# PDE by Laplace Transform

## Homework Statement

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

## 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

gabbagabbahey
Homework Helper
Gold Member
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

Thanks!

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

gabbagabbahey
Homework Helper
Gold Member
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

I finally solved it! Thanks very much!