How can I solve this without using integrating factor?


by ktklam9
Tags: factor, integrating, solve
ktklam9
ktklam9 is offline
#1
Nov15-12, 01:17 AM
P: 3
Let the Wronskian between the functions f and g to be 3e[itex]^{4t}[/itex], if f(t) = e[itex]^{2t}[/itex], then what is g(t)?

So the Wronskian setup is pretty easy

W(t) = fg' - f'g = 3e[itex]^{4t}[/itex]

f = e[itex]^{2t}[/itex]
f' = 2e[itex]^{2t}[/itex]

So plugging it in I would get:

e[itex]^{2t}[/itex]g' - 2e[itex]^{2t}[/itex]g = 3e[itex]^{4t}[/itex]

Which results in

g' - 2g = 3e[itex]^{2t}[/itex]

How can I solve for g without using integrating factor? Is it even possible? Thanks :)
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HallsofIvy
HallsofIvy is offline
#2
Nov15-12, 07:45 AM
Math
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Thanks
PF Gold
P: 38,894
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, [itex]g(t)= e^{2t+ c}= e^{2t}e^c= Ce^{2t}[/itex] 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 [itex]g= v(t)e^{2t}[/itex]. Then [itex]g'= v'(t)e^{2t}+ 2v(t)e^{2t}[/itex] so the equation becomes [itex]g'- 2g= v'(t)e^{2t}+ 2v(t)e^{2t}- 2v(t)e^{2t}= v'(t)e^{2t}= 3e^{2t}[/itex]. We can cancel the "[itex]e^{2t}[/itex]" terms to get [itex]v'(t)= 3[/itex] and, integrating, v(t)= 3t+ C. That gives the solution [itex]g(t)= v(t)e^{2t}= 3te^{2t}+ Ce^{2t}[/itex] where "C" can be any number.


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