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)?(adsbygoogle = window.adsbygoogle || []).push({});

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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# How can I solve this without using integrating factor?

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