Author Topic: TUT0801  (Read 304 times)

suyichen

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TUT0801
« on: October 11, 2019, 02:19:24 PM »
\noindent The wrouskian w of f and g is $3 e^{4 t}$,\\
and if $f(t)=e^{2 t}$, find $g(t)$.
$$
\begin{aligned} W(f, g) &=3 e^{4 t} \\ W(f, g) &=\left|\begin{array}{cc}{f(t)} & {g(t)} \\ {f^{\prime}(t)} & {g^{\prime}(t)}\end{array}\right| \\ &=\left|\begin{array}{cc}{e^{2 t}} & {g(t)} \\ {2 e^{2 t}} & {g^{\prime}(t)}\end{array}\right| \\ &=e^{2 t} g^{\prime}(t)-2 e^{2 t} g(t) \end{aligned}
$$
$$
\begin{array}{l}{\therefore e^{2 t} g^{\prime}(t)-2 e^{2 t} g(t)=3 e^{4 t}} \\ {\Rightarrow g^{\prime}(t)-2 g(t)=3 e^{2 t}}\end{array}
$$

We have $p(t)=-2$

$$
\mu=e^{\int p(t) d t}=e^{\int-2 d t}=e^{-2 t}
$$
$$
\therefore e^{-2 t} g^{\prime}(t)-2 e^{-2 t} g(t)=3
$$
$$
\begin{array}{l}
 {\Rightarrow \left(e^{-2 t} g(t)\right)^{\prime}=3}\\
 {\Rightarrow e^{-2 t} g(t)=\int 3 d t\qquad 3 t+c}\\
 {\Rightarrow g(t)=e^{2 t}(3 t+c)=3 e^{2 t} t+c e^{2 t}}
\end{array}
$$

check :
$$
g^{\prime}(t)=3 e^{2 t}+6 e^{2 t} t+2 c e^{2 t}
$$
$$
\begin{aligned} W(f, g) &=e^{2 t}\left(3 e^{2 t}+6 e^{2 t} t+2 c e^{2 t}\right)-2 e^{2 t}\left(3 e^{2 t} t+c e^{2 t}\right) \\ &=3 e^{4 t} \end{aligned}
$$

Therefore, $g(t)=3 t e^{2 t}+c e^{2 t}$