The solution of differential equation $x^{2} \frac{d y}{d x}=y^{2}+x y$ is

The solution of differential equation $x^{2} \frac{d y}{d x}=y^{2}+x y$ is
  1. $\frac{x}{y}+\log |x|=c$
  2. $\frac{y}{x}+\log |x|=c$
  3. $\frac{x}{y}-\log |x|=c$
  4. $\frac{y}{x}-\log |x|=c$

Solution

We have $x^{2} \frac{d y}{d x}=y^{2}+x y$ $\begin{array}{l} \therefore \frac{d y}{d x}=\frac{y^{2}+x y}{x^{2}} ...(1) \\ \therefore \text { Put } y=u x \\ \therefore \frac{d y}{d x}=u+x \frac{d u}{d x} \end{array}$ $\therefore$ Equation (1) becomes $u+x \frac{d u}{d x}=\frac{u^{2} x^{2}+x(u x)}{x^{2}} \Rightarrow u+x \frac{d u}{d x}=u^{2}+u$ $\begin{array}{l} \therefore \mathrm{x} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}^{2} \\ \therefore \int \frac{\mathrm{du}}{\mathrm{u}^{2}}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\ \therefore \frac{-1}{\mathrm{u}}=\log |\mathrm{x}|+\mathrm{c}_{1} \Rightarrow \quad \frac{-\mathrm{x}}{\mathrm{y}}=\log |\mathrm{x}|+\mathrm{c}_{1} \\ \therefore \frac{\mathrm{x}}{\mathrm{y}}+\log |\mathrm{x}|=\mathrm{c} \end{array}$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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