Q3 TUT0401

[Victor Ivrii]

If the Wronskian $W$ of $f$ and $g$ is $t^2e^t$ , and if $f(t)=t$, find $g(t)$.

[Yunqi(Yuki) Huang]

in the attachment

[Monika Dydynski]

If the Wronskian $W$ of $f$ and $g$ is $t^{2}e^{t}$, and if $f(t)=t$, find $g(t)$.


Suppose that $W(f,g)=t^{2}e^{t}$ and $f(t)=t \Rightarrow f'(t)=1$

Then from $W(f,g)=fg'-gf'$, we get a first order DE

 $$tg'-g\cdot 1=t^{2}e^{t}\tag{1}$$

Dividing both sides of $(1)$ by $t$ and multiplying by integrating factor, $\mu(t)=e^{\int{p(t)}dt}=\frac{1}{t}$, we  have


$$(\frac{1}{t} g)'=e^{t}$$


$$\int{(\frac{1}{t} g)'}=\int{e^{t}}dt$$

$$\frac{1}{t} g=e^{t}+c$$

$$g(t)=te^{t}+ct.$$

[Yunqi(Yuki) Huang]

$$
f(t)=t
$$
$$
So f'(t)=1
$$
$$
W=tg'(t)-g(t)=t^2e^t
$$
$$
g'(t)-\frac{1}{t}g(t)=te^t
$$
$$
p(t)=-\frac{1}{t}
$$
$$
u(t)=e^{\int1p(t)dt}
$$
$$
wherep(t)=-\frac{1}{t}
$$
$$
u(t)=\frac{1}{t}
$$
$$
\frac{1}{t}g(t)=\int e^tdt
$$
$$
g(t)=te^t+ct$$where let c=1
$$
$$
Thus g(t)=te^t+t