Constructing a 2nd order homogenous DE given fundamental solution

[sagamore4110]

Homework Statement

Given a set of fundamental solutions {e^x*sinx*cosx, e^x*cos(2x)}

Homework Equations

y''+p(x)y'+q(x)=0
det W(y₁,y₂) =Ce^-∫p(x)dx

The Attempt at a Solution

I took the determinant of the matrix to get
e^2x[cos(2x)cosxsinx-2sin(2x)sinxcosx-cos(2x)sinxcosx- cos²xcos(2x)+sin²xcos(2x)

Then using the identities sin²x+cos²x = 1 (for the last 2 terms) and sin(2x) = 2sinx*cosx (for the second term) and cancelling the 2 "cos(2x)cosxsinx" (first and third terms) I got
-e^2x(sin²x+cos(2x))

Setting this equal to Ce^-∫p(x)dx and trying to solve I got as far as
ln(-e^2x(sin²x+cos(2x))/C) = -∫p(x)dx

and now I'm a little bit stuck, I also don't know how to solve for q(x) here. Thanks for the help!

 

[Mark44]

Homework Statement

Given a set of fundamental solutions {e^x*sinx*cosx, e^x*cos(2x)}

Homework Equations

y''+p(x)y'+q(x)=0
det W(y₁,y₂) =Ce^-∫p(x)dx

The Attempt at a Solution

I took the determinant of the matrix to get
e^2x[cos(2x)cosxsinx-2sin(2x)sinxcosx-cos(2x)sinxcosx- cos²xcos(2x)+sin²xcos(2x)

Then using the identities sin²x+cos²x = 1 (for the last 2 terms) and sin(2x) = 2sinx*cosx (for the second term) and cancelling the 2 "cos(2x)cosxsinx" (first and third terms) I got
-e^2x(sin²x+cos(2x))

Setting this equal to Ce^-∫p(x)dx and trying to solve I got as far as
ln(-e^2x(sin²x+cos(2x))/C) = -∫p(x)dx

and now I'm a little bit stuck, I also don't know how to solve for q(x) here. Thanks for the help!

Your work seems like the long way around -- there's a much simpler way. Your fundamental set could also be $\{e^x\sin(2x), e^x\cos(2x) \}$. The fundamental set as given and this one both span identical function spaces, and are bases for the same space.

It's also helpful to recognize that yet another basis would suffice: $\{e^xe^{2ix}, e^xe^{-2ix} \} = \{e^{(1 + 2i)x}, e^{(1 - 2i)x} \}$. All three of these fundamental sets span the same solution space, and are bases for it.

Finally, if your fundamental set were $\{ e^{r_1t}, e^{r_2t}\}$, do you understand that $r_1$ and $r_2$ are solutions to the quadratic characteristic equation? If so, you can work backwards from the characteristic equation to the homogeneous diff. equation with very little work.

 

[sagamore4110]

Your work seems like the long way around -- there's a much simpler way. Your fundamental set could also be $\{e^x\sin(2x), e^x\cos(2x) \}$. The fundamental set as given and this one both span identical function spaces, and are bases for the same space.

It's also helpful to recognize that yet another basis would suffice: $\{e^xe^{2ix}, e^xe^{-2ix} \} = \{e^{(1 + 2i)x}, e^{(1 - 2i)x} \}$. All three of these fundamental sets span the same solution space, and are bases for it.

Finally, if your fundamental set were $\{ e^{r_1t}, e^{r_2t}\}$, do you understand that $r_1$ and $r_2$ are solutions to the quadratic characteristic equation? If so, you can work backwards from the characteristic equation to the homogeneous diff. equation with very little work.

I actually realized this shortly after posting the problem, but wouldn't sinx*cosx end up with the solution being 1/2*e^x*sin(2x)? That's the next part I got stuck on. I completed the problem using only 1+/-2i to get the quadratic in hopes for partial credit and got y''-2y'+3/2

 

[Mark44]

I actually realized this shortly after posting the problem, but wouldn't sinx*cosx end up with the solution being 1/2*e^x*sin(2x)?

Yes, it would, but it wouldn't matter. Constant multiples of the two original functions will still be a fundamental set (i.e., a basis for the solution space), so the constant 1/2 doesn't need to be included.

That's the next part I got stuck on. I completed the problem using only 1+/-2i to get the quadratic in hopes for partial credit and got y''-2y'+3/2

That's not the correct DE. Its characteristic equation would be $r^2 - 2r + 3/2 = 0$, and the roots of that equation aren't 1 +/- 2i.