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PDE by Laplace Transform 
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#1
May2209, 09:03 AM

P: 11

1. The problem statement, all variables and given/known data
Use the Laplace Transform to solve the PDE for u(x,t) with x>0 and t>0: x(du/dx) + du/dt = xt with IC: u(x,0) = 0 and BC: u(0,t) = 0 2. Relevant equations 3. The attempt at a solution After taking LT of the PDE wrt t, the PDE becomes x(dU/dx) + sU = x/(s^{2}) Integrating factor : I = exp([tex]\int(s/x)dx[/tex]) = x^{s} ODE becomes d/dx(Ux^{s}) = x^{s}/s^{2} Integrating both sides: U = x/(s^{3}+s) + A(s)/x^{s} then I don't know how to find A(s), if I use BC, the factor 1/0 will come out...or is there some other way to calculate the PDE with LT? thanks 


#3
May2309, 12:39 AM

P: 11

Thanks!



#4
May2409, 01:00 AM

P: 11

PDE by Laplace Transform
so U(x,s) = x/(s^{3}+s^{2})
but then I don't know how to do the inverse LT to get u(x,t) such that it fits the PDE... thanks 


#5
May2409, 01:25 AM

HW Helper
P: 5,004

[tex]\frac{1}{s^3+s}=\frac{1}{s(s^2+1)}=\frac{A}{s}+\frac{Bs}{s^2+1}+\frac{C }{s^2+1}[/tex] Solve for A,B and C 


#6
May2409, 01:47 AM

P: 11

I finally solved it! Thanks very much!



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