Question

The general solution of the differential equation $\frac{dy}{dx}+\frac{y}{x}$ = 1 is _____________.

Answer 1

Given: $\frac{dy}{dx}+\frac{y}{x}$ = 1

$$\begin{gathered} \frac{dy}{dx}+\frac{y}{x}=1 \\ \mathrm{Here},P=\frac{1}{x}\mathrm{and}Q=1 \\ \mathrm{Integrating}\mathrm{Factor}\left(\mathrm{I}.\mathrm{F}.\right)=e^{\int Pdx} \\ =e^{\int \frac{1}{x}dx} \\ =e^{\log x} \\ =x \\ \mathrm{Thus},\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\mathrm{is} \\ y\times \left(\mathrm{I}.\mathrm{F}.\right)=\int Q\times \left(\mathrm{I}.\mathrm{F}.\right)dx+C \\ \Rightarrow yx=\int xdx+C \\ \Rightarrow yx=\frac{x^2}{2}+C,\mathrm{where}C\mathrm{is}\mathrm{any}\mathrm{constant} \end{gathered}$$

Hence, the general solution of the differential equation $\frac{dy}{dx}+\frac{y}{x}$ = 1 is $\underline{yx=\frac{x^2}{2}+C}$.