Question

The solution of the differential equation $\frac{dy}{dx}+\frac{2y}{x}=0$ with y(1) = 1 is given by
(a) $y=\frac{1}{x^2}$

(b) $x=\frac{1}{y^2}$

(c) $x=\frac{1}{y}$

(d) $y=\frac{1}{x}$

Answer 1

(a) $y=\frac{1}{x^2}$

We have,

$$\begin{gathered} \frac{dy}{dx}+\frac{2y}{x}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-2y}{x} \\ \Rightarrow \frac{1}{2}\times \frac{1}{y}dy=\frac{-1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \frac{1}{2}\int \frac{1}{y}dy=-\int \frac{1}{x}dx \\ \Rightarrow \frac{1}{2}\log y=-\log x+\log C \\ \Rightarrow \log y^{\frac{1}{2}}+\log x=\log C \\ \Rightarrow \log \left(\sqrt{y}x\right)=\log C \\ \Rightarrow \sqrt{y}x=C.....\left(1\right) \\ \mathrm{As}\left(1\right)\mathrm{satisfies}y\left(1\right)=1,\mathrm{we}\mathrm{get} \\ 1=C \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \sqrt{y}x=1 \\ \Rightarrow y=\frac{1}{x^2} \end{gathered}$$