Question

Solve the following differential equations:
(i) $xy\frac{dy}{dx}=1+x+y+xy$

(ii) $y\left(1-x^2\right)\frac{dy}{dx}=x\left(1+y^2\right)$
(iii) $y{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}dx=\left(x{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}+y^2\right)dy,y\ne 0$
(iv) $\left(1+y^2\right){\tan }^{-1}xdx+2y\left(1+x^2\right)dy=0$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}, \\ xy\frac{dy}{dx}=1+x+y+xy \\ \Rightarrow xy\frac{dy}{dx}=\left(1+x\right)\left(1+y\right) \\ \Rightarrow \frac{y}{1+y}dy=\frac{\left(1+x\right)}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{y}{1+y}dy=\int \frac{\left(1+x\right)}{x}dx \\ \Rightarrow \int \frac{1+y-1}{1+y}dy=\int \frac{\left(1+x\right)}{x}dx \\ \Rightarrow \int dy-\int \frac{1}{1+y}dy=\int \frac{1}{x}dx+\int dx \\ \Rightarrow y-\log \left|1+y\right|=\log \left|x\right|+x+C \\ \Rightarrow y=\log \left|x\right|+\log \left|1+y\right|+x+C \\ \Rightarrow y=\log \left|x\left(1+y\right)\right|+x+C \\ \mathrm{Hence},y=\log \left|x\left(1+y\right)\right|+x+C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}, \\ y\left(1-x^2\right)\frac{dy}{dx}=x\left(1+y^2\right) \\ \Rightarrow \frac{y}{1+y^2}dy=\frac{x}{1-x^2}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}, \\ \int \frac{y}{1+y^2}dy=\int \frac{x}{1-x^2}dx \\ \mathrm{Substituting}1+y^2=t\mathrm{and}1-x^2=u \\ 2ydy=dt\mathrm{and}-2xdx=du \\ \therefore \frac{1}{2}\int \frac{1}{t}dt=\frac{-1}{2}\int \frac{1}{u}du \\ \Rightarrow \frac{1}{2}\log \left|t\right|=-\frac{1}{2}\log \left|u\right|+\log \mathrm{C} \\ \Rightarrow \frac{1}{2}\log \left|1+y^2\right|=-\frac{1}{2}\log \left|1-x^2\right|+\log \mathrm{C} \\ \Rightarrow \frac{1}{2}\left[\log \left|1+y^2\right|+\log \left|1-x^2\right|\right]=\log \mathrm{C} \\ \Rightarrow \log \left(\left|1+y^2\right|\left|1-x^2\right|\right)=2\log \mathrm{C} \\ \Rightarrow \left(1+y^2\right)\left(1-x^2\right)={\mathrm{C}}^2 \\ \Rightarrow \left(1+y^2\right)\left(1-x^2\right)={\mathrm{C}}_1,\mathrm{where}{\mathrm{C}}_1={\mathrm{C}}^2 \\ \mathrm{Hence},\left(1+y^2\right)\left(1-x^2\right)={\mathrm{C}}_1\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$

(iii)

$$\begin{gathered} y{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}dx=\left(x{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}+y^2\right)dy \\ \Rightarrow y{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}dx=x{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}dy+y^{2}dy \\ \Rightarrow y{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}dx-x{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}dy=y^{2}dy \\ \Rightarrow \left(ydx-xdy\right)e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}=y^{2}dy \\ \Rightarrow \frac{\left(ydx-xdy\right)}{y^2}{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}=dy \\ \Rightarrow e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}d\left(\frac{x}{y}\right)=dy \\ \Rightarrow \int e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}d\left(\frac{x}{y}\right)=\int dy \\ \Rightarrow e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}=y+C \end{gathered}$$
(iv)
$$\begin{gathered} \left(1+y^2\right){\tan }^{-1}xdx+2y\left(1+x^2\right)dy=0 \\ \Rightarrow \left(1+y^2\right){\tan }^{-1}xdx=-2y\left(1+x^2\right)dy \\ \Rightarrow \frac{{\tan }^{-1}x}{\left(1+x^2\right)}dx=-\frac{2y}{\left(1+y^2\right)}dy \\ \Rightarrow \int \frac{{\tan }^{-1}x}{\left(1+x^2\right)}dx=-\int \frac{2y}{\left(1+y^2\right)}dy \\ \Rightarrow \frac{{\left({\tan }^{-1}x\right)}^2}{2}=-\log \left|1+y^2\right|+C \\ \Rightarrow \frac{{\left({\tan }^{-1}x\right)}^2}{2}+\log \left|1+y^2\right|=C \end{gathered}$$