ODE by Laplace transform

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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


[itex]L(xy'') = -\frac{dL(y'')}{ds}[/itex]

[itex]L(4xy) = -\frac{4dL(y)}{ds}[/itex]

[itex]L(y'') = s²L(y) - sy(0) - y'(0) = s²L(y) -3s[/itex]

[itex]L(y') = sL(y) - sy(0) - y(0) = sL(y) - 3[/itex]

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

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

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

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

[itex]\frac{dL(y)}{L(y)} = \frac{sds}{s²+4}[/itex]

Integrating both sides

[itex]ln(L(y)) = \frac{ln(s²+4)}{2} + c[/itex]

[itex]L(y) = c\sqrt{s²+4}[/itex]

which won't lead me to the right answer.

I realized that if at (1) I use [itex]\frac{L(y)(s² + 4)}{ds} + sL(y) =0[/itex] 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
 
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Answers and Replies

  • #2
Samy_A
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xy'' + y' + 4xy = 0, y(0) = 3, y'(0) = 0

[itex]L(xy'') = -\frac{dL(y'')}{ds}[/itex]

[itex]L(4xy) = -\frac{4dL(y)}{ds}[/itex]

[itex]L(y'') = s²L(y) - sy(0) - y'(0) = s²L(y) -3s[/itex]

[itex]L(y') = sL(y) - sy(0) - y(0) = sL(y) - 3[/itex]

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

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

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

[itex]-\frac{L(y)(s²+4)}{ds} + sL(y) =0[/itex] (1)

[itex]\frac{dL(y)}{L(y)} = \frac{sds}{s²+4}[/itex]

Integrating both sides

[itex]ln(L(y)) = \frac{ln(s²+4)}{2} + c[/itex]

[itex]L(y) = c\sqrt{s²+4}[/itex]

which won't lead me to the right answer.

I realized that if at (1) I use [itex]\frac{L(y)(s² + 4)}{ds} + sL(y) =0[/itex] 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:
[itex]-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0[/itex]
to
[itex]-\frac{s²L(y)}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]

Check what you did with the ##-\frac{d(s²L(y))}{ds}## term.
 
  • #3
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Something went wrong from:
[itex]-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0[/itex]
to
[itex]-\frac{s²L(y)}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]

Check what you did with the ##-\frac{d(s²L(y))}{ds}## term.
It's supposed to be
[itex]-\frac{d(s²L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]
fixed it on the original post.
 
  • #4
Samy_A
Science Advisor
Homework Helper
1,242
510
It's supposed to be
[itex]-\frac{d(s²L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]
fixed it on the original post.
Ok, so now continue your derivation from that.
 
  • #5
11
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Ok, so now continue your derivation from that.
[itex]-\frac{d(s²L(y))}{ds} = -2sL(y) - \frac{s²d(L(y))}{ds}[/itex]
[itex]-2sL(y) - \frac{s²d(L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]
[itex] - \frac{s²d(L(y))}{ds} - sL(y) -\frac{4dL(y)}{ds} =0[/itex]
[itex] \frac{s²d(L(y))}{ds} + sL(y) +\frac{4dL(y)}{ds} =0[/itex]
[itex] \frac{d(L(y))(s²+4)}{ds} + sL(y) =0[/itex]

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

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