Question

If the Wronskian W of f and g is 3e^(4t) and if f(t)=e^(2t), find g(t).

Answer 1

Since the Wronskian is


`W(f,g)=fg'-gf'=3e^(4t)`

and


`f(t)=e^(2t)\text{ AND } f'(t)=2e^(2t)`

then


`g'e^(2t)-2ge^(2t)=3e^(4t)`

Divide by
`e^(2t)`


`g'-2g=3e^(2t)`

This is a nonhomogenous first order differential equation. We write this in the form


`g'+pg=h`

There is a really easy way to solve this. We introduce a dummy factor called the integrating factor. Let


`u'(t)=u(t)p \\ \\ug'+gu'=(gu)'=uh`

and


`g=(\int uh dt)/(u)`

The integrating factor is


`u(t)=e^(\int- 2dt+c) \\ \\u(t)=e^(-2t+c)`

Then,


`g(t)=(\int u(t)(3e^(2t))dt+k)/(u(t))=(\int e^(-2t+c)3e^(2t)dt+k)/(e^(-2t+c))=(3\int e^(c)dt+k)/(e^(-2t+c))=(3te^c+k)/(e^(-2t+c))`

So


`g(t)=3te^(2t)+ke^(2t)`