Question

sient terms in the gener
$x^{2}y+x(x + 2)y=e^{x}$

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, $\mu(x)=e^{\ln(e^{x}x^{2})}=x^{2}e^{x}$.

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

Multiply $y'+(1+\frac{2}{x})y=\frac{e^{x}}{x^{2}}$ by $\mu(x)=x^{2}e^{x}$:
$(x^{2}e^{x}y)'=e^{x}\cdot e^{x}=e^{2x}$.

Step4: Integrate both sides

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

Step5: Solve for y

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

Answer:

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