Solving ODE by Laplace Transform: Where Did I Go Wrong?

[roughwinds]

Homework Statement

Use Laplace transform to solve the following ODE

Homework Equations

xy'' + y' + 4xy = 0, y(0) = 3, y'(0) = 0

The Attempt at a Solution

L(xy'') = -\frac{dL(y'')}{ds}

L(4xy) = -\frac{4dL(y)}{ds}

L(y'') = s²L(y) - sy(0) - y'(0) = s²L(y) -3s

L(y') = sL(y) - sy(0) - y(0) = sL(y) - 3

-\frac{d(s²L(y)-3s)}{ds} + sL(y)-\frac{4dL(y)}{ds} =0

-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0

-\frac{d(s²L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0

-\frac{dL(y)(s²+4)}{ds} + sL(y) =0 (1)

\frac{dL(y)}{L(y)} = \frac{sds}{s²+4}

Integrating both sides

ln(L(y)) = \frac{ln(s²+4)}{2} + c

L(y) = c\sqrt{s²+4}

which won't lead me to the right answer.

I realized that if at (1) I use \frac{L(y)(s² + 4)}{ds} + sL(y) =0 instead I'll reach the right answer according to wolfram, but I can't figure out what I'm doing wrong to end up with that negative sign.

http://www.wolframalpha.com/input/?i=xy''+++y'+++4xy+=+0,+y(0)+=+3,+y'(0)+=+0

 

[Samy_A]

xy'' + y' + 4xy = 0, y(0) = 3, y'(0) = 0

L(xy'') = -\frac{dL(y'')}{ds}

L(4xy) = -\frac{4dL(y)}{ds}

L(y'') = s²L(y) - sy(0) - y'(0) = s²L(y) -3s

L(y') = sL(y) - sy(0) - y(0) = sL(y) - 3

-\frac{d(s²L(y)-3s)}{ds} + sL(y)-\frac{4dL(y)}{ds} =0

-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0

-\frac{s²L(y)}{ds} + sL(y) -\frac{4dL(y)}{ds} =0

-\frac{L(y)(s²+4)}{ds} + sL(y) =0 (1)

\frac{dL(y)}{L(y)} = \frac{sds}{s²+4}

Integrating both sides

ln(L(y)) = \frac{ln(s²+4)}{2} + c

L(y) = c\sqrt{s²+4}

which won't lead me to the right answer.

I realized that if at (1) I use \frac{L(y)(s² + 4)}{ds} + sL(y) =0 instead I'll reach the right answer according to wolfram, but I can't figure out what I'm doing wrong to end up with that negative sign.

http://www.wolframalpha.com/input/?i=xy''+++y'+++4xy+=+0,+y(0)+=+3,+y'(0)+=+0

Something went wrong from:
-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0
to
-\frac{s²L(y)}{ds} + sL(y) -\frac{4dL(y)}{ds} =0

Check what you did with the $-\frac{d(s²L(y))}{ds}$ term.

 

[roughwinds]

Something went wrong from:
-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0
to
-\frac{s²L(y)}{ds} + sL(y) -\frac{4dL(y)}{ds} =0

Check what you did with the $-\frac{d(s²L(y))}{ds}$ term.

It's supposed to be
-\frac{d(s²L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0
fixed it on the original post.

 

[Samy_A]

It's supposed to be
-\frac{d(s²L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0
fixed it on the original post.

Ok, so now continue your derivation from that.

 

[roughwinds]

Ok, so now continue your derivation from that.

-\frac{d(s²L(y))}{ds} = -2sL(y) - \frac{s²d(L(y))}{ds}
-2sL(y) - \frac{s²d(L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0
- \frac{s²d(L(y))}{ds} - sL(y) -\frac{4dL(y)}{ds} =0
\frac{s²d(L(y))}{ds} + sL(y) +\frac{4dL(y)}{ds} =0
\frac{d(L(y))(s²+4)}{ds} + sL(y) =0

Thanks, I initially solved as if s² wasn't part of the derivative.