Can u(x,y) be a product of single-variable functions for ∂²u/∂x∂y = u³?

[fredrick08]

can i just flip it?? then it will be the same as f... =ln(g(y))?

 

[Cyosis]

Yes you can flip one side, but then you also need to flip the other side.

 

[fredrick08]

ok so ln(g(y))=(ax+c)^-1??

 

[Cyosis]

No you are solving the differential equation g(y)/g'(y)=a , then g'(y)/g(y)=?

 

[fredrick08]

1/a? so ln(g(y))=(y/a)+c?? =>g(y)=e^((y/a)+c)

 

[Cyosis]

No. How can the integral of f'/f have a different value than g'/g. If g(y)/g'(y)=a then g'(y)/g(y)=1/a. Now integrate.

 

[fredrick08]

yes isn't that what i did? int(1/a)dy=(y/a)+c right?

 

[Cyosis]

Ugh sorry it's 4 am here I am getting sleepy! Yes you're right so what is the the total solution to the partial differential equation you began with?

 

[fredrick08]

u(x,y)=f(x)g(y)=Ae^(ax)*Be^(y/a)... but I am not sure... if i differentiate that, i don't thik it works out.

 

[Cyosis]

It's correct, just differentiate it and you will see it will turn out correctly.

 

[fredrick08]

ok omg thankyou very much... very much appreciated = ) have a good sleep

 

[Cyosis]

You're welcome, thanks and good night!