- #1
ktklam9
- 3
- 0
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 :)
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 :)