## How can I solve this without using integrating factor?

Let the Wronskian between the functions f and g to be 3e$^{4t}$, if f(t) = e$^{2t}$, then what is g(t)?

So the Wronskian setup is pretty easy

W(t) = fg' - f'g = 3e$^{4t}$

f = e$^{2t}$
f' = 2e$^{2t}$

So plugging it in I would get:

e$^{2t}$g' - 2e$^{2t}$g = 3e$^{4t}$

Which results in

g' - 2g = 3e$^{2t}$

How can I solve for g without using integrating factor? Is it even possible? Thanks :)

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Recognitions: Gold Member Science Advisor Staff Emeritus It's always possible to solve "some other way" but often much more difficult. This particular example, however, is a "linear equation with constant coefficients" which has a fairly simple solution method. Because it is linear, we can add two solutions to get a third so start by looking at g'- 2g= 0. g'= 2g give dg/g= 2dt and, integrating ln(g)= 2t+ c. Taking the exponential of both sides, $g(t)= e^{2t+ c}= e^{2t}e^c= Ce^{2t}$ where C is defined as ec. Now, we can use a method called "variation of parameters" because we allow that "C" in the previous solution to be a variable: let $g= v(t)e^{2t}$. Then $g'= v'(t)e^{2t}+ 2v(t)e^{2t}$ so the equation becomes $g'- 2g= v'(t)e^{2t}+ 2v(t)e^{2t}- 2v(t)e^{2t}= v'(t)e^{2t}= 3e^{2t}$. We can cancel the "$e^{2t}$" terms to get $v'(t)= 3$ and, integrating, v(t)= 3t+ C. That gives the solution $g(t)= v(t)e^{2t}= 3te^{2t}+ Ce^{2t}$ where "C" can be any number.