Question

$x^{2}y+x(x + 2)y=e^{x}$

Explanation:

Step1: Rewrite in standard form

First, rewrite the given first - order linear differential equation $x^{2}y'+x(x + 2)y=e^{x}$ in the standard form $y'+P(x)y = Q(x)$. Divide through by $x^{2}$:
$y'+\frac{x(x + 2)}{x^{2}}y=\frac{e^{x}}{x^{2}}$, which simplifies to $y'+(1+\frac{2}{x})y=\frac{e^{x}}{x^{2}}$. Here, $P(x)=1+\frac{2}{x}$ and $Q(x)=\frac{e^{x}}{x^{2}}$.

Step2: Find the integrating factor

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

Step3: Multiply the standard - form equation by the integrating factor

Multiply $y'+(1+\frac{2}{x})y=\frac{e^{x}}{x^{2}}$ by $\mu(x)=e^{x}x^{2}$:
$e^{x}x^{2}y'+e^{x}x(x + 2)y=e^{2x}$. The left - hand side is the derivative of the product $e^{x}x^{2}y$ by the product rule. That is, $(e^{x}x^{2}y)'=e^{2x}$.

Step4: Integrate both sides

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

Step5: Solve for $y$

Divide both sides by $e^{x}x^{2}$ to get $y=\frac{e^{x}}{2x^{2}}+\frac{C}{e^{x}x^{2}}$.

Answer:

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