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Hi. I'm trying to solve the heat equation with the initial boundary conditions:

[tex]u(0,t)=f_1(t)[/tex]

[tex]u(x_1,t)=f_2(t)[/tex]

[tex]u(x,0)=f(x)[/tex]

[tex]0<x<x_1[/tex]

[tex]t>0[/tex]

And the heat equation: [tex]\frac{\partial u}{\partial t}-k\frac{\partial^2 u}{\partial x^2}=0[/tex]

So when I make separation of variables I get:

[tex]\nu=X(x)T(t)[/tex]

[tex]\frac{T'(t)}{T(t)}=k\frac{X''(x)}{X(x)}[/tex]

Then I have to solve for X

[tex]kX''(x)-\lambda X(x)=0[/tex]

With the initial boundary conditions

[tex]X(0)=f_1(t)[/tex]

[tex]X(x_1)=f_2(t)[/tex]

And for T:

[tex]T'(t)-\lambda T(t)=0[/tex]

With initial value:

[tex]T(0)=f(x)[/tex]

How should I proceed from here? I'm not sure how to make this accomplish the boundary conditions.

Bye, thanks.

[tex]u(0,t)=f_1(t)[/tex]

[tex]u(x_1,t)=f_2(t)[/tex]

[tex]u(x,0)=f(x)[/tex]

[tex]0<x<x_1[/tex]

[tex]t>0[/tex]

And the heat equation: [tex]\frac{\partial u}{\partial t}-k\frac{\partial^2 u}{\partial x^2}=0[/tex]

So when I make separation of variables I get:

[tex]\nu=X(x)T(t)[/tex]

[tex]\frac{T'(t)}{T(t)}=k\frac{X''(x)}{X(x)}[/tex]

Then I have to solve for X

[tex]kX''(x)-\lambda X(x)=0[/tex]

With the initial boundary conditions

[tex]X(0)=f_1(t)[/tex]

[tex]X(x_1)=f_2(t)[/tex]

And for T:

[tex]T'(t)-\lambda T(t)=0[/tex]

With initial value:

[tex]T(0)=f(x)[/tex]

How should I proceed from here? I'm not sure how to make this accomplish the boundary conditions.

Bye, thanks.

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