Author Topic: TUT0402 quiz4  (Read 544 times)

Linqian Shen

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TUT0402 quiz4
« on: October 18, 2019, 10:18:57 PM »
Find the general solution of the given differential equation
$$
14 \cdot y^{\prime \prime}-y^{\prime}-2 y=\cosh (2 t)=\frac{1}{2} e^{2 t}+\frac{1}{2} e^{-2 t}
$$

$$
\begin{array}{c}{r^{2}-r-2=0} \\ {(r+1)(r-2)=0} \\ {r_{1}=-1 \quad r_{2}=2}\end{array}
$$
$$
y=c_{1} e^{-t}+c_{2} e^{2 t}
$$

$$
y_{1}(t)=A e^{2 t} $$
$$y_{1}(t)=A t e^{2 t}
$$
$$
\begin{array}{l}{y^{\prime}(t)=2 A t e^{2 t}+A e^{2 t}} \\ {y^{\prime \prime}(t)=4 A t e^{2 t}+2 A e^{2 t}+2 A e^{2 t}} \\ {4 A t e^{2 t}+2 A e^{2 t}+2 A e^{2 t}-2 A t e^{2 t}-A e^{2 t}-2 A t e^{2 t}=\frac{1}{2} e^{2 t}} \\ {e^{2 t}(4 A t+2 A+2 A-2 A t-A-2 A+)=\frac{1}{2} e^{2 t}}\end{array}
$$
$$
\begin{array}{l}{3 A=\frac{1}{2} \quad A=\frac{1}{6}} \\ {y_{1}(t)=\frac{1}{6}+e^{2 t}}\end{array}
$$

$$
\begin{array}{l}{y_{2}(t)=B e^{-2 t} \quad y_{2}^{\prime}(t)=-2 B e^{-2 t}} \\ {y_{2}^{\prime \prime}(t)=4 B e^{-2 t}}\end{array}
$$
$$
\begin{array}{c}{4 B e^{-2 t}+2 B e^{-2 t}-2 B e^{2 t}=\frac{1}{2} e^{-2 t}} \\ {4 B e^{-2 t}=\frac{1}{2} e^{-2 t}} \\ {4 B=\frac{1}{2}}\end{array}
$$
$$
B=\frac{1}{8} \quad y_{2}(t)=\frac{1}{8} e^{-2 t}
$$
$$
y(t)=c_{1} e^{-t}+c_{2} e^{2 t}+\frac{1}{6} t e^{2 t}+\frac{1}{8} e^{-2 t}
$$