MAT244--2019F > Quiz-4

TUT0303 Quiz4

(1/1)

Yiyang Huang:
Find the general solution for equation
$$y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t}$$

It's a non-homogeneous DE. To find complimentry solution,
$$y^{\prime \prime}+2 y^{\prime}+y=0$$

Assume that $y=e^{r t}$ is the solution of this eq
Then the characteristic equation is

$$\begin{array}{l}{r^{2}+2 r+1=0} \\ {r_{1}=r_{2}=-1 \quad \text { (repeated roots) }}\end{array}$$

Hence, $y_{c}(t)=c_{1} e^{-t}+c_{2} t e^{-t}$

To find the particular solution

y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t}

we assume
$$y_{p}(t)=A t^{2} e^{-t}$$

Then,
$$\begin{array}{l}{y_{p}^{\prime}(t)=2 A t e^{-t}-A t^{2} e^{-t}} \\ {y_{p}^{\prime \prime}(t)=2 A e^{-t}-4 A t e^{-t}+A t^{2} e^{-t}}\end{array}$$

substitute $y_{p}, y_{p}^{\prime}, y_{p}^{\prime \prime}$ into equation ( 1)

we get
\begin{aligned} 2 A e^{-t} &=2 e^{-t} \\ A &=1 \end{aligned}

so $y_{p}(t)=t^{2} e^{-t}$

Hence the general solution is
$$y=y_{c}(t)+y_{p}(t)=c_{1} e^{-t}+c_{2} t e^{-t}+t^{2} e^{-t}$$