Question

Find the values of λ for which y = e^(λx) satisfies the equation y + y' = y''

Answer 1

So far, this is what I have:
e^(λx) + λe^(λx) = 2λe^(λx)

The only value of λ that makes sense to me here is 1. Is that the only correct answer?

Answer 2

I don't think 1 works actually.
y'' gives you,\[\huge \lambda^2e^{(\lambda x)}\]Doesn't it? :O not 2lambda.

Answer 3

Rearranging the problem reveals that it's a second order Differential Equation,
\[\large y''-y'-y=0\]If we plug in y and it's derivatives, we get an equation like this.\[\huge \lambda^2e^{(\lambda x)}-\lambda e^{(\lambda x)}-e^{(\lambda x)}=0\]

Answer 4

Try to recall that an exponential can never be 0. So we can safely divide both sides by e^(lambda x) without any worries.

Giving us the characteristic equation:\[\huge \lambda^2-\lambda-1=0\]From here, we can solve for lambda by simply using the Quadratic Formula! :)

Answer 5

Hey, thanks! I stepped away from my computer for a moment, but now I understand perfectly!