General solution to a 2nd order differential

[EmmaLemming]

General solution to a 2nd order differential :(

Homework Statement

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

Homework Equations

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?

 

[I like Serena]

Homework Statement

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

Homework Equations

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?

Hi EmmaLemming!

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

 

[EmmaLemming]

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 :)

 

[I like Serena]

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?

 

[EmmaLemming]

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?

 

[I like Serena]

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)