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Differential Equations · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 20 questions.

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Question 13 Marks
Find a particular solution of the differential equation $(x+1) \frac{d y}{d x}=2 e^{-y}-1$ given that y = 0 when x = 0.
Answer
It is given that $(x+1) \frac{d y}{d x}=2 e^{-y}-1$ 
$\Rightarrow \frac{d y}{2 e^{-y}-1}=\frac{d x}{x+1}$ 
$\Rightarrow \frac{e^{y} d y}{2-e^{y}}=\frac{d x}{x+1}$ 
On integrating both sides, we get,
$\int \frac{e^{y} d y}{2-e^{y}}=\log |x+1|+\log C$ .....(i)
Let $2-e^{y}=t$ 
$\therefore \frac{d}{d y}\left(2-e^{y}\right)=\frac{d t}{d y}$ 
$\Rightarrow-\mathrm{e}^{\mathrm{y}}=\frac{\mathrm{dt}}{\mathrm{dy}}$ 
$\Rightarrow$ eydy = -dt
Substituting value in equation (i), we get,
$\int \frac{-\mathrm{dt}}{\mathrm{t}}=\log |\mathrm{x}+1|+\log \mathrm{c}$ 
$\Rightarrow$ -log|t| = log|C(x+1)|
$\Rightarrow$ -log|2 - ey| = log|C(x + 1)|
$\Rightarrow \frac{1}{2-\mathrm{e}^{\mathrm{y}}}=\mathrm{C}(\mathrm{x}+1)$ 
$\Rightarrow 2-\mathrm{e}^{\mathrm{y}}=\frac{1}{\mathrm{c}(\mathrm{x}+1)}$ ......(ii)
Now, at x = 0 and y = 0, equation (ii) becomes,
$\Rightarrow 2-1=\frac{1}{C}$ 
$\Rightarrow$ C = 1
Now, substituting the value of C in equation (ii), we get,
$\Rightarrow 2-\mathrm{e}^{\mathrm{y}}=\frac{1}{(\mathrm{x}+1)}$ 
$\Rightarrow \mathrm{e}^{\mathrm{y}}=2-\frac{1}{(\mathrm{x}+1)}$ 
$\Rightarrow \mathrm{e}^{\mathrm{y}}=\frac{2 \mathrm{x}+2-1}{(\mathrm{x}+1)}$ 
$\Rightarrow \mathrm{e}^{\mathrm{y}}=\frac{2 \mathrm{x}+1}{(\mathrm{x}+1)}$ 
$\Rightarrow \mathrm{y}=\log \left|\frac{2 \mathrm{x}+1}{\mathrm{x}+1}\right|,(\mathrm{x} \neq-1)$ 
Therefore, the required particular solution of the given differential equation is
$y=\log \left|\frac{2 x+1}{x+1}\right|,(x \neq-1)$
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Question 23 Marks
Find the general solution of $\frac{{dy}}{{dx}} + \left( {\sec x} \right)y = \tan x\left( {0 \leq x < \frac{\pi }{2}} \right)$
Answer
Given: Differential equation $\frac{{dy}}{{dx}} + \left( {\sec x} \right)y = \tan x$
Comparing with $\frac{{dy}}{{dx}} + Py = Q$, we have P = sec x and Q = tan x
$\therefore \int {Pdx = \int {\sec xdx = \log \left( {\sec x + \tan x} \right)} } $
$I.F = {e^{\int {Pdx} }} = {e^{\log \left( {\sec x + \tan x} \right)}} = \sec x + \tan x$
Solution is
$ = y\left( {I.F} \right) = \int {Q\left( {I.F} \right)} dx + c$
$\Rightarrow y\left( {\sec x + \tan x} \right) = \int {\tan x\left( {\sec x + \tan x} \right)} dx + c$
$\Rightarrow y\left( {\sec x + \tan x} \right) = \int {\left( {\sec x\tan x + {{\tan }^2}x} \right)} dx + c$
$\Rightarrow y\left( {\sec x + \tan x} \right) = \int {\left( {\sec x.\tan x + {{\sec }^2}x - 1} \right)} dx + c$
$\Rightarrow y\left( {\sec x + \tan x} \right) = \sec x + \tan x - x + c$
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Question 33 Marks
Find the general solution of $\frac{d y}{d x}+\frac{y}{x}=x^{2}$
Answer
It is given that $\frac{d y}{d x}+\frac{y}{x}=x^{2}$ 
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where, p = $\frac{1}{x}$ and Q = x2)
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}$ 
Thus, the solution of the given differential equation is given by the relation:
$y(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) \mathrm{d} \mathrm{x}+\mathrm{C}$ 
$\Rightarrow \mathrm{y}(\mathrm{x})=\int\left(\mathrm{x}^{2} \cdot \mathrm{x}\right) \mathrm{d} \mathrm{x}+\mathrm{C}$ 
$\Rightarrow x y=\int\left(x^{3}\right) d x+C$ 
$\Rightarrow x y=\frac{x^{4}}{4}+C$ 
Therefore, the required general solution of the given differential equation is $x y=\frac{x^{4}}{4}+C$
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Question 43 Marks
Find the general solution of $\frac{d y}{d x}+3 y=e^{-2 x}$
Answer
It is given that $\frac{d y}{d x}+3 y=e^{-2 x}$ 
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where, p = 3 and Q = e-2x)
Now, I.F. = $e^{\int p d x}=e^{\int 3 d x}=e^{3 x}$ 
Thus, the solution of the given differential equation is given by the relation:
$y(\mathrm{I} . \mathrm{F})=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) \mathrm{d} \mathrm{x}+\mathrm{C}$ 
$\Rightarrow \mathrm{ye}^{3 \mathrm{x}}=\int\left(\mathrm{e}^{-2 \mathrm{x}} \times \mathrm{e}^{3 \mathrm{x}}\right) \mathrm{d} \mathrm{x}+\mathrm{C}$ 
$\Rightarrow \mathrm{ye}^{3 \mathrm{x}}=\int \mathrm{e}^{\mathrm{x}} \mathrm{dx}+\mathrm{C}$ 
$\Rightarrow$ ye3x = ex + C
$\Rightarrow$ y = e-2x + Ce-3x
Therefore, the required general solution of the given differential equation is y = e-2x + Ce-3x 
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Question 53 Marks
Find the particular solution of the differential equation $\frac { d y } { d x } - 3 y \cot x = \sin 2 x$ , given that y = 2 when $x = \frac { \pi } { 2 }$.
Answer
Given,  $\frac { d y } { d x } - 3 y \cot x = \sin 2 x$ ...(i)
This is a linear differential equation of the form
$\frac { d y } { d x } + P y = Q$, here P = -3 cot x and Q = sin 2x
$\therefore \quad \mathrm { I } = e ^ { \int \mathrm { P } d x } = e ^ { - 3 \int \cot x d x }$
$\Rightarrow \quad \mathrm { IF } = e ^ { - 3 \log | \sin x | } = e ^ { \log | \sin x | ^ { - 3 } } = | \sin x | ^ { - 3 }$
$\therefore$ The general solution of differential equation is given by
$y \times \mathrm { IF } = \int ( \mathrm { IF } \times Q ) d x + C$
$\Rightarrow \quad y ( \sin x ) ^ { - 3 } = \int ( \sin x ) ^ { - 3 } ( \sin 2 x ) d x + C$
$= \int \frac { 2 \sin x \cos x } { \sin ^ { 3 } x } d x + C$
$\therefore \quad y ( \sin x ) ^ { - 3 } = \int \frac { 2 \cos x } { \sin ^ { 2 } x } d x + C$ ...(i)
Therefore,on putting sin x = t $\Rightarrow$ cos x dx = dt  in  Eq. (i), we get
$y ( \sin x ) ^ { - 3 } = 2 \int \frac { d t } { t ^ { 2 } } + C = 2 \times \frac { t ^ { - 1 } } { - 1 } + C$
$\Rightarrow \quad y ( \sin x ) ^ { - 3 } = - \frac { 2 } { t } + C$
$\Rightarrow \quad y ( \sin x ) ^ { - 3 } = \frac { - 2 } { \sin x } + C$ [put t = sin x]
$\Rightarrow$ y = -2 sin2 x + C sin3 x ...(ii)
Therefore,on putting $x = \frac { \pi } { 2 }$ and y = 2 in Eq. (ii), we get,
$2 = - 2 \sin ^ { 2 } \frac { \pi } { 2 } + C \sin ^ { 3 } \frac { \pi } { 2 } \Rightarrow 2 = - 2 \cdot 1 + C \cdot 1$
$\Rightarrow $ C = 4
$\therefore y = - 2 \sin ^ { 2 } x + 4 \sin ^ { 3 } x$
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Question 63 Marks
Find the particular solution of the differential equation $\left( 1 + x ^ { 2 } \right) \frac { d y } { d x } + 2 x y = \frac { 1 } { 1 + x ^ { 2 } }$, given that y = 0 when x = 1.
Answer
$\left(1+x^{2}\right) \frac{d y}{d x}+2 x y$ = $\frac{1}{1+x^{2}}$
Divide both sides by 1 + x2
$\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}$ = $\frac{1}{\left(1+x^{2}\right) \cdot\left(1+x^{2}\right)}$
$\frac{d y}{d x}+\left(\frac{2 x }{1+x^{2}}\right) y$ = $\frac{1}{\left(1+x^{2}\right)^2}$
Comparing with $\frac{d y}{d x}$ + Py = Q,
P = $\frac{2 x}{1+x^{2}}$ & Q = $\frac{1}{\left(1+x^{2}\right)^{2}}$
Finding Integrating factor:
IF = $e^{\int P d x}$
IF = $e^{\int \frac{2 x}{1+x^{2}}} d x$
Let 1 + x2 = t
Diff. w.r.t. x
2x = $\frac{dt}{d x} $
dx = $\frac{d t}{2 x}$
Thus, IF = $\mathrm{e}^{\int \frac{2 x}{t} \frac{d t}{2 x}}$
IF = $\mathrm{e}^{\int \frac{d t}{t}}$
IF = $\mathrm{e}^{\log |t|}$
IF = t
IF = 1 + x2
Solution of the differential equation:
y $\times$ I.F. = $\int {Q\, \times \,I.F\,dx} $
Putting values,
y $\times$ (1 + x2) = ${\int {\frac{1}{{(1 + {x^2})^2}}} (1 + {x^2})dx} $ 
y (1 + x2) = $\int{\frac{1}{{(1 + {x^2})}}} $ dx
y (1 + x2) = tan-1x + C ...(1)
Putting that y = 0 and x = 1,
0(1 + 12) = tan-1(1) + C
0 = $\frac{\pi}{4}$ + C
C = -$\frac{\pi}{4}$
Putting value of C in eq(1),
y(1 + x2) = tan-1x + C
y(1 + x2) = tan-1x - $\frac{\pi}{4}$
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Question 73 Marks
Find the particular solution of the differential equation $\frac { d y } { d x }$+ 2y tan x = sin x, given that y = 0 when x = $\frac{\pi}{3}$.
Answer
According to the question, $\frac { d y } { d x }$+ 2ytanx = sinx
Given equation is a linear differential equation in the form $\frac{dy}{dx}$+ Py = Q
Here, P = 2tanx and Q = sinx
Now, Integration Factor(IF) = $e ^ { \int P d x } = e ^ { \int 2 \tan x d x }$
$= e ^ { 2 \int \tan x d x } \\= e ^ { 2 \log | \sec x | }$
$= e ^ { \log \sec ^ { 2 } x } $
$= \sec ^ { 2 } x$$\left[ \because e ^ { \log f ( x ) } = f ( x ) \right]$
The solution of differential equation is given by
$y \cdot ( I F ) = \int ( \operatorname { IF } ) Q d x + C$
$\Rightarrow y \cdot \sec ^ { 2 } x = \int \sec ^ { 2 } x \cdot \sin x d x + C$
$\Rightarrow y \cdot \sec ^ { 2 } x = \int \frac { \sin x } { \cos ^ { 2 } x } d x + C$
$\Rightarrow y \cdot \sec ^ { 2 } x = \int \frac { \sin x } { \cos x } \cdot \frac { 1 } { \cos x } d x + C$
$\Rightarrow y \cdot \sec ^ { 2 } x = \int \tan x \cdot \sec x d x + C$
$\Rightarrow y \cdot \sec ^ { 2 } x = \sec x + C$
 Dividing with sec2x on both sides, we get
$\Rightarrow y = \frac { 1 } { \sec x } + \frac { C } { \sec ^ { 2 } x }$
$\Rightarrow$y = cosx + C.cos2x
According to the question, y = 0 when x = $\frac{\pi}{3}$
$0 = \cos \frac { \pi } { 3 } + C \cdot \cos ^ { 2 } \frac { \pi } { 3 }$
$\Rightarrow 0 = \frac { 1 } { 2 } + C \cdot \frac { 1 } { 4 }$
$\Rightarrow \frac { - 1 } { 2 } = \frac { C } { 4 }$
$\Rightarrow C = - 2$
Now, y = cosx + C.cos2x $\Rightarrow$ y = cos x - 2 cos2x
Therefore, the required particular solution is y = cos x - 2 cos2x
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Question 83 Marks
Find the general solution of $\left(x+3 y^{2}\right) \frac{d y}{d x}=y(y>0)$
Answer
It is given that $\left(x+3 y^{2}\right) \frac{d y}{d x}=y$ 
$\Rightarrow \frac{d y}{d x}=\frac{y}{x+3 y^{2}}$ 
$\Rightarrow \frac{d x}{d y}=\frac{x+3 y^{2}}{y}=\frac{x}{y}+3 y$ 
$\Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y$ 
This is equation in the form of $\frac{d x}{d y}+p x=Q$ (where, p = -$\frac{1}{y}$ and Q = 3y)
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{-\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{-\log \mathrm{y}}=\mathrm{e}^{\log \left(\frac{1}{\mathrm{y}}\right)}=\frac{1}{\mathrm{y}}$ 
Thus, the solution of the given differential equation is given by the relation:
$x(I. F )=\int(Q \times I . F ) d y+C$ 
$\Rightarrow \mathrm{x} \cdot \frac{1}{\mathrm{y}}=\int\left[3 \mathrm{y} \cdot \frac{1}{\mathrm{y}}\right] \mathrm{dy}+\mathrm{C}$ 
$\Rightarrow \frac{x}{y}=3 y+C$ 
$\Rightarrow$ x = 3y2 + Cy
Therefore, the required general solution of the given differential equation is x = 3y2 + Cy.
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Question 93 Marks
Find the general solution of $y d x+\left(x-y^{2}\right) d y=0$
Answer
It is given that $\mathrm{ydx}+\left(\mathrm{x}-\mathrm{y}^{2}\right) \mathrm{dy}=0$ 
$\Rightarrow \mathrm{ydx}=\left(\mathrm{y}^{2}-\mathrm{x}\right) \mathrm{dy}$ 
$\Rightarrow \frac{d x}{d y}=\frac{\left(y^{2}-x\right)}{y}=y-\frac{x}{y}$ 
$\Rightarrow \frac{d x}{d y}+\frac{x}{y}=y$ 
This is equation in the form of $\frac{d x}{d y}+p x=Q$ (where, p = $\frac{1}{y}$ and Q = y)
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdy}}=\mathrm{e}^{\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{\log \mathrm{y}}=\mathrm{y}$ 
Thus, the solution of the given differential equation is given by the relation:
$x(I . F .)=\int(Q \times I . F .) d y+C$ 
$\Rightarrow \mathrm{x} \cdot \mathrm{y}=\int[\mathrm{y} \cdot \mathrm{y}] \mathrm{d} \mathrm{y}+\mathrm{C}$ 
$\Rightarrow x . y=\int y^{2} d y+C$ 
$\Rightarrow x \cdot y=\frac{y^{3}}{3}+C$ 
$\Rightarrow x =\frac{y^{2}}{3}+\frac{C}{y}$ 
Therefore, the required general solution of the given differential equation is $x =\frac{y^{2}}{3}+\frac{c}{y}$.
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Question 103 Marks
In a bank, principal increases continuously at the rate of 5% per year. An amount of ₹ 1000 is deposited with this bank, how much will it worth after 10 years $\left( {{e^{0.5}} = 1.648} \right)$.
Answer
Let P be the principal (amount) at the end of t years.
According to the given condition, rate of increase of principal per year = 5% (of principal)
$ \Rightarrow \frac{{dP}}{{dt}} = \frac{5}{{100}} \times P$
$\Rightarrow \frac{{dP}}{{dt}} = \frac{P}{{20}}$
$\Rightarrow \frac{{dP}}{P} = \frac{{dt}}{{20}}$ [Separating variables]
Integrating both sides,
log P = $\frac{1}{20}$ t + c ...(i)
[Since P being principal > 0, hence $\log \left| P \right| = \log P$ ]
Now initial principal = ₹ 1000 (given), i.e., when t = 0 then P = 1000
Therefore, putting t = 0, P = 1000 in eq. (i), log 1000 = c
Putting $\log 1000 = c$ in eq. (i), log P = $\frac{1}{20}$ t + log 1000
$\Rightarrow$ log P - log 1000 = $\frac{1}{20}$t
$\Rightarrow$log $\frac{P}{1000}=\frac{1}{20} t$ ...(ii)
Now putting t = 10 years (given)
$\log \frac{P}{{100}} = \frac{1}{{20}} \times 10 = \frac{1}{2} = 0.5$
$\Rightarrow \frac{P}{{1000}} = {e^{0.5}}$ [$\because$ If x = t, then x = et
P = 1000 $\times$ 1.648 = ₹ 1648
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Question 113 Marks
In a bank, the principal increases continuously at the rate of r% per year. Find the value of r if Rs.100 double itself in 10 years (loge2 = 0.6931).
Answer
Let P be the principal at any time t. then,

$\frac{{dP}}{{dt}} = \frac{{rP}}{{100}} \Rightarrow \frac{{dP}}{{dt}} = \frac{P}{{100}}$

$ \Rightarrow \int {\frac{1}{P}dP = \int {\frac{r}{{100}}dt} } $

$ \Rightarrow \log P = \frac{r}{{100}}t + \log c$

$ \Rightarrow \log \frac{P}{c} = \frac{r}{{100}}t$

$\Rightarrow P = c{e^{\frac{r}{{100}}}}$
When P = 100 and t = 0., then, c = 100, Thus, we have:
$\Rightarrow P=100 \ {{e}^{{}^{r}\!\!\diagup\!\!{}_{100}\;}}$
Now, suppose t = T, when P = 100., then;
$\Rightarrow 200 = 100{e^{\frac{T}{{100}}}}$

$ \Rightarrow {e^{\frac{T}{{100}}}} = 2$

$ \Rightarrow T = 100\log 2$ = 100(0.6931) = 6.93%.

Which is the required solution

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Question 123 Marks
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).
Answer
Given that $\frac{dy}{dx}=2\frac{y+3}{x+4}$

$\frac{dy}{y+3}=\frac{2dx}{x+4}$

$\\\int \frac{dy}{y+3}=\int \frac{2dx}{x+4}$

$\\log|y+3|=2log|x+4|+logc$

Here, x= -2 and y = 1

 $log|1+3|=2log|-2+4|+logc$

$log4=log4+logc$

$logc=0$

$\log|y+3|=2log|x+4|$

$y+3=(x+4)^{2}$

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Question 133 Marks
For the differential equation $x y \frac{d y}{d x}=(x+2)(y+2)$, find the solution curve passing through the point (1, -1).
Answer
For this question, we need to find the particular solution at point(1, -1) for the given differential equation.
Given differential equation is
$\Rightarrow x y \frac{d y}{d x}=(x+2)(y+2)$ 
Separating variables,
$\Rightarrow \frac{y}{y+2} d y=\frac{(x+2) d x}{x}$ 
Or,
$\Rightarrow\left(1-\frac{2}{\mathrm{y}+2}\right) \mathrm{dy}=\left(1+\frac{2}{\mathrm{x}}\right) \mathrm{d} \mathrm{x}$ 
Integrating both sides,
$\Rightarrow \int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x$ 
$\Rightarrow \int d y-2 \int \frac{1}{y+2} d y=\int d x+2 \int \frac{1}{x} d x$ 
$\Rightarrow y-2 \log (y+2)=x+2 \log x+C$ 
Now separating like terms on each side,
$\Rightarrow$ y - x - c = 2 log x + 2 log (y + 2)
$\Rightarrow$ y - x - c = log x2 + log(y + 2)2
Or,
$\Rightarrow$ y - x - c = log{x2(y + 2)2} .....(i)
Now we are given that, the curve passes through (1, -1)
$\Rightarrow$-1 - 1 - c = log{1(-1 + 2)2}
$\Rightarrow$ -2 - c = log (1)
$\Rightarrow$ c = -2 + 0 ($\because$ log(1) = 0)
So c = -2
Putting the value of c in (i)
y - x - (-2) = log{x2(y + 2)2}
y - x + 2 = log{x2(y + 2)2}
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Question 143 Marks
Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x.
Answer
$\frac{{dy}}{{dx}} = {e^x}\sin x $

$\int {dy = \int {{e^x}} } \sin xdx $

$ y = \frac{1}{2}\left( {\sin x - \cos x} \right){e^x} + C$

When x = y = 0 , we get

$0 = \frac{1}{2}\left( {\sin 0 - \cos 0} \right){e^0} + C$

$\\C=\frac{1}{2}$

Hence, $y = \frac{1}{2}\left( {\sin x - \cos x} \right){e^x} + \frac{1}{2} $

$2y-1=\left( {\sin x - \cos x} \right){e^x} $

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Question 153 Marks
Find a solution of $ \left( x ^ { 3 } + x ^ { 2 } + x + 1 \right) \frac { d y } { d x } = 2 x ^ { 2 } + x$ which satisfy the condition y = 1 when x = 0.
Answer
According to the question,
Given differential equation is , 
$ \left( x ^ { 3 } + x ^ { 2 } + x + 1 \right) \frac { d y } { d x } = 2 x^2 + x$
$ \Rightarrow \quad \frac { d y } { d x } = \frac { 2 x ^ { 2 } + x } { x ^ { 3 } + x ^ { 2 } + x + 1 }$
$ \therefore \quad d y = \frac { 2 x ^ { 2 } + x } { x ^ { 3 } + x ^ { 2 } + x + 1 } d x$
On integrating both sides, we get
$ \int d y = \int \frac { 2 x ^ { 2 } + x } { x ^ { 3 } + x ^ { 2 } + x + 1 } d x$
$ \Rightarrow \quad y = \int \frac { 2 x ^ { 2 } + x } { x ^ { 2 } ( x + 1 ) + 1 ( x + 1 ) } d x + C_2$
$ \Rightarrow \quad y = \int \frac { 2 x ^ { 2 } + x } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) } d x$ ....(i)
By Using partial fractions,
$ \frac { 2 x ^ { 2 } + x } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) } = \frac { A } { x + 1 } + \frac { B x + C } { x ^ { 2 } + 1 }$ ...(ii)
$ \Rightarrow \frac { 2 x ^ { 2 } + x } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) } = \frac { A \left( x ^ { 2 } + 1 \right) + ( B x + C ) ( x + 1 ) } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) }$
$ \Rightarrow \quad 2 x ^ { 2 } + x = A \left( x ^ { 2 } + 1 \right) + ( B x + C ) ( x + 1 )$
$ \Rightarrow \quad 2 x ^ { 2 } + x = A \left( x ^ { 2 } + 1 \right) + B \left( x ^ { 2 } + x \right) + C ( x + 1 )$
On comparing the coefficients of x2, x and constant terms from both sides, we get
$A + B = 2 \ ; \\B + C = 1$
$A+C = 0$

$ \Rightarrow$ A = -C
On solving above equations, we get
$ A = \frac { 1 } { 2 } , B = \frac { 3 } { 2 } \text { and } C = \frac { - 1 } { 2 }$
On substituting the values of A, B and C in Eq. (ii), we get
$ \frac { 2 x ^ { 2 } + x } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) } = \frac { \frac { 1 } { 2 } } { x + 1 } + \frac { \frac { 3 } { 2 } x - \frac { 1 } { 2 } } { x ^ { 2 } + 1 }$
On integrating both sides w.r.t to x , we get
$ \int \frac { 2 x ^ { 2 } + x } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) } d x = \frac { 1 } { 2 } \int \frac { d x } { x + 1 }$ $ + \frac { 3 } { 2 } \int \frac { x } { x ^ { 2 } + 1 } d x - \frac { 1 } { 2 } \int \frac { d x } { x ^ { 2 } + 1 }$
$ \Rightarrow y = \frac { 1 } { 2 } \log | x + 1 | + I _ { 1 } - \frac { 1 } { 2 } \tan ^ { - 1 } x + C _ { 2 }$ ...(iii) [from Eq. (1)]
where $ I _ { 1 } = \frac { 3 } { 2 } \int \frac { x } { x ^ { 2 } + 1 } d x$
Put $ x ^ { 2 } + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac { d t } { 2 }$
$ \therefore \quad I _ { 1 } = \frac { 3 } { 4 } \int \frac { d t } { t } = \frac { 3 } { 4 } \log | t | + C _ { 1 }$
$ = \frac { 3 } { 4 } \log \left| x ^ { 2 } + 1 \right| + C _ { 1 }$
On putting the value of I1 in Eq. (iii), we get
$ y = \frac { 1 } { 2 } \log | x + 1 | + \frac { 3 } { 4 } \log \left| x ^ { 2 } + 1 \right| - \frac { 1 } { 2 } \tan ^ { - 1 } x + C$[where, C = C1 + C2]

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Question 163 Marks
Find the general solution of the differential equation ${e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}y\,dy = 0$
Answer
${e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}y\,dy = 0$

$\Rightarrow$${e^x}\tan ydx = - \left( {1 - {e^x}} \right){\sec ^2}y\,dy$

$\Rightarrow$$\int {\frac{{{e^x}}}{{1 - {e^x}}}dx = - \int {\frac{{{{\sec }^2}y}}{{\tan y}}} } dy$

$\Rightarrow$$ - \log (1 - {e^x}) = - \log \tan y +\log c$

$\Rightarrow\log \left( {\frac{{\tan y}}{{1 - {e^x}}}} \right) = \log c$

$\Rightarrow\frac{{\tan y}}{{1 - {e^x}}} = c$

$\Rightarrow\tan y = c(1 - {e^x})$

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Question 173 Marks
Find the equation of a curve passing through the point (-2, 3), given that the slope of the tangent to the curve at any point (x, y) is $\frac{2 x}{y^{2}}$.
Answer
We know that the slope of the tangent to a curve is given by $\frac{d y}{d x}$
so, $\frac{d y}{d x}=\frac{2 x}{y^{2}}$ ...(i)
Separating the variables, equation (i) can be written as
y2dy = 2x dx ...(ii)
Integrating both sides of equation (ii), we get
$\int$ y2dy = $\int$ 2x dx
or $\frac{y^{3}}{3}$ = x2 + C ...(iii)
Substituting x = -2, y = 3 in equation (iii), we get C = 5
Substituting the value of C in equation (iii), we get the equation of the required curve as
$\frac{y^{3}}{3}$ + x2 + 5 or y = (3x2 + 15)1/3
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Question 183 Marks
Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x2 + 1) dx (x ≠ 0).
Answer
The given differential equation can be expressed as $d y^{}=\left(\frac{2 x^{2}+1}{x}\right) d x^{}$
or $d y=\left(2 x+\frac{1}{x}\right) d x$ ...(i)
Integrating both sides of equation (i), we get $\int d y=\int\left(2 x+\frac{1}{x}\right) d x$
or y = x2 + log |x| + C ...(ii)
Equation (ii) represents the family of solution curves of the given differential equation but we are interested in finding the equation of a particular member of the family which passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (ii), we get C = 0.
Now substituting the value of C in equation (ii) we get the equation of the required curve as y = x2 + log |x|
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Question 193 Marks
Find the general solution of the differential equation $y d x-\left(x+2 y^{2}\right) d y=0$
Answer
The given differential equation can be written as 
$\frac{d x}{d y}-\frac{x}{y}=2 y$ 
This is a linear differential equation of the type $\frac{d x}{d y}+\mathrm{P}_{1} x=\mathrm{Q}_{1}$ where $P_{1}=-\frac{1}{y}$ and Q1 = 2y.
Therefore I.F = $e^{\int-\frac{1}{y} d y}=e^{-\log y}=e^{\log (y)^{-1}}=\frac{1}{y}$ 
Hence, the solution of the given differential equation is
$x \frac{1}{y}=\int(2 y)\left(\frac{1}{y}\right) d y+C$ 
or $\frac{x}{y}=\int(2 d y)+C$ 
or $\frac{x}{y}=2 y+C$ 
which is a general solution of the given differential equation.
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Question 203 Marks
Find the general solution of the differential equation $x \frac{d y}{d x}+2 y=x^{2}(x \neq 0)$
Answer
The given differential equation is 
$x \frac{d y}{d x}+2 y=x^{2}$ ......(i)
Dividing both sides of equation (i) by x, we get
$\frac{d y}{d x}+\frac{2}{x} y=x$ 
which is a linear differential equation of the type $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$, where $P=\frac{2}{x}$ and Q = x.
So, I.F = $e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}\left[\text { as } e^{\log f(x)}=f(x)\right]$ 
Therefore, solution of the given equation is given by
$y \cdot x^{2}=\int(x)\left(x^{2}\right) d x+\mathrm{C}=\int x^{3} d x+\mathrm{C}$ 
or, $y=\frac{x^{2}}{4}+C x^{-2}$ 
which is the general solution of the given differential equation.
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3 Marks - Maths STD 12 Science Questions - Vidyadip