# Thread: Solve sinxy + xycosxy + x^2cosxy(dy/dx)=0

1. ## Solve sinxy + xycosxy + x^2cosxy(dy/dx)=0

Find a first integral and thus a general solution of the differential equation

sinxy + xycosxy + x^2cosxy(dy/dx)=0

I divided throught by x^2cosxy but I couldn't use integrating factor. I don't know what a first integral is.

2. Try letting $\displaystyle u = xy$ and see how that goes.

3. Using brackets to make this readable would be nice...

4. yes sorry about, latex wasn't working.

Danny, whats a first integral?

5. Originally Posted by Danny
Try letting $\displaystyle u = xy$ and see how that goes.

I tried but ended up with (x^2tanu-u^2)/x^3+du/dx=0

6. Originally Posted by poirot
I tried but ended up with (x^2tanu-u^2)/x^3+du/dx=0
Hmmm...
sin(xy) + (xy)cos(xy) + x^2 * cos(xy) * y' = 0

Let u = xy, which implies y = u/x. Thus y' = ... = (1/x)*u' - (1/x^2)*u. So your original equation becomes
sin(u) + u*cos(u) + x^2 * cos(u) *[ (1/x)*u' - (1/x^2)*u ] = 0

Now simplify a bit.

-Dan

7. ok I got xsinxy= A

8. Yep - me too!

9. Originally Posted by poirot
ok I got xsinxy= A
Yes, that's it.

-Dan

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# solve (xycosxy sinxy)dx x2cosxydy=0

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