Author Topic: TUT0303 Quiz4  (Read 440 times)

Yiyang Huang

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TUT0303 Quiz4
« on: October 18, 2019, 02:00:05 PM »
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
\begin{equation}
y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t}
\end{equation}

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