Question

Find a particular solution of each of the following differential equations:
(i) $\left(1+x^2\right)\frac{dy}{dx}+2xy=\frac{1}{1+x^2};y=0,\mathrm{when}x=1$

(ii) (x + y) dy + (x − y) dx = 0; y = 1 when x = 1

(iii) x² dy + (xy + y²) dx = 0; y = 1 when x = 1

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}, \\ \left(1+x^2\right)\frac{dy}{dx}+2xy=\frac{1}{1+x^2} \\ \Rightarrow \frac{dy}{dx}+\frac{2x}{\left(1+x^2\right)}y=\frac{1}{{\left(1+x^2\right)}^2} \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{2x}{\left(1+x^2\right)} \\ Q=\frac{1}{{\left(1+x^2\right)}^2} \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{2x}{\left(1+x^2\right)}dx} \\ =e^{\log \left|1+x^2\right|} \\ =1+x^2 \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y\left(1+x^2\right)=\int \frac{1}{\left(1+x^2\right)}dx+C \\ \Rightarrow y\left(1+x^2\right)={\tan }^{-1}x+C.....\left(1\right) \\ \mathrm{Now}, \\ \mathrm{When}x=1,y=0 \\ \therefore 0\left(1+1\right)={\tan }^{-1}1+C \\ \Rightarrow C=-1 \\ \Rightarrow C=-\frac{\mathrm{\pi }}{4} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y\left(1+x^2\right)={\tan }^{-1}x-\frac{\mathrm{\pi }}{4} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}, \\ \left(x+y\right)dy+\left(x-y\right)dx=0 \\ \frac{dy}{dx}=\frac{y-x}{x+y} \\ \mathrm{Let}y=vx \\ \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \therefore v+x\frac{dv}{dx}=\frac{vx-x}{x+vx} \\ \Rightarrow x\frac{dv}{dx}=\frac{v-1}{1+v}-v \\ \Rightarrow \frac{xdv}{dx}=\frac{v-1-v-v^2}{1+v} \\ \Rightarrow x\frac{dv}{dx}=-\left(\frac{v^2+1}{1+v}\right) \\ \Rightarrow \frac{1+v}{v^2+1}dv=-\frac{1}{x}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1+v}{1+v^2}dy=-\int \frac{1}{x}dx \\ \int \frac{1}{1^2+v^2}dy+\frac{1}{2}\int \frac{2v}{1+v^2}=-\int \frac{1}{x}dx \\ \Rightarrow {\tan }^{-1}v+\frac{1}{2}\log \left(1+v^2\right)=-\log \left|x\right|+C \\ \Rightarrow 2{\tan }^{-1}v+\log \left(1+v^2\right)+2\log \left|x\right|=2C \\ \Rightarrow 2{\tan }^{-1}v+\log \left(1+v^2\right)x^2=k\mathrm{where},k=2C \\ \Rightarrow 2{\tan }^{-1}\frac{y}{x}+\log \left(1+\frac{y^2}{x^2}\right)x^2=k \\ \Rightarrow 2{\tan }^{-1}\frac{y}{x}+\log \left(x^2+y^2\right)=k.....\left(1\right) \\ \mathrm{Now}, \\ \mathrm{When}x=1,y=1 \\ \therefore 2{\tan }^{-1}1+\log \left(2\right)=k \\ \Rightarrow k=\frac{\mathrm{\pi }}{2}+\log 2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}k\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ 2{\tan }^{-1}\frac{y}{x}+\log \left(x^2+y^2\right)=\frac{\mathrm{\pi }}{2}+\log 2 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}, \\ {x}^{2}dy+\left(xy+y^2\right)dx=0 \\ \frac{dy}{dx}=-\left(\frac{xy+y^2}{x^2}\right) \\ \mathrm{Let}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \therefore v+x\frac{dv}{dx}=-\left(\frac{v{x}^2+v^2{x}^2}{x^2}\right) \\ \Rightarrow x\frac{dv}{dx}=-\left(v+v^2\right)-v \\ \Rightarrow x\frac{dv}{dx}=-2v-v^2 \\ \Rightarrow \frac{1}{v^2+2v}dv=-\frac{1}{x}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{v^2+2v}dy=-\int \frac{1}{x}dx \\ \int \frac{1}{v^2+2v+1-1}dy=-\int \frac{1}{x}dx \\ \int \frac{1}{{\left(v+1\right)}^2-{\left(1\right)}^2}dy=-\int \frac{1}{x}dx \\ \Rightarrow \frac{1}{2\times 1}\log \left|\frac{v+1-1}{v+1+1}\right|=-\log \left|x\right|+\log C \\ \Rightarrow \frac{1}{2}\log \left|\frac{v}{v+2}\right|=-\log \left|x\right|+\log C \\ \Rightarrow \log \left|\frac{v}{v+2}\right|=-2\log \left|x\right|+2\log C \\ \Rightarrow \log \left|\frac{v}{v+2}\right|+\log \left|x^2\right|=\log C^2 \\ \Rightarrow \log \left|\frac{v{x}^2}{v+2}\right|=\log C^2 \\ \Rightarrow \left|\frac{v{x}^2}{v+2}\right|=C^2 \\ \Rightarrow \left|\frac{v{x}^2}{v+2}\right|=k,\mathrm{where}k=2C \\ \Rightarrow \left|\frac{{\displaystyle \frac{y}{x}{x}^2}}{{\displaystyle \frac{y}{x}+2}}\right|=k \\ \Rightarrow \left|\frac{x^{2}y}{y+2x}\right|=k.....\left(1\right) \\ \mathrm{Now}, \\ \mathrm{When}x=1,y=1 \\ \therefore \left|\frac{1}{1+2}\right|=k \\ \Rightarrow k=\frac{1}{3} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}k\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \left|\frac{x^{2}y}{y+2x}\right|=\frac{1}{3} \\ \Rightarrow 3x^{2}y=\pm \left(y+2x\right) \\ \mathrm{But}y\left(1\right)=1\mathrm{does}\mathrm{not}\mathrm{satisfy}\mathrm{the}\mathrm{equation}3x^{2}y=-\left(y+2x\right). \\ \therefore 3x^{2}y=y+2x \\ \Rightarrow y=\frac{2x}{3x^2-1} \end{gathered}$$