Question

find the general solution of the given differential equation. (x^{2}y+x(x + 2)y=e^{x}). give the largest interval (i) over which the general solution is defined. determine whether there are any transient terms in the general solution.

Explanation:

Step1: Rewrite the differential equation

First, rewrite $x^{2}y'+x(x + 2)y=e^{x}$ as $y'+\frac{x + 2}{x}y=\frac{e^{x}}{x^{2}}$. This is in the first - order linear 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 the 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 the differential equation by the integrating factor

Multiply $y'+\frac{x + 2}{x}y=\frac{e^{x}}{x^{2}}$ by $\mu(x)=e^{x}x^{2}$ to get $(e^{x}x^{2}y)'=e^{2x}$.

Step4: Integrate both sides

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

Step5: Solve for y

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

To determine if there are transient terms, recall that a transient term is a term that approaches 0 as $x\to\pm\infty$. The term $\frac{C}{e^{x}x^{2}}$ is a transient term since $\lim_{x\to+\infty}\frac{C}{e^{x}x^{2}} = 0$ and $\lim_{x\to-\infty}\frac{C}{e^{x}x^{2}}=\pm\infty$ (but we usually consider the behavior as $x\to+\infty$ in the context of transient terms).

The general solution is $y=\frac{e^{x}}{2x^{2}}+\frac{C}{e^{x}x^{2}}$. The function is undefined at $x = 0$. The largest interval over which the general solution is defined is either $(-\infty,0)$ or $(0,+\infty)$.

Answer:

General solution: $y=\frac{e^{x}}{2x^{2}}+\frac{C}{e^{x}x^{2}}$; Transient term: $\frac{C}{e^{x}x^{2}}$; Largest interval: $(-\infty,0)$ or $(0,+\infty)$