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.
AnswerIt 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 $2-e^{y}=t$
$\therefore \frac{d}{d y}\left(2-e^{y}\right)=\frac{d t}{d y}$
$\Rightarrow-\mathrm{e}^{\mathrm{y}}=\frac{\mathrm{dt}}{\mathrm{dy}}$
$\Rightarrow e^ydy = -dt$
Substituting value in equation (i), we get,
$\int \frac{-\mathrm{dt}}{\mathrm{t}}=\log |\mathrm{x}+1|+\log \mathrm{c}$
$\Rightarrow -log|t| = log|C(x+1)|$
$\Rightarrow -log|2 - e^y| = log|C(x + 1)|$
$\Rightarrow \frac{1}{2-\mathrm{e}^{\mathrm{y}}}=\mathrm{C}(\mathrm{x}+1)$
$\Rightarrow 2-\mathrm{e}^{\mathrm{y}}=\frac{1}{\mathrm{c}(\mathrm{x}+1)}$ ......(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
$y=\log \left|\frac{2 x+1}{x+1}\right|,(x \neq-1)$
View full question & answer→Question 23 Marks
Find the general solution of $\frac{{dy}}{{dx}} + \left( {\sec x} \right)y = \tan x\left( {0 \leq x < \frac{\pi }{2}} \right)$
AnswerGiven: 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$
$\Rightarrow y\left( {\sec x + \tan x} \right) = \int {\tan x\left( {\sec x + \tan x} \right)} dx + c$
$\Rightarrow y\left( {\sec x + \tan x} \right) = \int {\left( {\sec x\tan x + {{\tan }^2}x} \right)} dx + c$
$\Rightarrow y\left( {\sec x + \tan x} \right) = \int {\left( {\sec x.\tan x + {{\sec }^2}x - 1} \right)} dx + c$
$\Rightarrow y\left( {\sec x + \tan x} \right) = \sec x + \tan x - x + c$
View full question & answer→Question 33 Marks
Find the general solution of $\frac{d y}{d x}+\frac{y}{x}=x^{2}$
AnswerIt 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 $Q = x^2$)
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 $x y=\frac{x^{4}}{4}+C$
View full question & answer→Question 43 Marks
Find the general solution of $\frac{d y}{d x}+3 y=e^{-2 x}$
AnswerIt 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:
$y(\mathrm{I} . \mathrm{F})=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{ye}^{3 \mathrm{x}}=\int\left(\mathrm{e}^{-2 \mathrm{x}} \times \mathrm{e}^{3 \mathrm{x}}\right) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{ye}^{3 \mathrm{x}}=\int \mathrm{e}^{\mathrm{x}} \mathrm{dx}+\mathrm{C}$
$\Rightarrow ye^{3x} = e^x + C$
$\Rightarrow y = e^{-2x} + Ce^{-3x}$
Therefore, the required general solution of the given differential equation is $y = e^{-2x} + Ce^{-3x}$
View full question & answer→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 $x = \frac { \pi } { 2 }$.
AnswerGiven, $\frac { d y } { d x } - 3 y \cot x = \sin 2 x$ ...(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]
$\Rightarrow y = -2 \sin^2 x + C \sin^3 x$ ...(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
$\therefore y = - 2 \sin ^ { 2 } x + 4 \sin ^ { 3 } x$
View full question & answer→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 $1 + x^2$
$\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}$ = $\frac{1}{\left(1+x^{2}\right) \cdot\left(1+x^{2}\right)}$
$\frac{d y}{d x}+\left(\frac{2 x }{1+x^{2}}\right) y$ = $\frac{1}{\left(1+x^{2}\right)^2}$
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 $1 + x^2= t$
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 = $1 + x^2$
Solution of the differential equation:
y $\times$ I.F. = $\int {Q\, \times \,I.F\,dx} $
Putting values,
y $\times (1 + x^2) = {\int {\frac{1}{{(1 + {x^2})^2}}} (1 + {x^2})dx} $
$y (1 + x^2) = \int{\frac{1}{{(1 + {x^2})}}} $ dx
$y (1 + x^2) = \tan^{-1}x + C$ ...(1)
Putting that y = 0 and x = 1,
$0(1 + 1^2) = \tan^{-1}(1) + C$
0 = $\frac{\pi}{4}$ + C
C = -$\frac{\pi}{4}$
Putting value of C in eq(1),
$y(1 + x^2) = \tan^{-1}x + C$
$y(1 + x^2) = \tan^{-1}x - \frac{\pi}{4}$
View full question & answer→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 = $\frac{\pi}{3}$.
AnswerAccording 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 $\sec^2x$ on both sides, we get
$\Rightarrow y = \frac { 1 } { \sec x } + \frac { C } { \sec ^ { 2 } x }$
$\Rightarrow y = cosx + C.\cos^2x$
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.\cos^2x \Rightarrow y = \cos x - 2 \cos^2x$
Therefore, the required particular solution is $y = \cos x - 2 \cos^2x$
View full question & answer→Question 83 Marks
Find the general solution of $\left(x+3 y^{2}\right) \frac{d y}{d x}=y(y>0)$
AnswerIt 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 = 3y^2 + Cy$
Therefore, the required general solution of the given differential equation is $x = 3y^2 + Cy$.
View full question & answer→Question 93 Marks
Find the general solution of $y d x+\left(x-y^{2}\right) d y=0$
AnswerIt 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}$.
View full question & answer→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)$.
AnswerLet $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 = e^t$
$P = 1000 \times 1.648 = ₹\ 1648$
View full question & answer→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 $(log_e2 = 0.6931).$
AnswerLet $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
View full question & answer→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).
AnswerGiven 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}$
View full question & answer→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).$
AnswerFor 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 x^2 + \log(y + 2)^2$
Or,
$\Rightarrow y - x - c = \log{x^2(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\{x^2(y + 2)^2\}$
$y - x + 2 = \log\{x^2(y + 2)^2\}$
View full question & answer→Question 143 Marks
Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y′ = e^x$ 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} $
View full question & answer→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 $x = 0$.
AnswerAccording 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 $x^2, x$ and constant terms from both sides, we get
$A + B = 2 \ ; \\B + C = 1$
$A+C = 0$$ \Rightarrow$ A = -C
On solving above equations, we get
$ A = \frac { 1 } { 2 } , B = \frac { 3 } { 2 } \text { and } C = \frac { - 1 } { 2 }$
On substituting the values of A, B and C in Eq. (ii), we get
$ \frac { 2 x ^ { 2 } + x } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) } = \frac { \frac { 1 } { 2 } } { x + 1 } + \frac { \frac { 3 } { 2 } x - \frac { 1 } { 2 } } { x ^ { 2 } + 1 }$
On integrating both sides w.r.t to x , we get
$ \int \frac { 2 x ^ { 2 } + x } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) } d x = \frac { 1 } { 2 } \int \frac { d x } { x + 1 }$ $ + \frac { 3 } { 2 } \int \frac { x } { x ^ { 2 } + 1 } d x - \frac { 1 } { 2 } \int \frac { d x } { x ^ { 2 } + 1 }$
$ \Rightarrow y = \frac { 1 } { 2 } \log | x + 1 | + I _ { 1 } - \frac { 1 } { 2 } \tan ^ { - 1 } x + C _ { 2 }$ ...(iii) [from Eq. (1)]
where $ I _ { 1 } = \frac { 3 } { 2 } \int \frac { x } { x ^ { 2 } + 1 } d x$
Put $ x ^ { 2 } + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac { d t } { 2 }$
$ \therefore \quad I _ { 1 } = \frac { 3 } { 4 } \int \frac { d t } { t } = \frac { 3 } { 4 } \log | t | + C _ { 1 }$
$ = \frac { 3 } { 4 } \log \left| x ^ { 2 } + 1 \right| + C _ { 1 }$
On putting the value of $I_1$ in Eq. (iii), we get
$ y = \frac { 1 } { 2 } \log | x + 1 | + \frac { 3 } { 4 } \log \left| x ^ { 2 } + 1 \right| - \frac { 1 } { 2 } \tan ^ { - 1 } x + C$[where, $C = C_1 + C_2$]
View full question & answer→Question 163 Marks
Find the general solution of the differential equation ${e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}y\,dy = 0$
Answer${e^x}\tan ydx + \left( {1 - {e^x}} \right){\sec ^2}y\,dy = 0$ $\Rightarrow$${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$
$\Rightarrow\log \left( {\frac{{\tan y}}{{1 - {e^x}}}} \right) = \log c$
$\Rightarrow\frac{{\tan y}}{{1 - {e^x}}} = c$
$\Rightarrow\tan y = c(1 - {e^x})$
View full question & answer→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 $\frac{2 x}{y^{2}}$.
AnswerWe 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
$y^2dy = 2x\ dx ...(ii)$
Integrating both sides of equation $(ii),$ we get
$\int y^2dy = \int 2x\ dx$
or $\frac{y^{3}}{3} = x^2 + C ...(iii)$
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
$\frac{y^{3}}{3} + x^2 + 5$ or $y = (3x^2 + 15)^{1/3}$
View full question & answer→Question 183 Marks
Find the equation of the curve passing through the point $(1, 1)$ whose differential equation is x dy = $(2x^2 + 1) dx (x \neq 0)$.
AnswerThe 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 $y = x^2+ \log |x| + C$ ...(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 $y = x^2 + \log |x|$
View full question & answer→Question 193 Marks
Find the general solution of the differential equation $y d x-\left(x+2 y^{2}\right) d y=0$
AnswerThe 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 $Q_1 = 2y$.
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.
View full question & answer→Question 203 Marks
Find the general solution of the differential equation $x \frac{d y}{d x}+2 y=x^{2}(x \neq 0)$
AnswerThe 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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