Question

The solution of the differential equation $\frac{dy}{dx}=\frac{x+y}{x}$ satisfying the condition $y\left(1\right)=1$ is

• A. $y=x\log x+x$
• B. $y=\log x+x$
• C. $y=x\log x+x^2$
• D. $y=x{e}^{x-1}$

Answer 1

The correct option is A

$y=x\log x+x$

Explanation for correct options

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

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

Let $xv=y$

$\Rightarrow x\frac{dv}{dx}+v=\frac{dy}{dx}$

$$\begin{gathered} x\frac{dv}{dx}+v=1+v \\ \Rightarrow dv=\frac{dx}{x} \end{gathered}$$

integrating both sides

$$\begin{gathered} \Rightarrow \int dv=\int \frac{dx}{x} \\ \Rightarrow v=\log \left(x\right)+C \end{gathered}$$

Put $v=\frac{y}{x}$

$$\begin{gathered} \Rightarrow \frac{y}{x}=\log \left(x\right)+C \\ \Rightarrow y=x\log \left(x\right)+Cx \end{gathered}$$

Since $\frac{dy}{dx}=\frac{x+y}{x}$ passes through $y\left(1\right)=1$

Put $x=1,y=1$

$$\begin{gathered} \Rightarrow 1=1\log \left(1\right)+C\times 1 \\ \Rightarrow C=1 \end{gathered}$$

$\Rightarrow y=x\log \left(x\right)+x$

Hence, option A is correct.