PDE by Laplace Transform

by freesnow
Tags: laplace, transform
 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/(s2) Integrating factor : I = exp($$\int(s/x)dx$$) = xs ODE becomes d/dx(Uxs) = xs/s2 Integrating both sides: U = x/(s3+s) + A(s)/xs 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
HW Helper
P: 5,004
 Quote by freesnow Integrating both sides: U = x/(s3+s) + A(s)/xs then I don't know how to find A(s), if I use BC, the factor 1/0 will come out...
Unless A(s)=0
 P: 11 Thanks!
P: 11

PDE by Laplace Transform

so U(x,s) = x/(s3+s2)

but then I don't know how to do the inverse LT to get u(x,t) such that it fits the PDE...

thanks
HW Helper
P: 5,004
 Quote by freesnow so U(x,s) = x/(s3+s2) but then I don't know how to do the inverse LT to get u(x,t) such that it fits the PDE... thanks
Just use partial fractions:

$$\frac{1}{s^3+s}=\frac{1}{s(s^2+1)}=\frac{A}{s}+\frac{Bs}{s^2+1}+\frac{C }{s^2+1}$$

Solve for A,B and C
 P: 11 I finally solved it! Thanks very much!

 Related Discussions Calculus & Beyond Homework 1 Engineering, Comp Sci, & Technology Homework 4 Calculus & Beyond Homework 2 Introductory Physics Homework 6 General Physics 1