Chapter 9: Differential Equations

Ans: The given differential equation is:

$\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)\frac{dy}{dx}=2{{x}^{2}}+x;\,\,y=1,\,x=0$.

Simplify the expression:

$\frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{x}^{2}}+x}{\left( {{x}^{3}}+x+{{x}^{2}}+1 \right)}$ 

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

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

Integrate both side:

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

Use partial fraction method to simplify the RHS:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$

$\Rightarrow 2{{x}^{2}}+x=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+1 \right)$ 

\[\Rightarrow 2{{x}^{2}}+x=\left( A+B \right){{x}^{2}}+\left( B+C \right)x+\left( A+C \right)\] 

By comparing coefficients:

$A+B=2$ 

$B+C=1$ 

$A+C=0$ 

Solving this we get:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{\left( \frac{1}{2} \right)}{x+1}+\frac{\left( \frac{3}{2} \right)x+\left( -\frac{1}{2} \right)}{{{x}^{2}}+1}$ 

$\Rightarrow \frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{1}{2}\left( \frac{1}{x+1}+\frac{3x-1}{{{x}^{2}}+1} \right)$ 

Rewriting the  integral(1):

$y=\frac{1}{2}\int \left( \frac{1}{x+1}+\frac{3x-1}{{{x}^{2}}+1} \right)dx$ 

\[\Rightarrow y=\frac{1}{2}\int \frac{1}{x+1}dx+\frac{1}{2}\int \frac{3x-1}{{{x}^{2}}+1}dx\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{2}\int \frac{x}{{{x}^{2}}+1}dx-\frac{1}{2}\int \frac{1}{{{x}^{2}}+1}dx\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\int \frac{2x}{{{x}^{2}}+1}dx-\frac{1}{2}{{\tan }^{-1}}x\] 

\[\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\log \left( {{x}^{2}}+1 \right)-\frac{1}{2}{{\tan }^{-1}}x+C\] 

For $y=1$ when $\text{x=0}$. 

\[1=\frac{1}{2}\log \left( 0+1 \right)+\frac{3}{4}\log \left( 0+1 \right)-\frac{1}{2}{{\tan }^{-1}}0+C\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is :

$\Rightarrow y=\frac{1}{2}\log \left( x+1 \right)+\frac{3}{4}\log \left( {{x}^{2}}+1 \right)-\frac{1}{2}{{\tan }^{-1}}x+1$.

Ans: The given differential equation is:

$x\left( {{x}^{2}}-1 \right)\frac{dy}{dx}=1$; $y=0$ when $x=2$ 

Simplify the expression:

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

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

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

Integrate both side:

$\int dy=\int \frac{dx}{x\left( x-1 \right)\left( x+1 \right)}$ ……(1)

Use partial fraction method to simplify the RHS:

$\frac{1}{x\left( x-1 \right)\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$

$\Rightarrow 1=A\left( {{x}^{2}}-1 \right)+Bx\left( x+1 \right)+Cx\left( x-1 \right)$ 

\[\Rightarrow 1=\left( A+B+C \right){{x}^{2}}+\left( B-C \right)x-A\] 

By comparing coefficients:

$A+B+C=0$ 

$B-C=0$ 

$-A=1$ 

Solving this we get:

$\frac{1}{x\left( x-1 \right)\left( x+1 \right)}=\frac{\left( -1 \right)}{x}+\frac{\left( \frac{1}{2} \right)}{x-1}+\frac{\left( \frac{1}{2} \right)}{x+1}$ 

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

Rewriting the  integral(1):

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

\[\Rightarrow y=-\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{x-1}dx+\frac{1}{2}\int \frac{1}{x+1}dx\] 

\[\Rightarrow y=-\log x+\frac{1}{2}\log \left( x-1 \right)+\frac{1}{2}\log \left( x+1 \right)+\log C\] 

\[\Rightarrow y=-\frac{2}{2}\log x+\frac{1}{2}\log \left( x-1 \right)+\frac{1}{2}\log \left( x+1 \right)+\frac{2}{2}\log C\] 

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

\[\Rightarrow y=\frac{1}{2}\log \left[ \frac{{{C}^{2}}\left( {{x}^{2}}-1 \right)}{{{x}^{2}}} \right]\] 

For $y=0$ when $\text{x=2}$. 

\[0=\frac{1}{2}\log \left[ \frac{{{C}^{2}}\left( {{2}^{2}}-1 \right)}{{{2}^{2}}} \right]\] 

\[\Rightarrow 0=\log \left[ \frac{3{{C}^{2}}}{4} \right]\] 

$\Rightarrow \frac{3{{C}^{2}}}{4}=1$ 

$\Rightarrow {{C}^{2}}=\frac{4}{3}$ 

Thus the required particular solution is :

$y=\frac{1}{2}\log \left[ \frac{4\left( {{x}^{2}}-1 \right)}{3{{x}^{2}}} \right]$.

Ans: The given differential equation is:

$\cos \left( \frac{dy}{dx} \right)=a\,\,\left( a\in R \right)$; $y=1$ when $x=0$

Simplify the expression:

$\cos \left( \frac{dy}{dx} \right)=a\,$ 

$\Rightarrow \frac{dy}{dx}={{\cos }^{-1}}a$ 

$\Rightarrow dy={{\cos }^{-1}}adx$ 

Integrate both side:

$\int dy=\int {{\cos }^{-1}}adx$ 

$\Rightarrow y={{\cos }^{-1}}a\int dx$ 

\[\Rightarrow y=x{{\cos }^{-1}}a+C\] 

For $y=1$ when $\text{x=0}$. 

\[1=0{{\cos }^{-1}}a+C\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is:

\[y=x{{\cos }^{-1}}a+1\].

\[\Rightarrow \frac{y-1}{x}={{\cos }^{-1}}a\]

\[\Rightarrow \cos \left( \frac{y-1}{x} \right)= a\]

Ans: The given differential equation is:

$\frac{dy}{dx}=y\tan x$; $y=1$ when $x=0$

Simplify the expression:

$\frac{dy}{dx}=y\tan x$

$\Rightarrow \frac{dy}{y}=\tan xdx$ 

Integrate both side:

$\int \frac{dy}{y}=\int \tan xdx$ 

$\Rightarrow \log y=\log \left( \sec x \right)+\log C$ 

\[\Rightarrow \log y=\log \left( C\sec x \right)\] 

$\Rightarrow y=C\sec x$ 

For $\text{y=1}$ when $\text{x=0}$. 

\[1=C\sec 0\] 

\[\Rightarrow C=1\] 

Thus the required particular solution is :

\[y=\sec x\]. 

Ans: The given differential equation is:

$y'={{e}^{x}}\sin x$

The curve passes through $\left( 0,0 \right)$.

Simplify the expression:

$\Rightarrow \frac{dy}{dx}={{e}^{x}}\sin x$

$\Rightarrow dy={{e}^{x}}\sin xdx$ 

Integrate both side:

$\int dy=\int {{e}^{x}}\sin xdx$ 

Use product rules for integration of RHS. Let:

$I=\int {{e}^{x}}\sin xdx$ 

$\Rightarrow I=\sin x\int {{e}^{x}}dx-\int \left( \cos x\int {{e}^{x}}dx \right)dx$ 

$\Rightarrow I={{e}^{x}}\sin x-\int {{e}^{x}}\cos xdx$ 

$\Rightarrow I={{e}^{x}}\sin x-\left( \cos x\int {{e}^{x}}dx+\int \left( \sin x\int {{e}^{x}}dx \right)dx \right)$ 

$\Rightarrow I={{e}^{x}}\sin x-{{e}^{x}}\cos x-\int \left( {{e}^{x}}\sin x \right)dx$ 

$\Rightarrow I={{e}^{x}}\sin x-{{e}^{x}}\cos x-I$ 

$\Rightarrow I=\frac{{{e}^{x}}}{2}\left( \sin x-\cos x \right)$ 

Thus integral will be:

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

Thus as the curve passes through $\left( \text{0,0} \right)$ 

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

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

$\Rightarrow C=\frac{1}{2}$ 

Thus the equation of the curve will be:

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

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

Ans: The given differential equation is:

$xy\frac{dy}{dx}=\left( x+2 \right)\left( y+2 \right)$

The curve passes through $\left( 1,-1 \right)$.

Simplify the expression:

\[\Rightarrow \left( \frac{y}{y+2} \right)dy=\frac{\left( x+2 \right)}{x}dx\]

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

Integrate both side:

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

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

$\Rightarrow y-2\log \left( y+2 \right)=x+2\log x+C$ 

$\Rightarrow y-x=2\log \left( y+2 \right)+2\log x+C$ 

$\Rightarrow \Rightarrow y-x=2\log \left[ x\left( y+2 \right) \right]+C$ 

$\Rightarrow y-x=\log \left[ {{x}^{2}}{{\left( y+2 \right)}^{2}} \right]+C$ 

Thus as the curve passes through $\left( \text{1,-1} \right)$ 

$\Rightarrow -1-1=\log \left[ {{\left( 1 \right)}^{2}}{{\left( -1+2 \right)}^{2}} \right]+C$

$\Rightarrow -2=\log 1+C$ 

$\Rightarrow C=-2$ 

Thus the equation of the curve will be:

$y-x=\log \left[ {{x}^{2}}{{\left( y+2 \right)}^{2}} \right]-2$ 

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

Ans: According to , the equation is given by:

$y\frac{dy}{dx}=x$ 

The curve passes through $\left( 0,-2 \right)$.

Simplify the expression:

\[\Rightarrow ydy=xdx\] 

Integrate both side:

$\int ydy=\int xdx$ 

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

$\Rightarrow {{y}^{2}}-{{x}^{2}}=2C$ 

Thus as the curve passes through $\left( \text{0,-2} \right)$ 

$\Rightarrow {{\left( -2 \right)}^{2}}-{{0}^{2}}=2C$

$\Rightarrow 4=2C$ 

$\Rightarrow C=2$ 

Thus the equation of the curve will be:

${{y}^{2}}-{{x}^{2}}=2\left( 2 \right)$ 

$\Rightarrow {{y}^{2}}-{{x}^{2}}=4$ .

Ans: Let the point of contact of the tangent be $\left( \text{x,y} \right)$.Then the slope of the segment joining point of contact and $\left( \text{-4,-3} \right)$:

$m=\frac{y+3}{x+4}$ 

According to  the for the slope of tangent $\frac{\text{dy}}{\text{dx}}$ it follows:

$\frac{dy}{dx}=2m$ 

$\Rightarrow \frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)$

Simplify the expression:

$\frac{dy}{dx}=2\left( \frac{y+3}{x+4} \right)$ 

$\frac{dy}{y+3}=\frac{2}{x+4}dx$ 

Integrate both side:

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

$\Rightarrow \log \left( y+3 \right)=2\log \left( x+4 \right)+\log C$ 

\[\Rightarrow \log \left( y+3 \right)=\log {{\left( x+4 \right)}^{2}}+\log C\] 

\[\Rightarrow \log \left( y+3 \right)=\log C{{\left( x+4 \right)}^{2}}\] 

\[\Rightarrow y+3=C{{\left( x+4 \right)}^{2}}\] 

Thus as the curve passes through $\left( \text{-2,1} \right)$ 

\[1+3=C{{\left( -2+4 \right)}^{2}}\]

$\Rightarrow 4=4C$ 

$\Rightarrow C=1$ 

Thus the equation of the curve will be:

\[y+3={{\left( x+4 \right)}^{2}}\] .

Ans: Let the volume of spherical balloon be $\text{V}$ and its radius $\text{r}$. Let the rate of change of volume be $\text{k}$.

$\frac{dV}{dt}=k$ 

$\Rightarrow \frac{d}{dt}\left( \frac{4}{3}\pi {{r}^{3}} \right)=k$ 

$\Rightarrow \frac{4}{3}\pi \frac{d}{dt}\left( {{r}^{3}} \right)=k$ 

$\Rightarrow \frac{4}{3}\pi \left( 3{{r}^{2}} \right)\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi {{r}^{2}}\frac{dr}{dt}=k$ 

$\Rightarrow 4\pi {{r}^{2}}dr=kdt$ 

Integrate both side:
$\int 4\pi {{r}^{2}}dr=\int kdt$ 

$\Rightarrow 4\pi \int {{r}^{2}}dr=kt+C$ 

$\Rightarrow \frac{4}{3}\pi {{r}^{3}}=kt+C$ 

At initial time, $\text{t=0}$ and $\text{r=3}$:

$\frac{4}{3}\pi {{3}^{3}}=k\left( 0 \right)+C$ 

$\Rightarrow C=36\pi $

At $\text{t=3}$ the radius $\text{r=6}$:

$\frac{4}{3}\pi \left( {{6}^{3}} \right)=k\left( 3 \right)+36\pi $ 

$\Rightarrow 3k=288\pi -36\pi $ 

$\Rightarrow k=84\pi $ 

Thus the radius-time relation can be given by:

$\frac{4}{3}\pi {{r}^{3}}=84\pi t+36\pi$ 

$\Rightarrow {{r}^{3}}=63t+27$ 

$\Rightarrow r={{\left( 63t+27 \right)}^{\frac{1}{3}}}$

The radius of balloon after $\text{t}$ seconds is given by: $r={{\left( 63t+27 \right)}^{\frac{1}{3}}}$ .

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Ans: Let the principal be $\text{p}$, according to :

$\frac{dp}{dt}=\left( \frac{r}{100} \right)p$ 

Simplify the expression:

$\frac{dp}{p}=\left( \frac{r}{100} \right)dt$ 

Integrate both side:
$\int \frac{dp}{p}=\int \left( \frac{r}{100} \right)dt$ 

$\Rightarrow \log p=\frac{rt}{100}+c$ 

\[\Rightarrow p={{e}^{\frac{rt}{100}+c}}\] 

\[\Rightarrow p=A{{e}^{\frac{rt}{100}}}\,\,\,\left( A={{e}^{c}} \right)\] 

At $\text{t=0}$, $\text{p=100}$:

\[100=A{{e}^{\frac{r\left( 0 \right)}{100}}}\] 

$\Rightarrow A=100$ 

Thus the principle and rate of interest relation:

$p=100{{e}^{\frac{rt}{100}}}$ 

At $\text{t=10}$, $\text{p=2}\times \text{100=200}$:

\[200=100{{e}^{\frac{r\left( 10 \right)}{100}}}\] 

$\Rightarrow 2={{e}^{\frac{r}{10}}}$ 

Take logarithm on both side:

$\log \left( {{e}^{\frac{r}{10}}} \right)=\log \left( 2 \right)$ 

$\Rightarrow \frac{r}{10}=0.6931$ 

$\Rightarrow r=6.931$ 

Thus the rate of interest $\text{r=6}\text{.931 }\!\!%\!\!\text{ }$.