Question

ysin2xdx=(1+y^2+cos^2x)dy

Answer 1

The general form of this equation: Pdx + Qdy = 0, where P = y sin2x, Q = - (1 + y^2 + cos^2x)

Let's check if exists smooth enough function u(x, y) such that: du = Pdx + Qdy

$\frac{dP}{dy} = sin2x$ and $\frac{dQ}{dx} = sin2x$ are equal, so we can state that such function exists.

$u = \int Pdx + f(y) = y\int sin2xdx + f(y) = -\frac12\cdot y\cdot cos2x + f(y)$

$=> \frac{du}{dy} = -\frac12\cdot cos2x + f'(y) = Q = - 1- y^2 - cos^2x$

$=> -cos^2x + \frac12 + f'(y) = -1 - y^2 - cos^2x => f'(y) = -\frac32 - y^2$

$=> Integrating: f(y) = -\frac{3y}{2} - \frac{y^3}{3} + C$

$=> u = -\frac12\cdot y\cdot cos2x - \frac{3y}{2} - \frac{y^3}{2} + C$

So, answer is this equation: $-\frac12\cdot y\cdot cos2x - \frac{3y}{2} - \frac{y^3}{2} + C = 0$