Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz3 => Topic started by: Victor Ivrii on October 12, 2018, 06:04:55 PM

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

in the attachment

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
$$(\frac{1}{t} g)'=e^{t}$$
$$\int{(\frac{1}{t} g)'}=\int{e^{t}}dt$$
$$\frac{1}{t} g=e^{t}+c$$
$$g(t)=te^{t}+ct.$$

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