Chapter 9: Differential Equations

Ans:
The given differential equation is:

 $\frac{dy}{dx}+y\cot x=4x\,\text{cosec}x$ 

It is linear differential equation of the form $\frac{\text{dy}}{\text{dx}}\text{+py=Q}$:

$p=\cot x$ 

$Q=4x\text{cosec}\,x$ 

Calculating integrating factor:

$I.F={{e}^{\int pdx}}$

$I.F={{e}^{\int \cot xdx}}$ 

$I.F={{e}^{\log \left| \sin x \right|}}$ 

$I.F=\sin x$ 

The general solution is given by:

$y\times I.F=\int \left( Q\times I.F \right)dx+C$

$\Rightarrow y\times \sin x=\int \left( 4x\cos \text{ec}x \right)\sin xdx+C$ 

$\Rightarrow y\sin x=4\int xdx+C$ 

$\Rightarrow y\sin x=4\left( \frac{{{x}^{2}}}{2} \right)+C$ 

$\Rightarrow y\sin x=2{{x}^{2}}+C$ 

As $\text{y=0}$ when $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$:

$0\times \sin \left( \frac{\pi }{2} \right)=2{{\left( \frac{\pi }{2} \right)}^{2}}+C$ 

$C=-2\left( \frac{{{\pi }^{2}}}{4} \right)$ 

$C=-\frac{{{\pi }^{2}}}{2}$ 

Thus the required particular solution is:

$y\sin x=2{{x}^{2}}-\frac{{{\pi }^{2}}}{2}$

Ans:
The given differential equation is:

 $\left( x+1 \right)\frac{dy}{dx}=2{{e}^{-y}}-1$

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

Integrate both side:

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

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \left( x+1 \right)+\log C$ ……(1) 

Evaluating LHS integral:

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\int \frac{{{e}^{y}}dy}{2-{{e}^{y}}}$ 

Put $\text{t=2-}{{\text{e}}^{\text{y}}}$:

$t=2-{{e}^{y}}$

$dt=-{{e}^{y}}dy$  

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=-\int \frac{dt}{t}$

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=-\log \left( t \right)$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \frac{1}{t}$ 

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \frac{1}{2-{{e}^{y}}}$ 

Back substituting in expression (1):

$\Rightarrow \int \frac{dy}{2{{e}^{-y}}-1}=\log \left( x+1 \right)+\log C$ 

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

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

As $\text{y=0}$ when $\text{x=0}$:

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

$\Rightarrow 2-1=\frac{1}{C}$ 

$\Rightarrow C=1$ 

Thus the required particular solution is:

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

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

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

$\Rightarrow {{e}^{y}}=\frac{2x+1}{x+1}$ 

$\Rightarrow y=\log \left( \frac{2x+1}{x+1} \right)$  

Thus for given conditions the particular solution is $\text{y=log}\left( \frac{\text{2x+1}}{\text{x+1}} \right)$ .

Ans: Given differential equation:

$\frac{ydx-xdy}{y}=0$

Divide both side by $\text{x}$ :

$\Rightarrow \frac{ydx-xdy}{xy}=0$

$\Rightarrow \frac{dx}{x}-\frac{dy}{y}=0$ 

Integrate both side:

$\Rightarrow \int \frac{dx}{x}-\int \frac{dy}{y}=0$ 

$\Rightarrow \log \left| x \right|-\log \left| y \right|=\log k$ 

$\Rightarrow \log \left| \frac{x}{y} \right|=\log k$ 

$\Rightarrow \frac{x}{y}=k$ 

$\Rightarrow y=Cx\,\,\,\left( C=\frac{1}{k} \right)$

Thus the correct option is (C)

Ans: The given differential equation is:

$\frac{dx}{dy}+{{P}_{1}}x={{Q}_{1}}$ 

It is a linear differential equation and its general solution is:

$x{{e}^{\int {{P}_{1}}dy}}=\int \left( {{Q}_{1}}{{e}^{\int {{P}_{1}}dy}} \right)dy+C$ 

With integrating factor $I.F={{e}^{\int {{P}_{1}}dy}}$.

Thus the correct option is (C).

Ans: The given differential equation is:
${{e}^{x}}dy+\left( y{{e}^{x}}+2x \right)dx=0$ 

$\Rightarrow {{e}^{x}}\frac{dy}{dx}+y{{e}^{x}}=-2x$ 

$\Rightarrow \frac{dy}{dx}+y=-2x{{e}^{-x}}$   

The given differential equation is of the form:

$\frac{dy}{dx}+Py=Q$ 

$\Rightarrow P=1$ 

$\Rightarrow Q=-2x{{e}^{-x}}$ 

Calculating integrating factor:
$I.F={{e}^{\int Pdx}}$

$\Rightarrow I.F={{e}^{\int dx}}$ 

$\Rightarrow I.F={{e}^{x}}$  

It is a linear differential equation and its general solution is:

$y\left( I.F \right)=\int \left( Q\times I.F \right)dx+C$

 $y\left( {{e}^{x}} \right)=\int \left( -2x{{e}^{-x}}\times {{e}^{x}} \right)dy+C$ 

$\Rightarrow y{{e}^{x}}=-2\int xdx+C$ 

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

$\Rightarrow y{{e}^{x}}+{{x}^{2}}=C$ 

Thus the correct answer is option (C).