Second-Order Differential Equation Help.

[Finhum]

Can you find functions p(x) and q(x) so that
e^x and sin(x) both solve the differencial equation:
y '' + p(x) y ' + q(x) y = 0 ?

Can anyone help me out with this?

 

[DrPhil]

Can you find functions p(x) and q(x) so that
e^x and sin(x) both solve the differencial equation:
y '' + p(x) y ' + q(x) y = 0 ?

Can anyone help me out with this?

We need to see your work! Don't know where you are getting stuck.

Have you set y(x) = A e^x + B sin(x) + C, and plugged that into the diff.eq.?

 

[Finhum]

I haven't really done any work, my problem was with how to start this problem.

So, taking your advice, (Thank you very much by the way.)

y(x) = Ae^x = Bsin(x) + c

y'' + p(x)y' + q(x)y = 0
(Ae^x - Bsin(x)) + p(x)(Ae^x + Bcos(x)) + q(x)(Ae^x + Bsin(x) + c) = 0

That's what I got,

i'm guessing it would not be too smart to distribute the q(x) & p(x) terms.
What would be the next step?

 

[DrPhil]

I haven't really done any work, my problem was with how to start this problem.

So, taking your advice, (Thank you very much by the way.)

y(x) = Ae^x + Bsin(x) + c

y'' + p(x)y' + q(x)y = 0
(Ae^x - Bsin(x)) + p(x)(Ae^x + Bcos(x)) + q(x)(Ae^x + Bsin(x) + c) = 0

That's what I got,

i'm guessing it would not be too smart to distribute the q(x) & p(x) terms.
What would be the next step?

Separate the exponential and the trig terms - and note that we can drop the C term because it is not a solution of the equation.

\(\displaystyle \displaystyle A\ e^x\left(1 + p(x) + q(x)\right) + B\left(-\sin{x} + p(x)\cos{x} + q(x)\sin{x}\right) = 0\)

To be true for all x, I think it is necessary that both terms be independently zero. That is,

\(\displaystyle 1 + p(x) + q(x) =0\),...........\(\displaystyle -\sin{x} + p(x)\cos{x} + q(x)\sin{x} = 0\)

Play with that for a while. For the moment, my answer to the original question is "No." Is that an allowed answer, if you can prove it?