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 we divide through by $x^{2}$ to get it in the form $y'+P(x)y = Q(x)$ where $P(x)=\frac{x + 2}{x}=1+\frac{2}{x}$ 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 through by integrating - factor

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

Step4: Integrate both sides

Integrate both sides with respect to $x$: $\int\frac{d}{dx}(e^{x}x^{2}y)dx=\int e^{2x}dx$. We know that $\int e^{2x}dx=\frac{1}{2}e^{2x}+C$. So, $e^{x}x^{2}y=\frac{1}{2}e^{2x}+C$.

Step5: Solve for y

Solve for $y$: $y=\frac{e^{x}}{2x^{2}}+\frac{C}{x^{2}e^{x}}$.

Answer:

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