Question

Find the particular solution of the differential equation $\frac{dy}{dx}=-4x{y}^2$ given that y = 1, when x = 0.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=-4x{y}^2 \\ \Rightarrow \frac{1}{y^2}dy=-4xdx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y^2}\mathrm{d}\mathrm{y}=-4\int x\mathrm{dx} \\ \Rightarrow -\frac{1}{y}=-4\times \frac{x^2}{2}+C \\ \Rightarrow -\frac{1}{y}=-2x^2+\mathrm{C}.....\left(1\right) \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{at}x=0,\mathit{}y=1. \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}\mathit{}x\mathrm{and}y\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{C}=-1 \\ \mathrm{Therefore},\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ -\frac{1}{y}=-2x^2-1 \\ \Rightarrow y=\frac{1}{2x^2+1} \\ \mathrm{Hence},\mathrm{y}=\frac{1}{2x^2+1}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$