# A simple differential equation question

1. Jan 29, 2012

### StephenD420

Hello everyone

I having a little trouble with this simple problem:
Show that x*dF(x,y)/dy = y*dF(x,y)/dx is F(x,y) = A*exp(b/2*(x^2+y^2)) where A and b are constants using separation of variables.

x*df/dy = y*df/dx
x= y* df/dx*dy/df
x = y* dy/dx
x*dx = y*dy
integrate both sides x^2/2 = y^2/2 + c

so what do I do now to find the answer??

Stephen

2. Jan 29, 2012

### cragar

so separation of variables allows you to guess that the solution is the product of a function of x and y. so try this and plug it in.

3. Jan 29, 2012

### StephenD420

sorry..I am not sure what you mean. Could you explain it or show me the first couple of steps as I do not understand what you mean that the answer is a product of x and y....I know that that is why I separated the x's and y's so x's were on one side and the y's were on the other side of the equals sign.
Please explain what you mean by that...

4. Jan 29, 2012

### cragar

like you would make the guess that F(x,y)=X(x)Y(y)
then find df/dy and df/dx and then plug this into your original equation.
so for example df/dx=dX/dx(Y(y))

5. Jan 29, 2012

### StephenD420

ok if F=h(x)g(y)
then dF/dx = g(y)dh/dx
and dF/dy = h(x)dg/dy
so
x*dF/dy = y*dF/dx becomes
x*h(x)*dg/dy = y*g(y)dh/dx
putting all x' on one side and y's on the other gives
x*h(x)*dx/dh = y*g(y)*dy/dg

now what????

6. Jan 30, 2012

### cragar

now subtract the y*g(y)*dy/dg to the other side. and then divide everything by
h(x) and g(y) now set each of the expressions being added together equal to a constant and then solve these 2 ode's.

7. Jan 30, 2012

### StephenD420

ok so
x*h(x)*dx/dh = y*g(y)*dy/dg
becomes when you subtract and divide by h(x) and g(x) gives
x*h(x)*dx/dh - y*g(y)*dy/dg = 0 becomes
x/g(x)*dx/dh -y/h(x) *dy/dg = 0

so now what?? I am confused as to what two equations you mean and how to go from here.
Do you mean
x*dx/dh=c1
y*dy/dg=c2
then
x^2/2 =c1*h(x) + C3
y^2/2 = c2*g(x) + C4
now what??
Thanks.
Stephen

8. Jan 30, 2012

### cragar

x/g(x)*dx/dh -y/h(x) *dy/dg = 0
x/g(x)*dx/dh=C and y/h(x) *dy/dg=D
so now you have x(dx/dh)=c(g(x)) and y(dy/dg)=D(h(y))
so now you have a derivative of a function equal to another number times itself.
this is usually an exponential function. I think i got the g(y) and h(x)'s mixed up
also maybe Google some examples of separation of variables

Last edited: Jan 30, 2012