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

Question

I am not sure how to complete this problem, I tried to solve it but I am lost.

Thanks in advance for the help!

Answer 1

Here's a solution,

Given, $W(f,g)=fg'-gf'=t^2e^t$ and $f(t)=t$

dividing $W$ by square of $f$ we obtain,

$$\frac{W(f,g)}{f^2}=\frac{t^2e^t}{t^2}$$ or, $$\frac{fg'-gf'}{f^2}=e^t$$ or, $$\frac{d}{dt}(\frac{g}{f})=e^t$$ Integrating, $$\frac{g}{f}=e^t+C$$ $C$ being constant of integration,

So, $$g=te^{t}+Ct$$ as, $f(t)=t$

Answer 2

We have $$ \begin{vmatrix}t & g(t)\\ 1 & g'(t)\end{vmatrix} = tg'(t) - g(t) = t^2e^t,\tag1 $$ which is a linear first order differential equation. Writing $(1)$ as $g'(t)-t^{-1}g(t) = te^t$, we compute the integrating factor $$ \mu(t) = e^{\int -t^{-1} \ \mathsf dt} = e^{-\log t} = t^{-1}. $$ It follows that $$ [t^{-1}g'(t)]' = [e^t]' $$ and hence $$ g(t) = te^t + Ct. $$