Question

$\frac{dy}{dx}=x^2+x-\frac{1}{x},x\ne 0$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=x^2+x-\frac{1}{x} \\ \Rightarrow dy=\left(x^2+x-\frac{1}{x}\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \int dy=\int \left(x^2+x-\frac{1}{x}\right)dx \\ \Rightarrow y=\frac{x^3}{3}+\frac{x^2}{2}-\log \left|x\right|+C \\ \mathrm{Clearly},y=\frac{x^3}{3}+\frac{x^2}{2}-\log \left|x\right|+C\mathrm{is}\mathrm{defined}\mathrm{for}\mathrm{all}x\in R\mathrm{except}x=0. \\ \mathrm{Hence},y=\frac{x^3}{3}+\frac{x^2}{2}-\log \left|x\right|+C,\mathrm{where}x\in R-\left\{0\right\},\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \end{gathered}$$