Question

Show that the equation is exact and find an implicit solution. y cos(xy) + 3x^2 + [x cos(xy) + 2y]y' = 0

Answer 1

We have

$\dfrac{\partial(y\cos(xy)+3x^2)}{\partial y}=\cos(xy)-xy\sin(xy)$

$\dfrac{\partial(x\cos(xy)+2y)}{\partial x}=\cos(xy)-xy\sin(xy)$

so the ODE is indeed exact. Then there's a solution of the form $f(x,y)=C$ such that

$\dfrac{\partial f}{\partial x}=y\cos(xy)+3x^2$

$\implies f(x,y)=\sin(xy)+x^3+g(y)$

Differentiating wrt $y$ gives

$\dfrac{\partial f}{\partial y}=x\cos(xy)+2y=x\cos(xy)+g'(y)$

$\implies g'(y)=2y\implies g(y)=y^2+C$

Then the solution to the ODE is

$f(x,y)=\boxed{\sin(xy)+x^3+y^2=C}$