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

• roughwinds
In summary: After thinking about it more, I realized that s² should be included in the derivative and re-did the equation.
roughwinds
Member warned about posting without the homework template

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

$ln(L(y)) = \frac{ln(s²+4)}{2} + c$

$L(y) = c\sqrt{s²+4}$

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

Last edited:
roughwinds said:
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

$ln(L(y)) = \frac{ln(s²+4)}{2} + c$

$L(y) = c\sqrt{s²+4}$

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
Samy_A said:
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.

roughwinds said:
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
Samy_A said:
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.

1. What is the Laplace transform?

The Laplace transform is a mathematical operation that converts a function of time into a function of complex frequency. It is commonly used in solving ordinary differential equations (ODEs) and is particularly useful for analyzing systems with inputs and outputs.

2. How does the Laplace transform work?

The Laplace transform involves integrating a function of time with a complex exponential function. This essentially translates the function from the time domain to the frequency domain. The inverse Laplace transform can then be used to convert the function back to the time domain.

3. What is the relationship between the Laplace transform and ODEs?

The Laplace transform is often used to solve ODEs because it converts the differential equation into an algebraic equation, which is typically easier to solve. This method is particularly useful for linear differential equations with constant coefficients.

4. What are the advantages of using the Laplace transform to solve ODEs?

One of the main advantages of using the Laplace transform is that it can handle a wide range of initial conditions and boundary conditions, making it a versatile tool for solving ODEs. It also allows for the use of techniques from complex analysis, which can simplify the solution process.

5. Are there any limitations to using the Laplace transform for ODEs?

While the Laplace transform is a powerful tool, it may not always be the best method for solving ODEs. In some cases, the inverse Laplace transform may be difficult to compute, and the solution may not be easily interpretable in the time domain. Additionally, the Laplace transform may not work for nonlinear or time-varying systems.

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