Author Topic: tUT5103 Quiz4  (Read 453 times)

Yingyingz

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tUT5103 Quiz4
« on: October 18, 2019, 02:00:01 PM »
$
\text{14.}\quad y^{\prime \prime}+4 y^{\prime}+4 y=0, y(-1)=2, y^{\prime}(-1)=1
$

The characteristic equation is,
$$
\begin{aligned} r^{2}+4 r+4 &=0 \\(r+2)^{2} &=0 \\ r_{1}=r_{2} &=-2 \end{aligned}
$$

since it has repeated roots, the general solution is:
$$
\begin{array}{l}{y(t)=c_{1} e^{-2 t}+c_{2} t e^{-2 t}} \\ {y^{\prime}(t)=-2 c_{1} e^{-2 t}+c_{2} e^{-2 t}-2 c_{2} t e^{-2 t}}\end{array}
$$

$\operatorname{plug}$ in $y(-1)=2, \quad y^{\prime}(-1)=1$

we get $$\left\{\begin{array}{l}{c_{1} e^{2}-c_{2} e^{2}=2} \\ {-2 c_{1} e^{2}+c_{2} e^{2}+2 c_{2} e^{2}=1}\end{array}\right.$$

$$
\therefore\left\{\begin{array}{l}{c_{1}=7 e^{-2}} \\ {c_{2}=5 e^{-2}}\end{array}\right.
$$

substitute $C_{2}$ and $C_{2}$ into $y(t)$

$$
=7 e^{-2} e^{-2 t}+5 e^{-2} t e^{-2 t}
$$

$\therefore$ The general solution is $y(t)=7 e^{-2(t+1)}+5 t e^{-2(t+1)}$