# Nontrivial Solution/Laplace Transform

• Dec 13th 2011, 08:48 AM
chaosier
Nontrivial Solution/Laplace Transform
So on this practice exam paper problem #5 (page 7) http://www.math.lsa.umich.edu/course...inalW10sol.pdf

I am told to find the Laplace Transform of:
tx''-3x'+x=0 where x(0)=0

Now I've always transformed tx'' in the following way:

Transform of tx'' -> -[(s^2)*X(s)-x'(0)]'
So I would get -[(2*s*X(s))+((s^2)*X'(s)))]

But the solution gives tx'' -> -(s*X(s)-x(0))'
Which gives -[(s*X'(s))+x(s)]

Why?

Even on my textbook: Differential Equations: Computing and Modelling by Edwards & Penney they are using the method that I have always used.
• Dec 13th 2011, 12:18 PM
FernandoRevilla
Re: Nontrivial Solution/Laplace Transform
Quote:

Originally Posted by chaosier
Transform of tx'' -> -[(s^2)*X(s)-x'(0)]'
So I would get -[(2*s*X(s))+((s^2)*X'(s)))]

You are right. Indeed:

$\mathcal{L}\{tx''(t)\}=-\frac{d}{ds}\mathcal{L}\{x''(t)\}=-\frac{d}{ds}(s^2X(s)-sx(0)-x''(0))=$

$-\frac{d}{ds}(s^2X(s)-x''(0))=-(2sX(s)+s^2X'(s))$
• Dec 13th 2011, 01:15 PM
chisigma
Re: Nontrivial Solution/Laplace Transform
Quote:

Originally Posted by chaosier
So on this practice exam paper problem #5 (page 7) http://www.math.lsa.umich.edu/course...inalW10sol.pdf

I am told to find the Laplace Transform of:
tx''-3x'+x=0 where x(0)=0

Now I've always transformed tx'' in the following way:

Transform of tx'' -> -[(s^2)*X(s)-x'(0)]'
So I would get -[(2*s*X(s))+((s^2)*X'(s)))]

But the solution gives tx'' -> -(s*X(s)-x(0))'
Which gives -[(s*X'(s))+x(s)]

Why?

Even on my textbook: Differential Equations: Computing and Modelling by Edwards & Penney they are using the method that I have always used.

Starting from the second order DE...

$t\ x^{''} -3\ x^{'} + x=0,\ x(0)=0$ (1)

... taking into account that is...

$\mathcal{L}\{t\ x^{''}\}= - \frac{d}{d s} \{s^{2}\ X(s) -s\ x(0) - x^{'}(0)\} = -s^{2}\ X^{'} (s)-2\ s\ X(s)$ (2)

... You arrive to write...

$s^{2}\ X^{'} +(5\ s -1)\ X=0$ (3)

Now You can find X(s) solving the first order DE (3), which is linear and relatively 'comfortable'...

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Marry Christmas from Serbia

$\chi$ $\sigma$