Question

The solution of the differential equation $\frac{dy}{dx}-\frac{y\left(x+1\right)}{x}=0$ is given by
(a) y = xe^{x + C}
(b) x = ye^x
(c) y = x + C
(d) xy = e^x + C

Answer 1

(a) y = xe^{x + C}

We have,
$$\begin{gathered} \frac{dy}{dx}-\frac{y\left(x+1\right)}{x}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{y\left(x+1\right)}{x} \\ \Rightarrow \frac{dy}{y}=\frac{\left(x+1\right)}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{dy}{y}=\int \frac{\left(x+1\right)}{x}dx \\ \Rightarrow \int \frac{dy}{y}=\int dx+\int \frac{1}{x}dx \\ \Rightarrow \log y=x+\log x+C \\ \Rightarrow \log y-\log x=x+C \\ \Rightarrow \log \left(\frac{y}{x}\right)=x+C \\ \Rightarrow \frac{y}{x}=e^{x+C} \\ \Rightarrow y=x{e}^{x+C} \end{gathered}$$