|v^{2}+v+1\right|+\frac{1}{2} \log x^{2}=\sqrt{3} \tan ^{-1}\left(\frac{2 v+1}{\sqrt{3}}\right)+C_{1}$
Replacing v by $\frac{y}{x}$ , we get
or $\frac{1}{2} \log \left|\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right|+\frac{1}{2} \log x^{2}$ = $\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+C_{1}$
or $\frac{1}{2} \log \left|\left(\frac{y^{2}}{x^{2}}+\frac{y}{x}+1\right) x^{2}\right|$ = $\sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+C_{1}$
or $\log \left|\left(y^{2}+x y+x^{2}\right)\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3} x}\right)+2 \mathrm{C}_{1}$
or $\log \left|\left(x^{2}+x y+y^{2}\right)\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3} x}\right)+C$
which is the general solution of the differential equation (i)
View full question & answer→