# General solution to a 2nd order differential

1. Apr 22, 2012

### EmmaLemming

General solution to a 2nd order differential :(

1. The problem statement, all variables and given/known data

What is the general solution of ∂2f(x,t)/∂x∂t = xt ?

2. Relevant equations

3. The attempt at a solution

I have no idea, I tried to follow an example out of the book but it was quite different to this question.

Do I need to replace x's and y's with u's and v's? If so, how?

2. Apr 22, 2012

### I like Serena

Re: General solution to a 2nd order differential :(

Hi EmmaLemming!

How about you try to integrate with respect to x, and afterward again with respect tot t?

3. Apr 22, 2012

### EmmaLemming

Re: General solution to a 2nd order differential :(

Ahh is it really that simple? Awesome :) I thought there was more to it than that.

I get, f(x,t) = 1/4(x2t2) + Ct + D = 1

Does that seem reasonable?

Do I need to do anything else?

Thank you for your help :)

4. Apr 22, 2012

### I like Serena

Re: General solution to a 2nd order differential :(

Basically that's it, except for your integration constants.
(I like simple. )

Your integration constants are not just any integration constants.
When you integrate with respect to x, you get an integration constant that can be any function of t.
Same for integrating with respect to t, where you would have to integrate the previous integration constant, and add an integration constant that is a function of x.

Btw, what is that "=1" that you appended?

5. Apr 22, 2012

### EmmaLemming

Re: General solution to a 2nd order differential :(

oh dear..

So do I have to find out what C and D are? ... How?

And I put "= 1" on the RHS because ∂2f(x,t)/∂x∂t = xt

I thought integration of 'xt' wrtx and then t equates to 1..?
Is that incorrect?

6. Apr 22, 2012

### I like Serena

Re: General solution to a 2nd order differential :(

I don't get what you mean about the "1".

Anyway, when integrating the first time wrt x, you should get:
∂f(x,t)/∂t = (1/2)x2t + c(t)

I'm writing c(t) to indicate that it is a function of t.
Differentiating wrt x will make any function of t disappear.

Next when you integrate wrt t, you effectively integrate c(t) to just another unknown function C(t).

So you get:
f(x,t) = (1/4)x2t2 + C(t) + D(x)