Question

The general solution of the differential equation $x\frac{dy}{dx}+2y=x^2\mathrm{is}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

Given: $x\frac{dy}{dx}+2y=x^2$


$$\begin{gathered} x\frac{dy}{dx}+2y=x^2 \\ \Rightarrow \frac{dy}{dx}+\frac{2y}{x}=\frac{x^2}{x} \\ \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x \\ \mathrm{Here},P=\left(\frac{2}{x}\right)\mathrm{and}Q=x \\ \mathrm{Integrating}\mathrm{factor}=e^{\int Pdx} \\ =e^{\int \left(\frac{2}{x}\right)dx} \\ =e^{2\log x} \\ =e^{\log x^2} \\ =x^2 \\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{is}: \\ y\times x^2=\int x\times x^{2}dx+C \\ \Rightarrow x^{2}y=\int x^{3}dx+C \\ \Rightarrow x^{2}y=\frac{x^4}{4}+C,\mathrm{where}C\mathrm{is}\mathrm{arbitrary}\mathrm{constant} \end{gathered}$$


Hence, the general solution of the differential equation $x\frac{dy}{dx}+2y=x^2\mathrm{is}\underline{x^{2}y=\frac{x^4}{4}+C}.$