Question

For each of the following differential equations, find a particular solution satisfying the given condition:
(i) $x\left(x^2-1\right)\frac{dy}{dx}=1,y=0\mathrm{when}x=2$

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

(iii) $\frac{dy}{dx}=y\tan x,y=1\mathrm{when}x=0$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}, \\ x\left(x^2-1\right)\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{x\left(x^2-1\right)} \\ \Rightarrow dy=\left\{\frac{1}{x\left(x^2-1\right)}\right\}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \left\{\frac{1}{x\left(x^2-1\right)}\right\}dx \\ \Rightarrow y=\int \left\{\frac{1}{x\left(x^2-1\right)}\right\}dx+C \\ \Rightarrow y=\int \left\{\frac{1}{x\left(x+1\right)\left(x-1\right)}\right\}dx+C.....\left(1\right) \\ \mathrm{Let}\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+1\right)\left(x-1\right)+Bx\left(x-1\right)+Cx\left(x+1\right) \\ \Rightarrow 1=A\left(x^2-1\right)+B\left(x^2-x\right)+C\left(x^2+x\right) \\ \Rightarrow 1=x^2\left(A+B+C\right)+x\left(-B+C\right)-A \\ \mathrm{Comparing}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ -A=1.....\left(2\right) \\ -B+C=0.....\left(3\right) \\ A+B+C=0.....\left(4\right) \\ \mathrm{Solving}\left(2\right),\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get} \\ A=-1 \\ B=\frac{1}{2} \\ C=\frac{1}{2} \\ \therefore \frac{1}{x\left(x+1\right)\left(x-1\right)}=\frac{-1}{x}+\frac{1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)} \\ \mathrm{Now},\left(1\right)\mathrm{becomes} \\ y=\int \left\{\frac{-1}{x}+\frac{1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}\right\}dx+C \\ \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 \left|x\right|+\frac{1}{2}\log \left|x-1\right|+\frac{1}{2}\log \left|x+1\right|+C \\ \Rightarrow y=\frac{1}{2}\log \left|x-1\right|+\frac{1}{2}\log \left|x+1\right|-\log \left|x\right|+C \\ \mathrm{Given}:\hspace{0.17em}y\left(2\right)=0 \\ \therefore 0=\frac{1}{2}\log \left|2-1\right|+\frac{1}{2}\log \left|2+1\right|-\log \left|2\right|+C \\ \Rightarrow C=\log \left|2\right|-\frac{1}{2}\log \left|3\right| \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}C,\mathrm{we}\mathrm{get} \\ y=\frac{1}{2}\log \left|x-1\right|+\frac{1}{2}\log \left|x+1\right|-\log \left|x\right|+\log \left|2\right|-\frac{1}{2}\log \left|3\right| \\ \Rightarrow 2y=\log \left|x-1\right|+\log \left|x+1\right|-2\log \left|x\right|+2\log \left|2\right|-\log \left|3\right| \\ \Rightarrow 2y=\log \left|x-1\right|+\log \left|x+1\right|-\log \left|x^2\right|+\log \left|4\right|-\log \left|3\right| \\ \Rightarrow 2y=\log \frac{\left(x-1\right)\left(x+1\right)}{x^2}-\left(\log \left|3\right|-\log \left|4\right|\right) \\ \Rightarrow y=\frac{1}{2}\log \frac{\left(x^2-1\right)}{x^2}-\frac{1}{2}\log \left(\frac{3}{4}\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}, \\ \cos \left(\frac{dy}{dx}\right)=a \\ \Rightarrow \frac{dy}{dx}={\cos }^{-1}a \\ \Rightarrow dy={\cos }^{-1}adx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int {\cos }^{-1}adx \\ \Rightarrow y=x{\cos }^{-1}a+C \\ \mathrm{Now}, \\ \mathrm{When}x=0,y=1 \\ \therefore 1=0+C \\ \Rightarrow C=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y=x{\cos }^{-1}a+1 \\ \Rightarrow \cos \left(\frac{y-1}{x}\right)=a \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=y\tan x \\ \Rightarrow \frac{1}{y}dy=\tan xdx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y}dy=\int \tan xdx \\ \Rightarrow \log y=\log \left|\sec x\right|+C....\left(1\right) \\ \mathrm{Now}, \\ \mathrm{When}x=0,y=1 \\ \therefore \log 1=\log 1+C \\ \Rightarrow C=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log y=\log \left|\sec x\right| \\ \Rightarrow y=\sec \mathit{}x \end{gathered}$$