PDE help, Green's Function

Hey Guys;

I'm solving PDE's with the use of Green's function where all the boundary conditions are homogeneous.
However, how do you solve ones in which we have non-homogeneous b.c's.

In case it helps, the particular PDE I'm looking at is:

[tex]y'' = -x^2[/tex]

[tex]y(0) + y'(0) = 4, y'(1)= 2 [/tex]

Thanks for any help.
 
That is an ODE, the general solution is

[tex]y = c_1 + c_2x - \frac{1}{12}x^4 [/tex]

The equations are:

[tex]y'(1) = c_2 - \frac{1}{3}(1)^3 = 1[/tex]

Therefore we have c_2 = 4/3. Therefore the second equation reads:

[tex]c_1 + c_2 = 4[/tex]

Therefore we have c _1 = 8/3.

Doing this problem with the method of Green functions, we begin with the piecewise solution c_1 + c_2x, x < x' , and b_1 + b_2 x, x > x'. The first boundary condition says that c_1 + c_2 = 4, and the second one says b_2 = 1. Continuity at x' yields c_1 + (4-c_1)x' = b_1 + x'. Finally, the jump condition on the first derivative yields y'(x')_right - y'(x')_left = 1 implies 1 - c_2 = 1, and so c_2 is zero, at which point I'm stuck too.
 
Hey Guys;

I'm solving PDE's with the use of Green's function where all the boundary conditions are homogeneous.
However, how do you solve ones in which we have non-homogeneous b.c's.

In case it helps, the particular PDE I'm looking at is:

[tex]y'' = -x^2[/tex]

[tex]y(0) + y'(0) = 4, y'(1)= 2 [/tex]

Thanks for any help.

Try making the substitution:
Y(x) = y(x) - 2x - 2.

Then
Y(0) = y(0) - 2 , Y'(0) = y'(0) - 2 and Y'(1) = y'(1) - 2.
These will make your bc homogeneous.
 
Thanks, that seems to work perfectly.
 

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