Chapter 9: Differential Equations

Ans: The given function is $\text{y=}{{\text{e}}^{\text{x}}}+1$ .

Take its derivative:

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

$\Rightarrow y'={{e}^{x}}$       ……(1)

Take the derivative of the above equation:

$\frac{d}{dx}\left( y' \right)=\frac{d}{dx}\left( {{e}^{x}} \right)$ 

$\Rightarrow y''={{e}^{x}}$ 

Using result from equation (1):

$y''-y'=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  }\!\!'\!\!\text{ -y }\!\!'\!\!\text{ =0}$.

Ans: The given function is $\text{y=}{{\text{x}}^{\text{2}}}\text{+2x+C}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( {{x}^{2}}+2x+C \right)$ 

$\Rightarrow y'=2x+2$

$\Rightarrow y'-2x-2=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{ -2x-2=0}$.

Ans: The given function is $\text{y=cos x + C}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( \cos x+C \right)$ 

$\Rightarrow y'=-\sin x$

$\Rightarrow y'+\sin x=0$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  + sin x = 0}$.

Ans: The given function is $\text{y = }\sqrt{1+{{x}^{2}}}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( \sqrt{1+{{x}^{2}}} \right)$ 

$y'=\frac{1}{2\sqrt{1+{{x}^{2}}}}\times \frac{d}{dx}\left( 1+{{x}^{2}} \right)$

$y'=\frac{2}{2\sqrt{1+{{x}^{2}}}}$ 

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

Multiply numerator and denominator by $\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}$:

$y'=\frac{1}{\sqrt{1+{{x}^{2}}}}\times \frac{\sqrt{1+{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}}$

Substitute $\text{y  = }\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}$ 

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

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  =}\frac{xy}{1+{{x}^{2}}}$.

Ans: The given function is $\text{y = Ax}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( Ax \right)$ 

$\Rightarrow y'=A$

Multiply by $\text{x}$ on both side:

$xy'=Ax$ 

Substitute $\text{y = Ax}$:

$\Rightarrow xy'=y$ 

Thus the given function is solution of differential equation $xy'=y\left( x\ne 0 \right)$.

Ans: The given function is $\text{y = xsin x}$ .

Take its derivative:

$\frac{dy}{dx}=\frac{d}{dx}\left( x\sin x \right)$ 

$\Rightarrow y'=\sin x\frac{d}{dx}\left( x \right)+x\frac{d}{dx}\left( \sin x \right)$

$\Rightarrow y'=\sin x+x\cos x$ 

Multiply by $\text{x}$ on both side:

$xy'=x\left( \sin x+x\cos x \right)$ 

$xy'=x\sin x+{{x}^{2}}\cos x$ 

Substitute $\text{y = xsin x}$:

$\Rightarrow xy'=y+{{x}^{2}}\cos x$ 

Use $\text{sin x = }\frac{\text{y}}{\text{x}}$ and substitute $\text{cos x}$:

$xy'=y+{{x}^{2}}\sqrt{1-{{\sin }^{2}}x}$ 

$xy'=y+{{x}^{2}}\sqrt{1-{{\left( \frac{y}{x} \right)}^{2}}}$ 

$\Rightarrow xy'=y+x\sqrt{{{y}^{2}}-{{x}^{2}}}$ 

Thus the given function is solution of differential equation $\text{xy }\!\!'\!\!\text{  = y + x}\sqrt{{{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}$.

Ans: The given function is $\text{xy = log y + C}$ .

Take derivative on both side:

$\frac{d}{dx}\left( xy \right)=\frac{d}{dx}\left( \log y \right)$ 

$\Rightarrow y\frac{d}{dx}\left( x \right)+x\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}$

$\Rightarrow y+xy'=\frac{1}{y}y'$ 

$\Rightarrow {{y}^{2}}+xy\,y'=y'$ 

Shift the $\text{y }\!\!'\!\!\text{ }$ terms on one side and take it common. 

$\Rightarrow \left( xy-1 \right)y'=-{{y}^{2}}$ 

$\Rightarrow y'=\frac{{{y}^{2}}}{1-xy}$ 

Thus the given function is solution of differential equation $\text{y }\!\!'\!\!\text{  = }\frac{{{\text{y}}^{\text{2}}}}{\text{1-xy}}$.

Ans: The given function is $\text{y - cos y = x}$ .

Take derivative on both side:

$\frac{dy}{dx}-\frac{d}{dx}\left( \cos y \right)=\frac{d}{dx}\left( x \right)$ 

$\Rightarrow y'+y'\sin y=1$

$\Rightarrow y'\left( 1+\sin y \right)=1$ 

$\Rightarrow y'=\frac{1}{1+\sin y}$ 

Multiply by \[\left( \text{ysin y + cos y + x} \right)\] on both side:

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{\left( \text{ysin y + cos y + x} \right)}{1+\sin y}\] 

Substitute $y=\cos y+x$ in the numerator:

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{\left( \text{ysin y + y} \right)}{1+\sin y}\] 

\[\left( \text{ysin y + cos y + x} \right)y'=\frac{y\left( \text{sin y + 1} \right)}{1+\sin y}\] 

$\Rightarrow \left( \text{ysin y + cos y + x} \right)y'=y$ 

Thus the given function is solution of differential equation$\left( \text{ysin y + cos y + x} \right)\text{y }\!\!'\!\!\text{  = y}$.

Ans: The given function is $\text{x + y = ta}{{\text{n}}^{-1}}y$ .

Take derivative on both side:

$\frac{d}{dx}\left( x+y \right)=\frac{d}{dx}\left( ta{{n}^{-1}}y \right)$ 

$1+y'=\left( \frac{1}{1+{{y}^{2}}} \right)y'$

$\Rightarrow y'\left[ \frac{1}{1+{{y}^{2}}}-1 \right]=1$ 

$\Rightarrow y'\left[ \frac{1-\left( 1+{{y}^{2}} \right)}{1+{{y}^{2}}} \right]=1$ 

$\Rightarrow y'\left[ \frac{-{{y}^{2}}}{1+{{y}^{2}}} \right]=1$ 

$\Rightarrow -{{y}^{2}}y'=1+{{y}^{2}}$ 

$\Rightarrow {{y}^{2}}y'+{{y}^{2}}+1=0$ 

Thus the given function is solution of differential equation ${{y}^{2}}y'+{{y}^{2}}+1=0$.

Ans: The given function is:

\[y\text{ }=\text{ }\sqrt{{{a}^{2}}-{{x}^{2}}}\,x\in \left( -a,a \right)\] .

Take derivative on both side:

$\frac{dy}{dx}=\frac{d}{dx}\left( \sqrt{{{a}^{2}}-{{x}^{2}}} \right)$ 

$\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{{{a}^{2}}-{{x}^{2}}}}\frac{d}{dx}\left( {{a}^{2}}-{{x}^{2}} \right)$

$\Rightarrow \frac{dy}{dx}=\frac{-2x}{2\sqrt{{{a}^{2}}-{{x}^{2}}}}$ 

$\Rightarrow \frac{dy}{dx}=\frac{-x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}$ 

Substitute $\text{y = }\sqrt{{{\text{a}}^{\text{2}}}\text{ - }{{\text{x}}^{\text{2}}}}$ 

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

$\Rightarrow x+y\frac{dy}{dx}=0$ 

Thus the given function is solution of differential equation:

$x+y\frac{dy}{dx}=0\left( y\ne 0 \right)$.