Question

verify the solution of the differential equation
solution: y=C1 e^(-x)cosx + C2 e^(-x)sinx
Differential equation: y'+y'+2y=0

Answer 1

Take the first and second derivatives of\[y = C_1 e^{-x}\cos x + C_2e^{-x}\sin x\]and plug them into the diffyq:

\[y'' + y' + 2y = 0\]

Answer 2

Are you positive you've copied the problem correctly? I think there's already one mistake in the differential equation...

Answer 3

I'm confusing between e^(-x) , cosx, and sinx I don't know how to differential

Answer 4

Before we worry about that, can you verify that you've typed the problem exactly correctly?
I'm suspicious that you wrote y' twice, for starters.

Answer 5

ohhhh yeahhh..it's a y''+y'....

Answer 6

Yes..I know how to verify that...but just got stuck about diff of e^(-x)cosx

Answer 7

Just apply the product rule.

\[(e^{-x}cosx)'=(e^{-x})'cosx+e^{-x}(cosx)'\]

Answer 8

I still don't think you have the correct differential equation, or maybe the correct solution there.

Answer 9

it was correctly!!!

Answer 10

http://www.wolframalpha.com/input/?i=y%27%27%2By%27%2B2y%3D0&t=crmtb01

Answer 11

Yeah, I'm still not convinced.

Here's what I think it should be: http://www.wolframalpha.com/input/?i=y%27%27%2By%27%2B2y%3D0&t=crmtb01

Answer 12

("yeah, I'm still not convinced" was aimed at @Miamy)

Answer 13

don't worry about it...thank ^^

Answer 14

Having Mathematica do the differentiation to eliminate any question of error on my part:

\[y = C_1 e^{-x}\cos (x) + C_2 e^{-x}\sin (x)\]\[y' = -C_1 e^{-x} \sin (x)-C_2 e^{-x} \sin (x)+C_1 \left(-e^{-x}\right) \cos (x)+C_2 e^{-x} \cos (x)\]\[y'' = 2 C_1 e^{-x} \sin (x)-2 C_2 e^{-x} \cos (x)\]

\[y'' + y' + 2y = 0\]\[(2 C_1 e^{-x} \sin (x)-2 C_2 e^{-x} \cos (x))+ \]\[(-C_1 e^{-x} \sin (x)-C_2 e^{-x} \sin (x)+C_1 \left(-e^{-x}\right) \cos (x)+C_2 e^{-x} \cos (x)) +\]\[ 2(C_1 e^{-x}\cos (x) + C_2 e^{-x}\sin (x)) = 0\]\[e^{-x}((2C_1 -C_1-C_2+2C_2)\sin(x)+(-2C_2-C_1+C_2+2C_1)\cos(x) )=0\]\[e^{-x} \left(C_1 (\sin (x)+\cos (x))+C_2 (\sin (x)-\cos (x))\right)=0\]

I think it is pretty self-evident that that last equation does not simplify to 0 on the left hand side.

Answer 15

Now, if we instead use \[y'' + 2y' + 2y =0\]
\[(2 C_1 e^{-x} \sin (x)-2 C_2 e^{-x} \cos (x))+\]\[
+2( -C_1 e^{-x} \sin (x)-C_2 e^{-x} \sin (x)+C_1 \left(-e^{-x}\right) \cos (x)+C_2 e^{-x} \cos (x))+\]\[
2( C_1 e^{-x}\cos (x) + C_2 e^{-x}\sin (x))=0\]
\[e^{-x}(( 2C_1-2C_1-2C_2+2C_2)\sin(x) + (-2C_2-2C_1+2C_2+2C_1)\cos(x)) = 0 \]\[e^{-x}(0\sin(x) + 0\cos(x)) = 0\]\[e^{-x}(0) = 0\checkmark\]

Answer 16

thank you so much ^^