Question

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

• A. $y=\frac{1}{x^2}$
• B. $y=\frac{1}{y^2}$
• C. $x=\frac{1}{y}$
• D. $x=y$

Answer 1

The correct option is A

$y=\frac{1}{x^2}$

Explanation for the correct option

Given: $\frac{dy}{dx}+\frac{2y}{x}=0$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=-\frac{2y}{x} \\ \Rightarrow -\frac{dy}{2y}=\frac{dx}{x} \end{gathered}$$

integrating both sides

$$\begin{gathered} \Rightarrow -\frac{1}{2}\int \frac{1}{y}dy=\int \frac{1}{x}dx \\ \Rightarrow -\frac{1}{2}\log \left(y\right)=\log \left(x\right) \\ \Rightarrow \log \left(\frac{1}{y}\right)=2\log \left(x\right) \\ \Rightarrow \log \left(\frac{1}{y}\right)=\log \left(x^2\right) \end{gathered}$$

taking anti-log to both sides

$$\begin{gathered} \Rightarrow \frac{1}{y}=x^2 \\ \Rightarrow y=\frac{1}{x^2} \end{gathered}$$

Hence option A is correct.