Question

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

Answer 1

Can you take the derivative of
\[y=\frac{2e}{3} + e^{-2x} \]
?

Answer 2

wouldn't it be: \[e^x - e^{-2x}\]

Answer 3

did you change the equation up there just now?

and if
\[y=\frac{2}{3} e^x + e^{-2x}\]
\[ y' = \frac{2}{3} e^x + -2e^{-2x}\]

Answer 4

Yes, I forgot the first x.

Answer 5

okay, phew, because that other equation was NOT working out ^_^

Answer 6

are you sure it's supposed to equal 2e^x?

Answer 7

yes. \[y' + 2y = 2e^x \]

Answer 8

oh, derp, I keep writing it down wrong :P

Answer 9

so it pretty much just falls out when you add y' + y ^_^

Answer 10

I got it! I was substituting y' for y on accident.

Answer 11

hooray diff eqs! ^_^