
#1
Nov1512, 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 :) 



#2
Nov1512, 07:45 AM

Math
Emeritus
Sci Advisor
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 e^{c}. 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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