Question

Solve the following differential equation:
(i) (xy² + 2x) dx + (x² y + 2y) dy = 0
(ii) $\mathrm{cosec}x\log y\frac{dy}{dx}+x^2{y}^2=0$

Answer 1

(i) (xy² + 2x) dx + (x² y + 2y) dy = 0

$$\begin{gathered} Wehave, \\ \left(x{y}^2+2x\right)dx+\left(x^{2}y+2y\right)dy=0 \\ \Rightarrow x\left(y^2+2\right)dx+y\left(x^2+2\right)dy=0 \\ \Rightarrow x\left(y^2+2\right)dx=-y\left(x^2+2\right)dy \\ \Rightarrow \frac{x}{\left(x^2+2\right)}dx=-\frac{y}{\left(y^2+2\right)}dy \\ Integratingbothsides,weget \\ \int \frac{x}{x^2+2}dx=-\int \frac{y}{y^2+2}\mathit{}dy \\ \Rightarrow \frac{1}{2}\int \frac{2x}{x^2+2}dx=-\frac{1}{2}\int \frac{2y}{y^2+2}\mathit{}dy \\ \Rightarrow \frac{1}{2}log\left|x^2+2\right|=-\frac{1}{2}log\left|y^2+2\right|+logC \\ \Rightarrow \frac{1}{2}log\left|x^2+2\right|+\frac{1}{2}log\left|y^2+2\right|=logC \\ \Rightarrow log\left|x^2+2\right|+log\left|y^2+2\right|=2logC \\ \Rightarrow log\left(\left|x^2+2\right|\left|y^2+2\right|\right)=log{C}^2 \\ \Rightarrow \left(\left|x^2+2\right|\left|y^2+2\right|\right)=C^2 \\ \Rightarrow \left(x^2+2\right)\left(y^2+2\right)=K \\ \Rightarrow y^2+2=\frac{K}{x^2+2} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{cosec}x\log y\frac{dy}{dx}+x^2{y}^2=0 \\ \Rightarrow \mathrm{cosec}x\log y\frac{dy}{dx}=-x^2{y}^2 \\ \Rightarrow \frac{1}{y^2}\log ydy=-\frac{x^2}{\mathrm{cosec}x}dx \\ \Rightarrow \frac{1}{y^2}\log ydy=-x^2\sin xdx \\ \Rightarrow \int \frac{1}{y^2}\log ydy=-\int x^2\sin xdx \end{gathered}$$
$$\begin{gathered} \Rightarrow -\frac{\log y}{y}+\int \frac{1}{y}\times \frac{1}{y}=-\left[-x^2\cos x+\int 2x\cos xdx\right]+C \\ \Rightarrow -\frac{\log y}{y}-\frac{1}{y}=-\left[-x^2\cos x+2x\sin x-2\int \sin xdx\right]+C \\ \Rightarrow -\left(\frac{1+\log y}{y}\right)=-\left[-x^2\cos x+2x\sin x+2\cos xdx\right]+C \\ \Rightarrow -\left(\frac{1+\log y}{y}\right)-x^2\cos x+2\left(x\sin x+\cos x\right)=C \end{gathered}$$