Question

Solve the differential equation $\left(y+3x^2\right)\frac{dx}{dy}=x$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(y+3x^2\right)\frac{dx}{dy}=x \\ \Rightarrow \frac{dy}{dx}=\frac{y+3x^2}{x} \\ \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=3x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=-\frac{1}{x}\mathrm{and}Q=3x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{-\int \frac{1}{x}dx} \\ =e^{-\log x} \\ =\frac{1}{x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\frac{1}{x},\mathrm{we}\mathrm{get} \\ \frac{1}{x}\left(\frac{dy}{dx}-\frac{1}{x}y\right)=\frac{1}{x}3x \\ \Rightarrow \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=3 \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \frac{1}{x}y=3\int dx+C \\ \Rightarrow \frac{y}{x}=3x+C \\ \mathrm{Hence},\frac{y}{x}=3x+C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$