Question

$x\frac{dy}{dx}+1=0;y\left(-1\right)=0$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+1=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-1}{x} \\ \Rightarrow dy=\left(\frac{-1}{x}\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \int dy=\int \left(\frac{-1}{x}\right)dx \\ \Rightarrow y=-\log \left|x\right|+C.....\left(1\right) \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}y\left(-1\right)=0. \\ \therefore 0=-\log \left|-1\right|+C \\ \Rightarrow C=0 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y=-\log \left|x\right| \\ \mathrm{Hence},y=-\log \left|x\right|\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \end{gathered}$$