Question

1. $y = 2y + x^{2}+5$

Explanation:

Step1: Identify the type of ODE

This is a first - order linear ordinary differential equation of the form $y' - 2y=x^{2}+5$, where $P(x)= - 2$ and $Q(x)=x^{2}+5$.

Step2: Find the integrating factor

The integrating factor $\mu(x)=e^{\int P(x)dx}=e^{\int - 2dx}=e^{-2x}$.

Step3: Multiply the ODE by the integrating factor

We get $e^{-2x}y'-2e^{-2x}y=(x^{2}+5)e^{-2x}$. The left - hand side is the derivative of the product $e^{-2x}y$, i.e., $(e^{-2x}y)'=(x^{2}+5)e^{-2x}$.

Step4: Integrate both sides

$\int(e^{-2x}y)'dx=\int(x^{2}+5)e^{-2x}dx$. First, find $\int x^{2}e^{-2x}dx$ using integration by parts twice. Let $u = x^{2},dv=e^{-2x}dx$, then $du = 2xdx,v=-\frac{1}{2}e^{-2x}$. $\int x^{2}e^{-2x}dx=-\frac{1}{2}x^{2}e^{-2x}+\int xe^{-2x}dx$. For $\int xe^{-2x}dx$, let $u = x,dv=e^{-2x}dx$, then $du=dx,v = -\frac{1}{2}e^{-2x}$, and $\int xe^{-2x}dx=-\frac{1}{2}xe^{-2x}+\frac{1}{2}\int e^{-2x}dx=-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C_1$. Also, $\int 5e^{-2x}dx=-\frac{5}{2}e^{-2x}+C_2$. So, $\int(x^{2}+5)e^{-2x}dx=-\frac{1}{2}x^{2}e^{-2x}-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}-\frac{5}{2}e^{-2x}+C=-\frac{1}{2}x^{2}e^{-2x}-\frac{1}{2}xe^{-2x}-\frac{11}{4}e^{-2x}+C$.
Since $\int(e^{-2x}y)'dx=e^{-2x}y$, we have $e^{-2x}y=-\frac{1}{2}x^{2}e^{-2x}-\frac{1}{2}xe^{-2x}-\frac{11}{4}e^{-2x}+C$.

Step5: Solve for y

$y =-\frac{1}{2}x^{2}-\frac{1}{2}x-\frac{11}{4}+Ce^{2x}$.

Answer:

$y =-\frac{1}{2}x^{2}-\frac{1}{2}x-\frac{11}{4}+Ce^{2x}$