How can I solve the partial differentiation equation provided?

[Lavace]

Homework Statement

http://img94.imageshack.us/img94/3853/physicse.jpg

The Attempt at a Solution

I kept y fixed, and so I ended up with the following equation:

Integ[dU/U] = Integ[x]

And we end up with: U(x,y) = e^x * g(y)

To solve g(y), we sub the solution into the 2nd PDE provided to give:

d/dy[e^x * g(y)] = y[e^x * g(y)]

Dividing through by e^x: d/dy [g(y)] = y*g(y)

I was stuck at this point, so took a peek at the answers to find the lecturer wrote:
=> ln[g(y)] = 1/2*y^2 + c

How did he come to that? I can't solve this equation, could someone please help me out?

Thank you very much!

 

[Lavace]

I have also tried working the other way round but still, no joy.

Any help?

 

[n1person]

I don't think that is quite the right answer,

Cause couldn't for the first step anything like g(y)e^x + h(y) work?

So pluging that into the second equation you get g'(y)e^x+h'(y) = y(g(y)e^x+0)
so h'(y)= c, and g'(y)=yg(y), you can solve from there.

 

[HallsofIvy]

Homework Statement

http://img94.imageshack.us/img94/3853/physicse.jpg

The Attempt at a Solution

I kept y fixed, and so I ended up with the following equation:

Integ[dU/U] = Integ[x]

And we end up with: U(x,y) = e^x * g(y)

To solve g(y), we sub the solution into the 2nd PDE provided to give:

d/dy[e^x * g(y)] = y[e^x * g(y)]

Dividing through by e^x: d/dy [g(y)] = y*g(y)

This is a separable equation.
dg/g= ydy

Integrate both sides.

I was stuck at this point, so took a peek at the answers to find the lecturer wrote:
=> ln[g(y)] = 1/2*y^2 + c

How did he come to that? I can't solve this equation, could someone please help me out?

Thank you very much!

 

[HallsofIvy]

I don't think that is quite the right answer,

Cause couldn't for the first step anything like g(y)e^x + h(y) work?

No, it wouldn't. The derivative of that, with respect to x, is g(y)e^x, NOT U(x,y)= g(y)e^x+ h(y). What Lavace did was correct.

So pluging that into the second equation you get g'(y)e^x+h'(y) = y(g(y)e^x+0)
so h'(y)= c, and g'(y)=yg(y), you can solve from there.