# Partial Differential Equation

x(δu/δx)-(1/2)y(δu/δy)=0

By first looking for a separable solution of the form u(x, y)=X(x)Y(y), find the general solution of the equation given above.

Determine the u(x,y) which satisfies the boundary condition u(1,y)=1+siny

For the separable form I have u(x, y)=A(x^c)(y^2c), could someone please show me how to do the rest of it.

Thank you.

arildno
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Write sin(y) as a power series.

arildno
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Each choice of c gives you a solution of your diff.eq.
Since your diff.eq is linear, a sum of such solutions is also a solution of your diff.eq.

How do I get from u(x, y)=A(x^c)(y^2c) to the general solution?

Thank you.

arildno
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All right:
A series solution of your diff.eq is:
$$u(x,y)=\sum_{n=1}^{\infty}A_{n}x^{c_{n}}y^{2c_{n}}$$,
whereby follows:
$$u(1,y)=\sum_{n=1}^{\infty}A_{n}y^{2c_{n}}$$
and $A_{n},c_{n}$ are constants.

Now, how can you fit this expression for u(1,y) to the given boundary condition?

All right:
A series solution of your diff.eq is:
$$u(x,y)=\sum_{n=1}^{\infty}A_{n}x^{c_{n}}y^{2c_{n}}$$,
whereby follows:
$$u(1,y)=\sum_{n=1}^{\infty}A_{n}y^{2c_{n}}$$
and $A_{n},c_{n}$ are constants.

Now, how can you fit this expression for u(1,y) to the given boundary condition?

arildno
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Well, use my first hint in post 2.

Well, use my first hint in post 2.

Ok if you could please show me the COMPLETE working I would really appreciate it... Thank you.

Well, use my first hint in post 2.

Ok if you could please show me the COMPLETE working I would really appreciate it... Thank you.

arildno
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Do you know what a power series is? Do you know what a power series is? Yes lol, I just can't do this question, could you please show me the working for it... In fact can you do it yourself?

HallsofIvy
Homework Helper
If you are not capable of doing basic algebra, you should not be attempting partial differential equations!

(Yes, I can do it myself! That's not really the point is it? You have been told exactly HOW to solve your equation, yet you have not even TRIED to apply what you have been told.)

arildno