Question

find the general solution of the given differential equation. $x^{2}y+x(x + 2)y=e^{x}$

Explanation:

Step1: Rewrite in standard form

First, rewrite the given differential equation $x^{2}y'+x(x + 2)y=e^{x}$ as $y'+\frac{x + 2}{x}y=\frac{e^{x}}{x^{2}}$. Here, $P(x)=\frac{x + 2}{x}=1+\frac{2}{x}$ and $Q(x)=\frac{e^{x}}{x^{2}}$.

Step2: Find integrating - factor

The integrating factor $\mu(x)=e^{\int P(x)dx}$. Calculate $\int P(x)dx=\int(1+\frac{2}{x})dx=x + 2\ln|x|=\ln(e^{x}x^{2})$. So, $\mu(x)=e^{x}x^{2}$.

Step3: Multiply by integrating - factor

Multiply the standard - form equation by $\mu(x)$: $(e^{x}x^{2}y)'=e^{x}\cdot\frac{e^{x}}{x^{2}}\cdot x^{2}=e^{2x}$.

Step4: Integrate both sides

Integrate both sides with respect to $x$: $e^{x}x^{2}y=\int e^{2x}dx$. Since $\int e^{2x}dx=\frac{1}{2}e^{2x}+C$, then $y=\frac{e^{x}}{2x^{2}}+\frac{C}{e^{x}x^{2}}$.

Answer:

$y=\frac{e^{x}}{2x^{2}}+\frac{C}{e^{x}x^{2}}$