Thread: PDE

**Altair** · May 9th 2009, 12:36 AM

$u_y + 2yu = 0$

I have to solve this as ODE...How can it be done ? u is a function of x and y but I do not have any idea how to transform this equation into derivative of one variable only

**mr fantastic** · May 9th 2009, 03:03 AM

Originally Posted by Altair

$u_y + 2yu = 0$

I have to solve this as ODE...How can it be done ? u is a function of x and y but I do not have any idea how to transform this equation into derivative of one variable only

Assume a seperable solution: $U = X(x) Y(y)$ .

This leads to $\frac{dY}{dy} + 2yY = 0$ which is seperable.

You end up with $U = e^{-y^2} X(x)$ .

**HallsofIvy** · May 9th 2009, 11:05 AM

Originally Posted by Altair

$u_y + 2yu = 0$

I have to solve this as ODE...How can it be done ? u is a function of x and y but I do not have any idea how to transform this equation into derivative of one variable only

Or, since x appears neither explictely nor as a variable of differentiation, ignore the fact that it is a partial differential equation!
The problem $\frac{du}{dy}+ 2yu= 0$ can be written as $\frac{du}{dy}= -2yu$ and is a separable equation: $\frac{du}{u}= -2ydy$ . Integrating, $ln(u)= -y^2+ C$ so, taking the exponential of both sides, $u= e^{-y^2+ C}= e^Ce^{-y^2}= C'e^{-y^2}$ .

Finally, use the fact that we have been ignoring x- that is, treating it like a constant: the constant C' might be any function of x. If we call that "F(x)" then we have $u(x,y)= F(x)e^{-y^2}$ where F(x) is an arbitrary function of x, exactly the solution mr. fantastic got.

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