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MATHS 5 Marks Questions

Differential Equations · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 9 questions.

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Question 15 Marks
Solve the differential equation $\frac{d y}{d x}+y \tan x=y^2$ $\sec x$.
Answer
The given differential equation is
$
\begin{array}{r}
\frac{d y}{d x}+y \tan x=y^2 \sec x \\
\Rightarrow \frac{1}{y^2} \frac{d y}{d x}+\frac{1}{y} \tan x=\sec x
\end{array}
$
Let $\frac{1}{y}=z \Rightarrow-\frac{1}{y^2} \frac{d y}{d x}=\frac{d z}{d x}$
Substituting
$
\begin{aligned}
-\frac{d z}{d x}+(\tan x) z & =\sec x \\
- & -\frac{d z}{d x}-(\tan x) z=\sec x
\end{aligned}
$
This is a linear differential equation comparing which with $\frac{d z}{d x}+ P z= Q$
$
\begin{aligned}
P & =-\tan x, Q=-\sec x \\
\text { I.F. }=e^{\int P d x} & =e^{\int(-\tan x) d x} \\
& =e^{-\int \tan x d x}=e^{-\log (\sec x)} \\
& =e^{-\log (\sec x)}=\cos x
\end{aligned}
$
Hence the solution of differential equation (3) is
$
\begin{aligned}
z(IF) & =\int(IF) Q d x+c \\
\Rightarrow \quad z(\cos x) & =\int(-\cos x) \cdot(-\sec x) d x+c \\
& =-\int 1 d x+c \\
\Rightarrow \quad z(\cos x) & =-x+c \\
\Rightarrow \quad \frac{1}{y} \cos x & =-x+c \\
\Rightarrow \quad \frac{\cos x}{y}+x & =c
\end{aligned}
$
Ans.
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Question 25 Marks
Find a particular solution of the differential equation $\left(1+ x ^2\right) d y+2 x y\ d x=\cot x\ d x,$ given that $y=0$ of $x=\frac{\pi}{2}$.
Answer
The given differential equation is
$\left(1+x^2\right) d y+2 x y\ d x =\cot x\ d x$
$\Rightarrow \left(1+x^2\right) d y =(\cot x-2 x y) d x$
$\Rightarrow \left(1+x^2\right) \frac{d y}{d x} =-2 x y+\cot x$
or  $\left(1+x^2\right) \frac{d y}{d x}+2 x y =\cot x$
or $ \frac{d y}{d x}+\frac{2 x}{1+x^2} y =\frac{\cot x}{1+x^2},$
which is a linear differential equation.
$\therefore$ here $P =\frac{2 x}{1+x^2}$ and $Q =\frac{\cot x}{1+x^2}$
Hence
$\text { I.F. } =e^{\int Pdx}=e^{\int \frac{2 x}{1+x^2} dx}=e^{\log \left(1+x^2\right)}$
$ =\left(1+x^2\right)$
Therefore the solution of the equation will be :
$y .(\text { I.F. })=\int Q .(I . F .) dx$
$\Rightarrow y\left(1+x^2\right)=\int \frac{\cot x}{\left(1+x^2\right)} \cdot\left(1+x^2\right) dx$
$\Rightarrow y\left(1+x^2\right)=\int \cot x \ dx$
$\therefore y\left(1+x^2\right)=\log |\sin x|+c$
Given that $y=0$, if $x=\frac{\pi}{2}$
$\text { 0. } \begin{aligned}
\left(1+\frac{\pi^2}{2}\right) & =\log \left|\sin \frac{\pi}{2}\right|+c \\
0 & =0+c \quad \therefore c=0 \end{aligned}$
Hence a particular solution of the given differential equation:
$\left(1+x^2\right) d y+2 x y\ d x=\cot \ x\ d x$
$y\left(1+x^2\right)=\log |\sin x|+0$
$\Rightarrow y\left(1+x^2\right)=\log |\sin x| $
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Question 35 Marks
Solve the following differential equation : $\left(x^2+1\right) \frac{d y}{d x}+2 x y=\sqrt{x^2+4}$
Answer
The given differential equation is
$\left(x^2+1\right) \frac{d y}{d x}+2 x y$
$=\sqrt{x^2+4}$
$\Rightarrow \frac{d y}{d x}+\frac{2 x}{x^2+1} y$
$=\frac{\sqrt{x^2+4}}{x^2+1}$
Which is a linear differential equation.
Here $ P =\frac{2 x}{x^2+1}$ and $ Q =\frac{\sqrt{x^2+4}}{x^2+1}$
Therefore integrating factor
$\text { I.F. } =e^{\int P d x}$
$ =e^{\int \frac{2 x}{1+x^2} d x}$
Let $1+x^2=t $
$\therefore 2 x \ d x=d t$
$\therefore \text { I.F. }=e^{\int \frac{d t}{t}}=e^{\log t}=t$
Putting the value of $t$
$\text { I.F. }=\left(1+x^2\right)$
Hence the required solution is
$\text { y.I.F. }=\int(I . F .)(Q) d x+C$
$\Rightarrow y \cdot\left(1+x^2\right)=\int \frac{\left(1+x^2\right) \sqrt{x^2+4}}{x^2+1} d x+C$
$\Rightarrow y \cdot\left(1+x^2\right)=\int \sqrt{(x)^2+(2)^2} d x+C$
${\left(\because \int \sqrt{x^2+a^2} d x=\frac{x}{2} \sqrt{x^2+a^2}+\frac{a^2}{2} \log \right.)}$
$\left.\left|\left(x+\sqrt{x^2+a^2}\right)\right|+C\right)$
$\Rightarrow y\left(x^2+1\right)=\frac{x}{2} \sqrt{x^2+4}+\frac{4}{2} \log$
$\left|\left(x+\sqrt{x^2+4}\right)\right|+C$
$\Rightarrow y\left(x^2+1\right)=\frac{1}{2} x \sqrt{x^2+4}+2 \log$
$\left|\left(x+\sqrt{x^2+4}\right)\right|+C$
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Question 45 Marks
Solve the following differential equation : $\sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x}=0$
Answer
The given differential equation is
$\sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x}=0$
$\Rightarrow \sqrt{1\left(1+x^2\right)+y^2\left(1+x^2\right)}+x y \frac{d y}{d x}=0$
$\Rightarrow \sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x y \frac{d y}{d x}=0$
$\Rightarrow \sqrt{1+x^2} \sqrt{1+y^2}+x y \frac{d y}{d x}=0$
$\text { or } x y \frac{d y}{d x}=-\sqrt{1+x^2} \sqrt{1+y^2} \ldots (1)$
Separating the variable of equation $(1)$
$\frac{y d y}{\sqrt{1+y^2}}=\frac{-\sqrt{1+x^2}}{x} d x \ldots (2)$
Integrating both sides of equation $(2)$
or $\int \frac{y d y}{\sqrt{1+y^2}}=-\int \frac{\sqrt{1+x^2}}{x^2} x d x$
Let $1+y^2=u^2$ and $1+x^2=v^2$
$\therefore 2 y d y=2 u d u$ and $2 x d x=2 v d v$
$\Rightarrow y d y=u d u$ and $x d x=v d v$
$\therefore \int \frac{u d u}{u} =-\int \frac{\sqrt{v^2} \cdot v d v}{v^2-1}$
$\Rightarrow \int d u =-\int \frac{v^2}{v^2-1} d v$
$\Rightarrow \int d u =-\int\left(1+\frac{1}{v^2-1}\right) d v$
Integrating
$\Rightarrow u =-v-\frac{1}{2} \log \left|\frac{v-1}{v+1}\right|+C$
$\therefore \sqrt{1+y^2} =-\sqrt{1+x^2}-\frac{1}{2} \log \left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C$
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Question 55 Marks
Find a particular solution of the differential equation $\frac{ dy }{ dx }+2 y \cot x=4 x \operatorname{cosec} x$ $(x \neq 0)$, given that $y=0$, when $x=\frac{\pi}{2}$.
Answer
The given differential solution is
$\frac{d y}{d x}+2 y \cot x=4 x \operatorname{cosec} x,(x \neq 0)$
The above differential equation is a linear differential equation.
Hence comparing the given differential equation with $\frac{d y}{d x}+ P y= Q$,
$P =2 \cot x$ and $Q =4 x \operatorname{cosec} x$
Therefore $\int P d x=\int 2 \cot x d x=2 \int \cot x d x$
$
\int P d x=2 \log \sin x=\log \sin ^2 x$
$
\text {Hence} \quad \text{ I.F. }=e^{\int P d x}=e^{\log \sin ^2 x}=\sin ^2 x
$
Therefore the solution of the given differential equation is:
$
\begin{array}{rlrl}
& y \times \text { I.F. } =\int Q \times \text { I.F. } d x+C \\
\Rightarrow & y \times \sin ^2 x =\int 4 x \operatorname{cosec} x \times \sin ^2 x d x+C \\
\Rightarrow & y \sin ^2 x =4 \int x \sin x d x+C \\
\Rightarrow & y \sin ^2 x =4\left[-x \cos x+\int 1 \cdot \cos x d x\right]+C \\
\Rightarrow & y \sin ^2 x =-4 x \cos x+4 \sin x+C
\end{array}
$
Given that $y=0$, when $x=\frac{\pi}{2}$
$
\begin{array}{rlrl}
& 0 =0+4 \sin \frac{\pi}{2}+C \\
& \Rightarrow \quad 0 =0+4+C \quad \therefore \quad C=-4
\end{array}
$
Hence the particular solution of the given differential equation is :
$
\begin{array}{rlrl}
& y \sin^2 x =-4 x \cos x+4 \sin x-4 \\
\Rightarrow & y =\frac{-4 x \cos x}{\sin ^2 x}+\frac{4 \sin x}{\sin ^2 x}-\frac{4}{\sin ^2 x} \\
\Rightarrow & y =-4 x \cot x \cdot \operatorname{cosec} x+4 \operatorname{cosec} x \\
& -4 \operatorname{cosec}^2 x
\end{array}
$
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Question 65 Marks
Solve the following differential equation :
$
x d y+\left(y-x^3\right) d x=0
$
Answer
The given differential equation is
or $ \ \ \ \ \
x d y+\left(y-x^3\right) d x=0
$
or $\quad x \frac{d y}{d x}+y=x^3$
or $\quad \frac{d y}{d x}+\frac{1}{x} y=x^2$
$
x \frac{d y}{d x}+y-x^3=0
$
Comparing equation (1) with the linear differential equation $\frac{d y}{d x}+ P y= Q$
Here $P =\frac{1}{x}$ and $Q =x^2$
Integrating factor $
\text { I.F. }=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x
$
Hence the required solution of the differential equation will be :
$
\begin{aligned}
y \times \text { I.F. } & =\int(I . F) Q d x+C \\
\Rightarrow \quad y \times x & =\int(x) \cdot x^2 d x+C \\
\Rightarrow \quad y x & =\int x^3 d x+C \\
\Rightarrow \quad & =\frac{x^4}{4}+C \\
\Rightarrow \quad y & =\frac{1}{4} x^3+\frac{C}{x}
\end{aligned}
$
This is the general solution of the given differential equation.
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Question 95 Marks
Solve the following differential equation : $\left(y+3 x^2\right) \frac{d x}{d y}=x$
Answer
From the given differential equation,
$\left(y+3 x^2\right) \frac{d x}{d y} =x$
 or $\frac{d x}{d y} =\frac{x}{y+3 x^2}$
or  $\frac{d y}{d x} =\frac{y+3 x^2}{x}$
$\Rightarrow \frac{d y}{d x} =\frac{y}{x}+3 x$
$\Rightarrow \frac{d y}{d x}+\left(-\frac{y}{x}\right) =3 x$
Comparing equation $(1)$ with linear differential equation
$\frac{d y}{d x}+P y=Q,$
Here $P =-\frac{1}{x}$, and $Q =3 x$
$\therefore$ Integrating factor
$\text { I.F. } =e^{\int Pd \ d x}$
$=e^{-\int \frac{1}{x} d x}=e^{-\log x}$
$=e^{\log (x)^{-1}}=(x)^{-1}$
$ =\frac{1}{x}$
Hence the required solution will be :
$y . \text { I.F. } =\int((I . F . Q) d x+C$
$\Rightarrow y \times \frac{1}{x} =\int \frac{1}{x} \times 3 x\ d x+C$
$\Rightarrow \frac{y}{x}=\int 3 d x+C$
$\Rightarrow \frac{y}{x}=3 x+C $
$\therefore y=3 x^2+C x$
which is the general solution of the given differential equation.
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