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How does one evaluate the Laplace of functions like $t^2\frac{d^2y}{dt^2}$ ?

I wanted to solve a differential equation using Laplace Transform resembling: $$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 5$$

MATLAB Provides me the answer as :

C6*cos(log(t)) + C5*sin(log(t)) + 5

Can someone give me a derivation for this?

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2 Answers 2

up vote 1 down vote accepted

Use ${\cal L} \{ t^n f(t)\}=(-1)^n {d^n\over ds^n} F(s)$ and ${\cal L} \{ f''(t)\}=s^2F(s)-sf(0)-f'(0)$.

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So, $x^2\frac{d^2y}{dx^2}$ would be $\frac{d^2(s^2F(s) - sf(0) -f'(0))}{ds^2}$ ? I did this, I seemed to have done something silly then. I'll recheck. –  Inquest Feb 7 '12 at 15:06
    
@Nunoxic I don't think Laplace transforms are the way to proceed here. I would use Jon's approach and then use Laplace transforms to solve the equation he obtains. –  David Mitra Feb 7 '12 at 15:52
    
@Nunoxic BTW, the associated homogeneous form of your equation is a Cauchy-Euler equation –  David Mitra Feb 7 '12 at 15:57
    
I am just fiddling around. I don't really need to solve the equation using LT but was wondering if I could. –  Inquest Feb 7 '12 at 15:59

Take a new variable $x=e^t$.Then,

$$\frac{d}{dx}=\frac{d}{dt}\frac{dt}{dx}=\frac{d}{dt}e^{-t}$$

and

$$\frac{d^2}{dx^2}=\frac{d^2}{dt^2}e^{-2t}$$

and so your equation just becomes

$$\frac{d^2y}{dt^2}+\frac{dy}{dt}+y=5$$

that can be solved by combinations of sine and cosine.

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How does Laplace come into the picture? I necessarily need to solve it using Laplace. –  Inquest Feb 7 '12 at 15:10
    
Yes, you can use Laplace on the last equation. But, I think that the hint by David could be more helpful. –  Jon Feb 7 '12 at 15:18

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