Author Topic: TUT0303 Quiz1  (Read 446 times)

Yiyang Huang

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TUT0303 Quiz1
« on: September 27, 2019, 02:00:01 PM »
Find the general solution of the equation
$$
y^{\prime}-y=2 t e^{2 t}, \quad y(0)=1
$$
$$
p(t)=-1
$$
$$
\begin{aligned} \mu &=e^{\int p(t) d t} \\ &=e^{\int-1 d t} \\ &=e^{-t} \end{aligned}
$$
$$
\begin{aligned} e^{-t} y^{\prime}-e^{-t} y &=2 t e^{t} \\\left(e^{-t} y\right)^{\prime} &=2 t e^{t} \\ e^{-t} y &=\int 2 t e^{t} d t \end{aligned}
$$
$$
\begin{array}{ll}{u=2 t} & {v=\int d v=\int e^{t} d t=e^{t}} \\ {d u=2} & {d v=e^{t}}\end{array}
$$
$$
e^{-t} y=2 t e^{t}-\int e^{t} \cdot 2 d t\text{(integration by parts)}
$$
$$
\begin{aligned} e^{-t} y &=2 t e^{t}-2 e^t+c \\ &=2 e^{t}(t-1)+c \\ y &=2 e^{2 t}(t-1)+C e^{t} \end{aligned}
$$
$$\text{substitute}\quad  y(0)=1$$
$$
\begin{aligned} 1 &=2 e^{0}(0-1)+c e^{0} \\ 1 &=-2+c \\ & c=3 \end{aligned}
$$
$$
y=2 e^{2 t}(t-1)+3 e^{t}
$$