Question 14 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 24 Marks
Prove that x2 - y2 = c(x2 + y2)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 }$(where C1 = C$^2$).
View full question & answer→Question 34 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 44 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 54 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 64 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 74 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 84 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 94 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 = sec2 x and Q = sec2 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
$\Rightarrow \frac{d}{d x}(\tan x)=\frac{d t}{d x}$
$\Rightarrow \sec ^{2} x=\frac{d t}{d x}$
$\Rightarrow$ sec2xdx = dt
Thus, the equation (i) becomes,
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\tan \mathrm{x}}=\int\left(\mathrm{e}^{\mathrm{t}} \cdot \mathrm{t}\right) \mathrm{d} \mathrm{t}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\tan \mathrm{x}}=\int\left(\mathrm{t} \cdot \mathrm{e}^{\mathrm{t}}\right) \mathrm{dt}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\mathrm{tanx}}=\mathrm{t} \cdot \int \mathrm{e}^{\mathrm{t}} \mathrm{dt}-\int\left(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t}) \cdot \int \mathrm{e}^{\mathrm{t}} \mathrm{dt}\right) \mathrm{dt}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{e}^{\mathrm{tanx}}=\mathrm{t} \cdot \mathrm{e}^{\mathrm{t}}-\int \mathrm{e}^{\mathrm{t}} \mathrm{dt}+\mathrm{c}$
$\Rightarrow$ tetanx = (t – 1)et + C
$\Rightarrow$ tetanx = (tanx – 1)etanx + C
$\Rightarrow$ y = (tanx -1) + C e-tanx
Therefore, the required general solution of the given differential equation is
y = (tanx -1) + C e-tanx.
View full question & answer→Question 104 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 + Cex
$\Rightarrow$ x + y - 4 = Cex
Now, it is given that curve passes through (0, 2).
Thus, equation (ii) becomes:
0 + 2 - 4 = C e0
$\Rightarrow$ - 2 = C
$\Rightarrow$ C = -2
Substituting C = -2 in equation (ii), we get,
x + y - 4 =-2ex
$\Rightarrow$ y = 4 - x - 2ex
Therefore, the required general solution of the given differential equation is
y = 4 - x - 2ex
View full question & answer→Question 114 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) + Cex
$\Rightarrow$ x + y + 1 = Cex .......(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 = ex
Therefore, the required general solution of the given differential equation is
x + y + 1 = ex
View full question & answer→Question 124 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 + Cey
$\Rightarrow$ x + y + 1 = Cey
Therefore, the required general solution of the given differential equation is
x + y + 1 = Cey
View full question & answer→Question 134 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 144 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 154 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
$f(\mathrm{kx}, \mathrm{ky})=\frac{\mathrm{ky}-\mathrm{kx} \sin \left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)}{\mathrm{kx}}$
$f(\mathrm{kx}, \mathrm{ky})=\frac{\mathrm{k}}{\mathrm{k}} \cdot \frac{\mathrm{y}-\mathrm{x} \sin \left(\frac{\mathrm{y}}{\mathrm{x}}\right)}{\mathrm{x}}$
= k0.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 164 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
$f(\mathrm{kx}, \mathrm{ky})=\frac{\left\{\mathrm{kx} \cos \left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)+\mathrm{kysin}\left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)\right\} \mathrm{ky}}{\left\{\mathrm{kysin}\left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)-\mathrm{kx} \cos \left(\frac{\mathrm{ky}}{\mathrm{kx}}\right)\right\} \mathrm{kx}}$
$f(\mathrm{kx}, \mathrm{ky})=\frac{\mathrm{k}^{2}}{\mathrm{k}^{2}} \cdot \frac{\left\{\mathrm{x} \cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\mathrm{y} \sin \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\} \mathrm{y}}{\left\{\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\mathrm{x} \cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right\} \mathrm{x}}$
= k0.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 174 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 = eC
which is the required solution.
View full question & answer→Question 184 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
$f(k x, k y)=\frac{k^{2} x^{2}-2 k^{2} y^{2}+k x k y}{k^{2} x^{2}}$
$f(k x, k y)= \frac{x^{2}-2 y^{2}+x y}{x^{2}}$
= k0.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 194 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
$f(k x, k y)=-\frac{k^{2} x^{2}-k^{2} y^{2}}{2 k^{2} x y}$
$f(k x, k y)=-\frac{k^{2}}{k^{2}} \cdot \frac{x^{2}-y^{2}}{2 x y}$
= k0.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 + v2 = 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 + v2) = -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 204 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
$f(k x, k y)=\frac{k x+k y}{k x-k y}$
$f(k x, k y)=\frac{x+y}{x-y}$
= k0.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 214 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 224 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
$f(k x, k y)=\frac{2 k x k y+(k y)^{2}}{2(k x)^{2}}$
= $\frac{\mathrm{k}^{2}}{\mathrm{k}^{2}} \cdot \frac{2 \mathrm{xy}+\mathrm{y}^{2}}{2 \mathrm{x}^{2}}$
= k0.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 234 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
$f(k x, k y)=\frac{k y}{k x}-cosec \left(\frac{k y}{k x}\right)$
= $\frac{y}{x}-cosec \left(\frac{y}{x}\right)$
= k0.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 244 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 sin2($\frac yx$) − y]dx + xdy = 0
xdy = -[x sin2($\frac{y}{x}$) − 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) = $\frac{-\left(t x \sin ^{2}(\frac{t y} {t x})-t y\right)}{t x}$
= t0 F(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$\frac{dv}{dx}$ = -sin2v + v
$\Rightarrow$ x$\frac{dv}{dx}$ = -sin2v + v - v
$\Rightarrow$ x$\frac{dv}{dx}$ = -sin2v
$\Rightarrow$ $\large\frac{dv}{\sin ^2 v}=-\frac{dx}{x}$
$\Rightarrow$ cosec2v dv $=\large\frac{-dx}{x}$
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 e1 = c
Hence ecot($\frac{y}{x}$) = ex
$\Rightarrow$ $e^{\cot (x / y)-1}$ = x
View full question & answer→Question 254 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
x2dy + (xy + y2)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
$f(k x, k y)=-\frac{\left(k x k y+k^{2} y^{2}\right)}{k^{2} x^{2}}$
= $\frac{\mathrm{k}^{2}}{\mathrm{k}^{2}} \cdot-\frac{\left(\mathrm{xy}+\mathrm{y}^{2}\right)}{\mathrm{x}^{2}}$
= k0.f(x,y)
Therefore, the given differential equation is homogeneous.
x2dy + (xy + y2)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
$C^{2}=\frac{1}{1+2}=\frac{1}{3}$
$\therefore \frac{x^{2} y}{y+2 x}=\frac{1}{3}$
3x2y = y + 2x
y + 2x = 3x2y
The required solution of the differential equation.
View full question & answer→Question 264 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
$f(\mathrm{kx}, \mathrm{ky})=-\frac{(\mathrm{kx}-\mathrm{ky})}{(\mathrm{kx}+\mathrm{ky})}$ = k0.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 274 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
$\mathrm{f}(\mathrm{kx}, \mathrm{ky})=\frac{-\mathrm{e}^{\mathrm{kx} / \mathrm{ky}}\left(1-\frac{\mathrm{kx}}{\mathrm{ky}}\right)}{\left(1+\mathrm{e}^{\mathrm{kx} / \mathrm{ky}}\right)}$
$=\frac{-\mathrm{e}^{\mathrm{x} / \mathrm{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right)}{\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right)}$
= k0f(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 ev + v = t
(ev + 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(ev + v)
$\therefore$ log(ev + 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
yex/y + x = C
x + yex/y = C
The required solution of the differential equation.
View full question & answer→Question 284 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
$f(k x, k y)=\frac{(k x)^{2}+(k y)^{2}}{(k x)^{2}+k x . k y}$ = k0.f(x,y)
Therefore, the given differential equation is homogeneous.
(x2 + xy)dy = (x2 + y2)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 294 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 304 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 314 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(x2 - 1) + B(x2 + x) + C(x2 - x)
Therefore, on comparing the coefficients of x2, 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 324 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 = $\frac{t}{20}$ + C1
or P = $e^{\frac{t}{20}} \cdot e^{\mathrm{C}_{1}}$
or P = $\mathrm{C} e^{\frac{t}{20}}$ (where eC1 = 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 et/20
Let t years be the time required to double the principal. Then 2000 = 1000 et/20 $\Rightarrow$ t = 20 loge2.
View full question & answer→Question 334 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 344 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 + x2 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 354 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 364 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 x2 - y2 = cx.
AnswerWe have, $ \frac { d y } { d x } = \frac { x ^ { 2 } + y ^ { 2 } } { 2 x y }$
Clearly, each of the function x2 + y2 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|v2 - 1| = log x - log C1
$\Rightarrow - \log \left| v ^ { 2 } - 1 \right| - \log x = - \log C _ { 1 }$
$\Rightarrow \quad \log \left| x \left( v ^ { 2 } - 1 \right) \right| = \log C _ { 1 } \Rightarrow x \left( v ^ { 2 } - 1 \right) = C _ { 1 }$
$\Rightarrow \quad x \left( \frac { y ^ { 2 } } { x ^ { 2 } } - 1 \right) = C _ { 1 }$ $\Rightarrow x \left( \frac { y ^ { 2 } - x ^ { 2 } } { x ^ { 2 } } \right) = C _ { 1 }$
$\Rightarrow \quad \frac { y ^ { 2 } - x ^ { 2 } } { x } = C _ { 1 }$ $\Rightarrow x ^ { 2 } - y ^ { 2 } = - C _ { 1 } x$
$ \Rightarrow \quad x ^ { 2 } - y ^ { 2 } = C x$ $ \left[ \because C = - C _ { 1 } \right]$
View full question & answer→Question 374 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 384 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 394 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→