Question 12 Marks
Solve the differential equation $y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y(y \neq 0)$
AnswerIt is given that $\text { ye }^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y$
$\Rightarrow \text { ye }^{\frac{x}{y}} \frac{d x}{d y}=x e^{\frac{x}{y}}+y^{2}$
$\Rightarrow \mathrm{e}^{\frac{x}{y}}\left[y \cdot \frac{d x}{d y}-x\right]=y^{2}$
$\Rightarrow e^{\frac{x}{y}} \cdot \frac{\left[y \cdot \frac{d x}{d y}-x\right]}{y^{2}}=1$ .......(i)
Let $e^{\frac{x}{y}}=z$
Differentiating it w.r.t. y, we get,
$\frac{d}{d y}\left(e^{\frac{x}{y}}\right)=\frac{d z}{d y}$
$\Rightarrow e^{\frac{x}{y}} \cdot \frac{d}{d y}\left(\frac{x}{y}\right)=\frac{d z}{d y}$
$\Rightarrow e^{\frac{x}{y}} \cdot\left[\frac{y \cdot \frac{d x}{d y}-x}{y^{2}}\right]=\frac{d z}{d y}$ ......(ii)
From equation (i) and equation (ii), we get,
$\frac{\mathrm{d} \mathrm{z}}{\mathrm{dy}}=1$
$\Rightarrow$ dz = dy
On integrating both sides, we get,
z = y + C
$\Rightarrow \mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}}=\mathrm{y}+\mathrm{C}$
View full question & answer→Question 22 Marks
Find the particular solution of the differential equation $(1 + e^{2x})dy + (1 + y^2)e^x dx = 0,$ given that $y = 1$, when $x = 0.$
AnswerGiven differential equation is,$(1 + e^{2x})dy + (1 + y^2)e^x dx = 0$
Above equation may be written as
$\frac { d y } { 1 + y ^ { 2 } } = \frac { - e ^ { x } } { 1 + e ^ { 2 x } } d x$
On integrating both sides, we get
$\int \frac { d y } { 1 + y ^ { 2 } } = - \int \frac { e ^ { x } } { 1 + e ^ { 2 x } } d x$
On putting $e^x = t \Rightarrow e^x dx = dt$ in RHS, we get
$\tan ^ { - 1 } y = - \int \frac { 1 } { 1 + t ^ { 2 } } d t$
$\Rightarrow \quad \tan ^ { - 1 } y = - \tan ^ { - 1 } t + C$
$\Rightarrow \quad \tan ^ { - 1 } y = - \tan ^ { - 1 } \left( e ^ { x } \right) + C ...(i) [$put $t = e^x]$
Also, given that $y = 1,$ when $x = 0.$
On putting above values in Eq. (i), we get
$\tan^{-1}1 = -\tan^{-1}(e^0) + C$
$\Rightarrow \quad \tan ^ { - 1 }1 = - \tan ^ { - 1 } 1 + C \quad \left[ \because e ^ { 0 } = 1 \right]$
$\Rightarrow \quad 2 \tan ^ { - 1 } 1 = C$
$\Rightarrow \quad 2 \tan ^ { - 1 } \left( \tan \frac { \pi } { 4 } \right) = C$
$\Rightarrow \quad C = 2 \times \frac { \pi } { 4 } = \frac { \pi } { 2 }$
On putting $C = \frac { \pi } { 2 }$ in Eq. (i), we get
$\tan ^ { - 1 } y = - \tan ^ { - 1 } e ^ { x } + \frac { \pi } { 2 }$
$\Rightarrow \quad y = \tan \left[ \frac { \pi } { 2 } - \tan ^ { - 1 } \left( e ^ { x } \right) \right] = \cot \left[ \tan ^ { - 1 } \left( e ^ { x } \right) \right]$
$= \cot \left[ \cot ^ { - 1 } \left( \frac { 1 } { e ^ { x } } \right) \right] \left[ \because \tan ^ { - 1 } x = \cot ^ { - 1 } \frac { 1 } { x } \right]$
$\therefore \quad y = \frac { 1 } { e ^ { x } }$
which is the required solution.
View full question & answer→Question 32 Marks
Find the equation of the curve passing through the point $\left( {0,\frac{\pi }{4}} \right)$whose differential equation is $\sin x\cos ydx + \cos x\sin ydy = 0$
Answer$\sin x.\cos ydx = - \cos x.\sin ydy$ $\frac{sin x}{cos x}dx = - \frac{sin y}{cos y} dy $
$\int {\tan xdx = - \int {\tan y\,dy} } $
$\log \left( {\sec x} \right) = - \log \left( {\sec y} \right) + \log c$
$\log \left( {\sec x.\sec y} \right) = \log c$
$\sec x.\sec y = c$ ......(i)
When $x = 0,y= \pi /4$
$c = \sqrt 2 $
put the value of c in eq (i)
$\sec x.\sec y = \sqrt 2 $
View full question & answer→Question 42 Marks
Find the general solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$
AnswerIt is given that $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$
$\Rightarrow \frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$
$\Rightarrow \frac{\mathrm{dy}}{\sqrt{1-\mathrm{y}^{2}}}=-\frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^{2}}}$
On integrating, we get,
$\Rightarrow \sin^{-1} y + \sin^{-1} x = C$
View full question & answer→Question 52 Marks
Verify that the function $x^{2}=2 y^{2} \log y$ (implicit or explicit) is a solution of the differential equation $\left(x^{2}+y^{2}\right) \frac{d y}{d x}-x y=0$
AnswerIt is given that $x^2 = 2y^2 \log y$
Now, differentiating both sides w.r.t. x, we get,
$2 x=2 \cdot \frac{d}{d x}\left(y^{2} \log y\right)$
$\Rightarrow \mathrm{x}=\left[2 \mathrm{y} \cdot \log \mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{2} \cdot \frac{1}{\mathrm{y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right]$
$\Rightarrow \mathrm{x}=\frac{\mathrm{dy}}{\mathrm{dx}}(2 \mathrm{ylogy}+\mathrm{y})$
$\Rightarrow \frac{d y}{d x}=\frac{x}{y(1+2 \log y)}$
Now, substituting the value of $\frac{d y}{d x}$ in the LHS of the given differential equation, we get,
$\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{xy}=\left(2 \mathrm{y}^{2} \log \mathrm{y}+\mathrm{y}^{2}\right) \cdot \frac{\mathrm{x}}{\mathrm{y}(1+2 \log \mathrm{y})}-\mathrm{xy}$
$= \mathrm{y}^{2}(1+2 \log \mathrm{y}) \cdot \frac{\mathrm{x}}{\mathrm{y}(1+2 \log \mathrm{y})}-\mathrm{xy}$
$= xy - xy$
$= 0$
Therefore, the given function is the solution of the corresponding differential equation.
View full question & answer→Question 62 Marks
Verify that the function $y=x \sin 3 x$ (implicit or explicit) is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x=0$
AnswerIt is given that y = x sin 3x
Now, differentiating both sides w.r.t. x , we get,
$\frac{d y}{d x}=\frac{d}{d x}(x \sin 3 x)=\sin 3 x+x . \cos 3 x .3$
$\Rightarrow \frac{d y}{d x}=\sin 3 x+3 x \cos 3 x$
Now, again differentiating both sides w.r.t. x, we get,
$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(\sin 3 x)+3 \frac{d}{d x}(x \cos 3 x)$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=3 \mathrm{cos} 3 \mathrm{x}+3[\cos 3 \mathrm{x}+\mathrm{x}(-\sin 3 \mathrm{x}) \cdot 3]$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=6 \cos 3 x-9 x \sin 3 x$
Now, substituting the value of $\frac{d^{2} y}{d x^{2}}$ in the LHS of the given differential equation, we get,
$\frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x$
= (6.cos 3x - 9x sin 3x) + 9x sin 3x - 6 cos 3x
= 0 = RHS
Therefore, the given function is the solution of the corresponding differential equation.
View full question & answer→Question 72 Marks
Verify that the function $y=e^{x}(a \cos x+b \sin x)$ (implicit or explicit) is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$
AnswerIt is given that $y = e^x(a. \cos x + b \sin x) = a.e^x \cos x + b.e^x \sin x$
Now, differentiating both sides w.r.t. x, we get,
$\frac{d y}{d x}=a \frac{d}{d x}\left(e^{x} \cos x\right)+b \frac{d}{d x}\left(e^{x} \sin x\right)$
$\Rightarrow \frac{d y}{d x}=a\left(e^{x} \cos x-e^{x} \sin x\right)+b \cdot\left(e^{x} \sin x+e^{x} \cos x\right)$
$\Rightarrow \frac{d y}{d x}=(a+b) e^{x} \cos x+(b-a) e^{x} \sin x$
Now, again differentiating both sides w.r.t. x, we get,
$\frac{d^{2} y}{d x^{2}}=(a+b) \cdot \frac{d}{d x}\left(e^{x} \cos x\right)+(b-a) \frac{d}{d x}\left(e^{x} \sin x\right)$
= $(a+b) \cdot\left[e^{x} \cos x-e^{x} \sin x\right]+(b-a)\left[e^{x} \sin x+e^{x} \cos x\right]$
= $\mathrm{e}^{\mathrm{x}}[\mathrm{a.cosx}-\mathrm{a.sinx}+\mathrm{b.cosx}-\mathrm{b.sinx}+\mathrm{b.sinx}+\mathrm{b.cosx}-\mathrm{a.sinx}-\mathrm{a.cosx}]$
= $\left[2 \mathrm{e}^{\mathrm{x}}(\mathrm{b.cosx}-\mathrm{a.sinx})\right]$
Now, Substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the given differential equations, we get,
LHS = $=\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y$
$= 2e^x(b \cos x - a \sin x) -2e^x[(a + b) \cos x + (b - a ) \sin x] + 2e^x(a \cos x + b \sin x)$
$=e^x[(2b \cos x - 2a \sin x) - (2a \cos x + 2b \cos x) - (2b \sin x - 2a \sin x) + (2a \cos x + 2b \sin x)]$
$= e^x[(2b - 2a - 2b + 2a) \cos x] + e^x[(-2a - 2b + 2a + 2b) \sin x]$
$= 0 =$ RHS.
Therefore, the given function is the solution of the corresponding differential equation.
View full question & answer→Question 82 Marks
Verify that the function $x y=a e^{x}+b e^{-x}+x^{2}$ (implicit or explicit) is a solution of the differential equation $x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0$
AnswerIt is given that $xy = a e^x + b e^{-x} + x^2$
Now, differentiating both sides w.r.t. x, we get,
$\Rightarrow y+x \cdot \frac{d y}{d x}=a \cdot e^{x}-b e^{-x}+2 x$ ...(i)
Now, Again differentiating both sides w.r.t. x, we get,
$\Rightarrow \frac{d y}{d x}+\frac{d y}{d x}+x \cdot \frac{d^{2} y}{d x^{2}}=a e^{x}+b e^{-x}+2$ ...(ii)
Now, Using Equations. (i) and (ii), we get,
LHS $x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2$
= $a e^{x}+b e^{-x}+2-\left[a e^{x}+b e^{-x}+x^{2}\right]+x^{2}-2$
= $\begin{aligned} &a e^{x}+b e^{x}+2 -a e^{x}-b e^{-x}-x^{2}+x^{2}-2 \end{aligned}$
= 0
$\Rightarrow$ LHS = RHS.
Therefore, the given function is the solution of the corresponding differential equation.
View full question & answer→Question 92 Marks
Find the general solution of the differential equation $\frac{d y}{d x}=\sin ^{-1} x$
Answer$Given\: \frac{d y}{d x}=\sin ^{-1} x$
Separating variables,
$\Rightarrow dy = \sin^{-1} x dx$
Integrating both sides,
$\Rightarrow \int d y=\int \sin ^{-1} x d x$
Now to integrate $sin^{-1} x$ we have to multiply it by 1
because,
$\left.\because(\left\{\int \mathrm{u} . \mathrm{v} \mathrm{dx}=\mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{u}\right)\left(\int \mathrm{vdx}\right) \mathrm{dx}\right\}\mathrm{}\right)$
So,
$\Rightarrow$ y = $\int 1 . \sin ^{-1} x d x$
Let u be $\sin^{-1} x$ and $v$ be $1$
We can take the values of u and v from the formula (I.L.A.T.E)
$\therefore {y}=\left\{\sin ^{-1} x \int 1 \cdot dx-\int\left(\frac{d}{dx} \sin ^{-1} x\right)\left(\int 1 \cdot dx\right) d x\right\}$
$\Rightarrow {y}=x \sin ^{-1} x-\int \frac{x}{\sqrt{1-x^{2}}} d x$
$\Rightarrow$ let $1 - x^2 = t$
$\Rightarrow$ - 2x dx = dt
$\Rightarrow xdx = -\frac{\mathrm{dt}}{2}$
$\Rightarrow y = x \sin^{-1} x + \int \frac{1}{2 \sqrt{t}} d t$
$\Rightarrow y = x \sin^{-1} x + \frac{1}{2} \int t^{-\frac{1}{2}} d t$
$\Rightarrow y = x \sin^{-1}x + \frac{1}{2} \sqrt{t}+c$
putting the value of t,
$\Rightarrow y = x \sin^{-1} x + \sqrt{1-x^{2}}+c$
View full question & answer→Question 102 Marks
Find the general solution of the differential equation $x^{5} \frac{d y}{d x}=-y^{5}$
AnswerGiven differential equation is: $x^{5} \frac{d y}{d x}=-y^{5}$
Separating the variables we get
$ \frac{d y}{y^{5}}=\frac{-d x}{x^{5}}$
Integrating both sides,
$\Rightarrow \int \frac{d y}{y^{5}}= -\int \frac{d x}{x^{5}}$,
$\Rightarrow \int \mathrm{y}^{-5} \mathrm{dy}= -\int \mathrm{x}^{-5} \mathrm{dx}$
$\Rightarrow$ $\frac{y^{-4}}{-4}=-\frac{x^{-4}}{-4}+C_{1}$
$\Rightarrow$ $-y^{-4}=x^{-4}+4 c_{1}$
$\Rightarrow x^{-4}+y^{-4}=c$
View full question & answer→Question 112 Marks
Find the general solution of the differential equation $y \log y d x-x d y=0$
AnswerGiven Differential Equation is: y log y dx - x dy = 0
$\Rightarrow$ (y log y) dx = xdy
Separating variables,
$\Rightarrow \frac{d x}{x}=\frac{d y}{y \log y}$
Integrating both sides,
$\Rightarrow \int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{\mathrm{dy}}{\mathrm{ylogy}}$
$\Rightarrow \log x +\log c=\int \frac{\mathrm{dy}}{\mathrm{ylogy}}$
Let log y = t $\Rightarrow \frac{1}{y} d y=d t$
$\Rightarrow \log x + \log c=\int \frac{d t}{t}$
$\Rightarrow$ log x + log c = log t
or $\log (\log y)=\log x+\log c$
$\Rightarrow \log (\log y)=\log c x$
$\Rightarrow \begin{aligned} &\log y=c x \Rightarrow y=e^{c x} \end{aligned}$
View full question & answer→Question 122 Marks
Find the general solution of the differential equation $\frac{{dy}}{{dx}} = \left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)$
AnswerGiven: Differential equation $\frac{{dy}}{{dx}} = \left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)$ $\Rightarrow dy = \left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)dx$
$\Rightarrow \frac{{dy}}{{1 + {y^2}}} = \left( {1 + {x^2}} \right)dx$ [Separating variables]
Integrating both sides,
$\Rightarrow \int {\frac{1}{{1 + {y^2}}}} dy = \int {\left( {{x^2} + 1} \right)} dx$
$\Rightarrow {\tan ^{ - 1}}y = \frac{{{x^3}}}{3} + x + c$
View full question & answer→Question 132 Marks
Find the general solution of the differential equation $\left( {{e^x} + {\text{ }}{e^{ - x}}} \right){\text{ }}dy{\text{ }} - {\text{ }}\left( {{e^x} - {\text{ }}{e^{ - x}}} \right){\text{ }}dx{\text{ }} = {\text{ }}0$
Answer$({e^x} + {e^{ - x}})dy = ({e^x} - {e^{ - x}})dx$ $\int {dy} = \int {\frac{{({e^x} - {e^{ - x}})}}{{({e^x} + {e^{ - x}})}}} dx $ Since $\int {\frac{{f'(x)}}{{f(x)}}} dx = \ ln f(x) +c$
$y = \log \left| {({e^x} + {e^{ - x}})} \right| + C$
View full question & answer→Question 142 Marks
Find the general solution of the differential equation $\sec^2x \tan y\ dx + \sec^2y \tan x\ dy = 0$
AnswerGiven Differential equation is, $\sec^2x \tan y\ dx + \sec^2y \tan x\ dy = 0$
Dividing by tan x tan y, we have
$\frac{{{{\sec }^2}x}}{{\tan x}}dx + \frac{{{{\sec }^2}y}}{{\tan y}}dy = 0$
$ \Rightarrow\int\frac{{{{\sec }^2}x}}{{\tan x}}dx + \int\frac{{{{\sec }^2}y}}{{\tan y}}dy = 0$
$\Rightarrow \log \left| {\tan x} \right| + \log \left| {\tan y} \right| = \log c$ ...$\left[ {\because \int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}} dx = \log \left| {f\left( x \right)} \right|} \right]$
$ \Rightarrow \log \left| {\tan x\tan y} \right| = \log c$
$\Rightarrow \left| {\tan x\tan y} \right| = c$
$\Rightarrow \tan x\tan y = \pm c = C$ ... $\left[ {\because \,\left| t \right| = a\left( {a \geqslant 0} \right) \Rightarrow t = \pm a} \right]$
View full question & answer→Question 152 Marks
Find the general solution of the differential equation $\frac{d y}{d x}+y=1~(y \neq 1)$
AnswerGiven equation is: $ \frac{d y}{d x}$ + y = 1
$\Rightarrow$ dy = (1 - y) dx
Separating variables
$\Rightarrow \frac{\mathrm{dy}}{1-\mathrm{y}}$ = dx
$\Rightarrow \int \frac{d y}{1-y}=\int d x$
$\Rightarrow \int \frac{d y}{y-1}=\int-d x$
$\Rightarrow$ $\log (y-1)=-x+c$
$\Rightarrow$ $y-1=e^{(-x+c)}$
$\Rightarrow$ $y=e^{-x} \times e^{c}+1$
Putting $e^{c}=A$
$y=A e^{-x}+1$ is the general solution.
View full question & answer→Question 162 Marks
Find the general solution of the differential equation $\frac{d y}{d x}=\sqrt{4-y^{2}}(-2<y<2)$
AnswerHere, we have
$ \frac{d y}{d x}=\sqrt{4-y^{2}}$
$\Rightarrow \frac{d y}{\sqrt{4-y^{2}}}$ = dx
Integrating throughout, we get
$ \int \frac{\mathrm{dy}}{\sqrt{4-\mathrm{y}^{2}}}=\int \mathrm{d} \mathrm{x}$
$\Rightarrow \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}$
$\Rightarrow \sin ^{-1} \frac{y}{2}$ = x + c
$\Rightarrow \frac{y}{2}=\sin (x+c)$
$\Rightarrow y=2 \sin (x+c)$
$\therefore ~y=2 \sin (x+c) $is the general solution of the given differential equation.
View full question & answer→Question 172 Marks
Find the equation of a curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
AnswerGiven that $ y\frac{dy}{dx}=x $ $ydy=xdx$
$\int ydy=\int xdx$
$\\\frac{y^{2}}{2}=\frac{x^{2}}{2}+c$
When x = 0 and y = 2, we get
$\frac{-2^{2}}{2}=\frac{0^{2}}{2}+c$
c = 2
$\frac{y^{2}}{2}=\frac{x^{2}}{2}+2$
$y^{2}-x^{2}=4$
View full question & answer→Question 182 Marks
Find a solution of $\frac{{dy}}{{dx}} = y\tan x$ which satisfy the condition y = 1 when x = 0.
AnswerGiven: Differential equation $\frac{{dy}}{{dx}} = y\tan x$ $\Rightarrow dy = y\tan xdx$
$\Rightarrow \frac{{dy}}{y} = \tan xdx$ [Separating variables]
Integrating both sides,
$\Rightarrow \int {\frac{1}{y}} dy = \int {\tan xdx} $
$\Rightarrow \log \left| y \right| = \log \left| {\sec x} \right| + \log \left| c \right|$
$ \Rightarrow \log \left| y \right| = \log \left| {c\sec x} \right|$
$\Rightarrow \left| y \right| = \left| {c\sec x} \right|$
$\Rightarrow y = \pm c\sec x$
$\Rightarrow y = C\sec x$ where $ \pm c = C$…(i)
Now putting y = 1 and x = 0 in eq. (i), we get 1 = C sec 0 = C
Putting C = 1 in eq. (i), we get the required general solution
$ \Rightarrow y = \sec x$
View full question & answer→Question 192 Marks
Find a solution of $\cos \left( {\frac{{dy}}{{dx}}\;} \right) = a\;(a\; \in \;R)$ which satisfy the condition y = 1 when x = 0.
Answer$\frac{dy}{dx}=cos^{-1}a $
$ \int dy=cos^{-1}a\int dx $
$ y=xcos^{-1}a+c$
When y = 1 , x = 0 , then 1=0 $cos^{-1}a+c$ $c=1$
$\therefore y=xcos^{-1}a+1$
$\therefore \frac{y-1}{x}=cos^{-1}a$
View full question & answer→Question 202 Marks
Find the general solution of the differential equation $\frac{{dy}}{{dx}} = \frac{{1 - \cos x}}{{1 + \cos x}}$
AnswerGiven: Differential equation $\frac{{dy}}{{dx}} = \frac{{1 - \cos x}}{{1 + \cos x}}$
$ \Rightarrow dy = \frac{{1 - \cos x}}{{1 + \cos x}}dx$
Integrating both sides, $\int {dy = \int {\frac{{2{{\sin }^2}\frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}dx} } $
$ \Rightarrow y = \int {{{\tan }^2}\frac{x}{2}dx} $
$ \Rightarrow y= \int {\left( {{{\sec }^2}\frac{x}{2} - 1} \right)} dx$
$\Rightarrow y= \frac{{\tan \frac{x}{2}}}{{\frac{1}{2}}} - x + c$
$\Rightarrow y = 2\tan \frac{x}{2} - x + c$
View full question & answer→Question 212 Marks
Verify that the function $x + y = \tan^{-1}y$ (explicit or implicit) is a solution of differential equation $y^2y' + y^2 + 1 = 0$.
AnswerGiven: $x + y = \tan^{-1}y$ …(i)
To prove:y given by eq. (i) is a solution of differential equation $y^2y' + y^2 + 1 = 0$ …(ii)
Proof: Differentiating both sides of eq. (i) w.r.t x we have
$1 + y' = \frac{1}{{1 + {y^2}}}y'$
$ \Rightarrow \left( {1 + y'} \right)\left( {1 + {y^2}} \right) = y'$
$ \Rightarrow 1 + {y^2} + y' + y'{y^2} = y'$
$ \Rightarrow {y^2}y' + {y^2} + 1 = 0$
Hence, function given by eq. (i) is a solution of $y^2y' + y^2 + 1 = 0$
View full question & answer→Question 222 Marks
Verify that the function y - cos y = x (explicit or implicit) is a solution of differential equation (y sin y + cos y + x)y' = y
AnswerGiven: y - cos y = x ....(i) To prove: y given by eq. (i) is a solution of differential equation
(y sin y + cosy + x)y' = y ....(ii)
Proof: Differentiating both sides of eq. (i) w.r.t x, we have
y' + (sin y)y' = 1
$\Rightarrow y'\left( {1 + \sin y} \right) = 1$
$ \Rightarrow y'=\frac{1}{{1 + \sin y}}$ ....(iii)
Putting the value of x from eq. (i) and value of y' from eq. (iii) in L.H.S. of eq. (ii),
(y sin y + cos y + x)y'
$\Rightarrow \left( {y\sin y + \cos y + y - \cos y} \right)\frac{1}{{1 + \sin y}}$
$ \Rightarrow \left( {y\sin y + y} \right)\frac{1}{{1 + \sin y}}$
$ \Rightarrow y\left( {\sin y + 1} \right)\frac{1}{{1 + \sin y}} = y$ = R.H.S. of (ii)
Hence, Function given by eq. (i) is a solution of (y sin y + cos y + x)y' = y.
View full question & answer→Question 232 Marks
Verify that the function xy = log y + C (explicit or implicit) is a solution of differential equation $y' = \frac{{{y^2}}}{{1 - xy}}\left( {xy \ne 1} \right)$
AnswerGiven: xy = log y + C …(i) To prove:y given by eq. (i) is a solution of differential equation $y' = \frac{{{y^2}}}{{1 - xy}}$ ....(ii)
Proof: Differentiating both sides of eq. (i) w.r.t x, we have
$xy' + y\left( 1 \right) = \frac{1}{y}y' + 0$
$ \Rightarrow xy' - \frac{{y'}}{y} = - y$
$ \Rightarrow y'\left( {x - \frac{1}{y}} \right) = - y$
$ \Rightarrow y'\left( {\frac{{xy - 1}}{y}} \right) = - y$
$\Rightarrow y'\left( {xy - 1} \right) = - {y^2}$
$ \Rightarrow y' = \frac{{ - {y^2}}}{{xy - 1}}$
$ \Rightarrow y' = \frac{{ - {y^2}}}{{ - \left( {1 - xy} \right)}} = \frac{{{y^2}}}{{1 - xy}}$
Hence, function (implicit) given by eq. (i) is a solution of $y' = \frac{{{y^2}}}{{1 - xy}}$.
View full question & answer→Question 242 Marks
Verify that the function y = x sin x (explicit or implicit) is a solution of differential equation $xy' = y + x\sqrt {{x^2} - {y^2}} $ $\left( {x \ne 0\,\,and\,\,x > y\,or\,x < - y\,} \right)$
AnswerGiven: y = x sin x …(i)
To prove:y given by eq. (i) is a solution of differential equation $xy' = y + x\sqrt {{x^2} - {y^2}} $ ...(ii)
Proof: From eq. (i),
$\frac{{dy}}{{dx}}\left( { = y'} \right) = x\frac{d}{{dx}}\sin x + \sin x\frac{d}{{dx}}x$
= x cos x + sin x
L.H.S. of eq. (ii)
$= xy' = x(x \cos x + \sin x) = x^2 \cos x + x \sin x$
R.H.S. of eq. (ii)
$ = y + x\sqrt {{x^2} - {y^2}} $
$ = x\sin x + x\sqrt {{x^2} - {x^2}{{\sin }^2}x} $ [from eq. (i)]
$ = x\sin x + x\sqrt {{x^2}\left( {1 - {{\sin }^2}x} \right)} $
$= x\sin x + x\sqrt {{x^2}{{\cos }^2}x} $
$= x\sin x + x.x\cos x$
$= x\sin x + {x^2}\cos x$
$= {x^2}\cos x + x\sin x$
$\therefore$ L.H.S. = R.H.S
Hence, y given by eq. (i) is a solution of $xy' = y + x\sqrt {{x^2} - {y^2}} $.
View full question & answer→Question 252 Marks
Verify that the function y = Ax (explicit or implicit) is a solution of differential equation $xy' = y\left( {x \ne 0} \right)$
AnswerGiven: y = Ax …(i)
To prove:y given by eq. (i) is a solution of differential equation $xy' = y\left( {x \ne 0} \right)$…(ii)
Proof: From eq. (i)
y' = A(1) = A
L.H.S. of eq. (ii)
= xy' = xA = Ax = y = R.H.S. of eq. (ii)
$\therefore$ given by eq. (i) is a solution of differential equation $xy' = y\left( {x \ne 0} \right)$.
View full question & answer→Question 262 Marks
Verify that the function $y = \sqrt {1 + {x^2}}$ (explicit or implicit) is a solution of differential equation $y' = \frac{{xy}}{{1 + {x^2}}}$
AnswerGiven: $y = \sqrt {1 + {x^2}} $ …(i)
To prove:y is a solution of the differential equation $y' = \frac{{xy}}{{1 + {x^2}}}$ …(ii)
Proof: From eq. (i),
$y' = \frac{d}{{dx}}\sqrt {1 + {x^2}} = \frac{d}{{dx}}{\left( {1 + {x^2}} \right)^{1/2}}$
$= \frac{1}{2}{\left( {1 + {x^2}} \right)^{ - 1/2}}\frac{d}{{dx}}\left( {1 + {x^2}} \right) = \frac{1}{2}{\left( {1 + {x^2}} \right)^{ - 1/2}}.2x$
$ = \frac{x}{{\sqrt {1 + {x^2}} }}$ …(iii)
Now R.H.S. of eq. (ii)
$ = \frac{{xy}}{{1 + {x^2}}}$
$= \frac{x}{{1 + {x^2}}}\sqrt {1 + {x^2}} $ [From eq. (i)]
$ = \frac{x}{{\sqrt {1 + {x^2}} }} = y'$
$\therefore$ L.H.S. = R.H.S
Hence, y given by eq. (i) is a solution of $y' = \frac{{xy}}{{1 + {x^2}}}$.
View full question & answer→Question 272 Marks
Verify that the function y = cos x + C (explicit or implicit) is a solution of differential equation y' + sin x = 0.
AnswerGiven: y = cos x + C ....(i)
To prove: y is a solution of the differential equation y' + sin x = 0 ...(ii)
Proof: From eq. (i),
y' = -sin x
L.H.S. of eq. (ii),
y' + sin x = -sin x + sin x = 0 = R.H.S.
Hence, y given by eq. (i) is a solution of y' + sin x = 0.
View full question & answer→Question 282 Marks
Verify that the function $y = x^2 + 2x + C$ (explicit or implicit) is a solution of differential equation $y' - 2x - 2 = 0$
AnswerGiven: $y = x^2 + 2x + C ...(i)$
To prove: y is a solution of the differential equation $y' - 2x - 2 = 0 ...(ii)$
Proof:From, eq. (i),
$y' = 2x + 2$
L.H.S. of eq. (ii),
$= y' - 2x - 2$
$= (2x + 2) - 2x - 2$
$= 2x + 2 - 2x - 2 = 0 =$ R.H.S.
Hence, y given by eq. (i) is a solution of $y' - 2x - 2 = 0.$
View full question & answer→Question 292 Marks
Verify that the function $y = \sqrt {{a^2} - {x^2}} , $ $x\in \left( { - a,a} \right)$ (explicit or implicit) is a solution of differential equation $x + y\frac{{dy}}{{dx}} = 0\left( {y \ne 0} \right)$
AnswerGiven: $y = \sqrt {{a^2} - {x^2}} , x\in \left( { - a,a} \right)$ …(i)
To prove:y given by eq. (i) is a solution of differential equation $x + y\frac{{dy}}{{dx}} = 0$ …(ii)
Proof: From eq. (i),
$\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{a^2} - {x^2}} }}\left( { - 2x} \right)$
$ = \frac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}$ …(iii)
Putting the values of y and $\frac{{dy}}{{dx}}$ from eq. (i) and (iii) in L.H.S. of eq. (ii),
$ = x + y\frac{{dy}}{{dx}}$
$= x + \sqrt {{a^2} - {x^2}} \left( {\frac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}} \right)$
= x - x = 0 = R.H.S. of eq. (ii)
Hence, function given by eq. (i) is a solution of $x + y\frac{{dy}}{{dx}} = 0$.
View full question & answer→Question 302 Marks
Verify that the function $y = e^x + 1$(explicit or implicit) is a solution of differential equation $y" - y' = 0$
AnswerIt is given that $y = e^x + 1$
Now, differentiating both sides w.r.t. x, we get,
$\frac{d y}{d x}=y^\prime=\frac{d}{d x}\left(e^{x}\right)=e^x$ ...(i)
Now, Again, differentiating both sides w.r.t. x, we get,
$\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^{x}\right)$
$\Rightarrow y'' = e^x=y'$ ...[Using (i)]
$\Rightarrow y'' = y'$
$\Rightarrow y'' - y'=0$
View full question & answer→Question 312 Marks
Find the particular solution of the differential equation $\frac{d y}{d x} = -4xy^2$ given that $y = 1$, when $x = 0$
AnswerIf y ≠ 0, the given differential equation can be written as
$\frac{d y}{y^{2}}$ = -4x dx ...(i)
Integrating both sides of equation (i), we get
$\int \frac{d y}{y^{2}}=-4 \int x d x$
or $-\frac{1}{y} = -2x^2 + C$
or y = $\frac{1}{2 x^{2}-\mathrm{C}}$ ...(ii)
Substituting y = 1 and x = 0 in equation (ii), we get, C = - 1
Now substituting the value of C in equation (ii), we get the particular solution of the given differential equation as $y=\frac{1}{2 x^{2}+1}$
View full question & answer→Question 322 Marks
Find the general solution of the differential equation $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$
AnswerThe given differential equation is $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$
Since $1 + y^2 \ne 0$, therefore separating the variables, the given differential equation can be written as
$\frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}}$ ...(i)
Integrating both sides of equation (i), we get
$\int \frac{d y}{1+y^{2}}=\int \frac{d x}{1+x^{2}}$
or $\tan^{-1} y = \tan^{-1}x + C$
which is the general solution of equation (i)
View full question & answer→Question 332 Marks
Find the general solution of the differential equation $\frac{d y}{d x}=\frac{x+1}{2-y}$, (y $\neq$ 2)
AnswerWe have $\frac{d y}{d x}=\frac{x+1}{2-y}$ ...(i)
Separating the variables in equation (i), we get
$(2 - y) dy = (x + 1) dx ...(ii)$
Integrating both sides of equation (ii), we get
$\int(2 - y) dy = \int(x + 1) dx$
or $2y - \frac{y^{2}}{2}=\frac{x^{2}}{2} + x + C_1$
or $x^2 + y^2 + 2x - 4y + 2 C_1 = 0$
or $x^2 + y^2 + 2x - 4y + C = 0,$ where $C = 2C_1$
Which is the general solution of equation (i)
View full question & answer→Question 342 Marks
Verify that the function y = a cos x + b sin x, where, a, b $\in$ R is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}$ + y = 0
AnswerWe have to show y = a cosx + b sinx ...(i) is a solution of $\frac{d^{2} y}{d x^{2}} + y = 0$ ....(ii)
Differentiating both sides of equation (i) with respect to x, successively, we get
$\frac{d y}{d x}$ = -a sinx + b cosx
$\frac{d^{2} y}{d x^{2}}$ = -a cosx - b sinx
Substituting the values of $\frac{d^{2} y}{d x^{2}}$ and y in equation (ii), we get
L.H.S = (-a cosx - b sinx) + (a cosx + b sinx) = 0 = R.H.S.
Therefore, the given function is a solution of the given differential equation.
View full question & answer→Question 352 Marks
Verify that the function $y = e^{-3x}$ is a solution of the differential equation $\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} - 6y = 0$
AnswerGiven function is $y = e^{-3x}$
Differentiating both sides of equation with respect to x , we get
$\frac{d y}{d x}=-3 e^{-3x}$ ...(i)
Now, differentiating (1) with respect to x, we have$\frac{d^{2} y}{d x^{2}}=9 e^{-3 x}$
Substituting the values of $\frac{d^{2} y}{d x^{2}}, \frac{d y}{d x}$ and y in the given differential equation, we get
$L.H.S. = 9.e^{-3x} + (-3.e^{-3x}) - 6.e^{-3x} = 9.e^{-3x} - 9.e^{-3x} = 0 = R.H.S.$
Therefore, the given function is a solution of the given differential equation.
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