# Thread: [SOLVED] Solve the following differential equation?

1. ## [SOLVED] Solve the following differential equation?

Solve the following differential equation:
$3x^2y^2 + \cos (xy) - xy \sin (xy) + \frac{dy}{dx}\{2x^3y - x^2\sin (xy)\} = 0$

2. Originally Posted by fardeen_gen
Solve the following differential equation:
$3x^2y^2 + \cos (xy) - xy \sin (xy) + \frac{dy}{dx}\{2x^3y - x^2\sin (xy)\} = 0$
$\frac{dy}{dx}\{2x^3y - x^2\sin (xy)\} = xy \sin (xy) -3x^2y^2 - \cos (xy) +xy \sin (xy)$

$\frac{dy}{dx} = \frac{xy \sin (xy) -3x^2y^2 - \cos (xy) }{2x^3y - x^2\sin (xy)}$

$\int \frac{dy}{dx} dx = \int\frac{xy \sin (xy) -3x^2y^2 - \cos (xy) }{2x^3y - x^2\sin (xy)} dx$

$\int dy = \int\frac{xy \sin (xy) -3x^2y^2 - \cos (xy) }{2x^3y - x^2\sin (xy)} dx$

$y+c = \int\frac{xy \sin (xy) -3x^2y^2 - \cos (xy) }{2x^3y - x^2\sin (xy)} dx$

There's my lazy contribution!

3. The solution is:
Spoiler:
$x(x^2y^2 + \cos xy) = c$

I am unable to find it! Anyone?

4. Originally Posted by fardeen_gen
Solve the following differential equation:
$3x^2y^2 + \cos (xy) - xy \sin (xy) + \frac{dy}{dx}\{2x^3y - x^2\sin (xy)\} = 0$
Note that the differential equaiton is exact( you can check the partials) i.e there is a function $z(x,y)$ such that

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

so

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

Now we take the partial integral with respect to x and get

$z=x^3y^2+\frac{\sin(xy)}{y}+x\cos(xy)-\frac{\sin(xy)}{y}+g(y)$

Now we take the partial with respect to y to get

$\frac{\partial z}{\partial y}=2x^3y-xy\sin(xy)+g'(y)$

Now we see that $g'(y)=0$

So $g(y)=c$

So we get an implict solution

$x^3y^2+x\cos(xy)=c$