Question

Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x² + 1) dx, x ≠ 0.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ xdy=\left(2x^2+1\right)dx \\ \Rightarrow dy=\left(\frac{2x^2+1}{x}\right)dx \\ \Rightarrow dy=\left(2x+\frac{1}{x}\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \left(2x+\frac{1}{x}\right)dx \\ \Rightarrow y=x^2+\log \left|x\right|+C.....\left(1\right) \\ \mathrm{Now}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{passes}\mathrm{through}\left(1,1\right) \\ \mathrm{Therefore},\mathrm{when}x=1,y=1 \\ \therefore 1=1+0+C \\ \Rightarrow C=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y=x^2+\log \left|x\right| \end{gathered}$$