Question

Solve the differential equation $\left(x+2y^2\right)\frac{dy}{dx}=y$, given that when x = 2, y = 1.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(x+2y^2\right)\frac{dy}{dx}=y \\ \Rightarrow \frac{dx}{dy}=\frac{1}{y}\left(x+2y^2\right) \\ \Rightarrow \frac{dx}{dy}-\frac{1}{y}x=2y.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dx}{dy}+Px=Q \\ \mathrm{where} \\ P=-\frac{1}{y} \\ Q=2y \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdy} \\ =e^{-\int \frac{1}{y}dy} \\ =e^{-\log y}=\frac{1}{y} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\frac{1}{y},\mathrm{we}\mathrm{get} \\ \frac{1}{y}\left(\frac{dx}{dy}-\frac{1}{y}x\right)=\frac{1}{y}\times 2y \\ \Rightarrow \frac{1}{y}\frac{dx}{dy}-\frac{1}{y^2}x=2 \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}y,\mathrm{we}\mathrm{get} \\ x\frac{1}{y}=\int 2dy+C \\ \Rightarrow x\frac{1}{y}=2y+C \\ \Rightarrow x=2y^2+Cy.....\left(2\right) \\ \mathrm{Now}, \\ y=1\mathrm{at}x=2 \\ \therefore 2=2+C \\ \Rightarrow C=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ x=2y^2 \\ \mathrm{Hence},x=2y^2\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$