Thread: 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.

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:

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...

(1)

... taking into account that is...

(2)

... You arrive to write...

(3)

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




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