Question

Verify that y1=e^(2x) and y2=e^(5x) are solutions of y"-7y'+10y=0 on (-infinity, infinity).

Answer 1

just find the derivative (1st and 2nd) (by using the chain for exponent)
and plug the solutions in

Answer 2

\(\large\color{black}{ \displaystyle y_1=e^{2x}}\)

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tell me what is:

1. \(\large\color{black}{ \displaystyle \frac{d}{dx}\left( e^{2x}\right)}\)


2. \(\large\color{black}{ \displaystyle \frac{d^2}{dx^2}\left( e^{2x}\right)}\)
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Answer 3

I got it!

Answer 4

oh, very nice:)

Answer 5

Thanks a lot!

Answer 6

and in any case, you just need to plug those (simple) derivatives into your differential equation

Answer 7

yw

Answer 8

I love e^(something) btw:)