Show that y= (2/3)e^x + e^-2x is a solution of the differential equation y' + 2y=2e^x

Question

bahrom7893 · Answer

just take the derivative and substitute

anonymous · Answer

Is this correct for the derivative y'=2/3e^x+1/e^2?

bahrom7893 · Answer

wait is it:
e^(-2x) or [e^(-2)]*x

anonymous · Answer

its e^(-2x)

bahrom7893 · Answer

okay then the derivative was wrong..

anonymous · Answer

fml.

bahrom7893 · Answer

y= (2/3)e^x + e^-2x
y' = (2/3)e^x + e^(-2x) * (-2)= 2 [ (e^x/3) - e^(-2x) ]

anonymous · Answer

Lovely Stony Brook university textbook does not have the answer in the back.

anonymous · Answer

ahhh I forgot to do chain rule didnt I

bahrom7893 · Answer

not only that u just did the differentiation as if it were:
e^(-2) * x, but x is in the power.. (e^u)' = e^u (du/dx)

anonymous · Answer

bloody rusty at derivatives, last semester all we did was integrate hardly any deriving at all

anonymous · Answer

So im assuming after I plug that in I get 2e^x?

bahrom7893 · Answer

You should get 2e^x once u plug in y' and y where they belong.

anonymous · Answer

thanks man

bahrom7893 · Answer

np.. I remember how i was killing myself with ODEs last semester haha.. I'm killin myself with lin algebra now..

anonymous · Answer

Linear algebra will be after differential equations for me. Btw if you wouldnt mind can you attach a file or something to break down plugging in the y and the y'?