Question

Dx + (x cot y+ sin y)dy=0 ,solve the DE

Answer 1

$\mathrm dx+(x\cot y+\sin y)\,\mathrm dy=0$

Multiply both sides by $\sin y$:

$\sin y\,\mathrm dx+(x\cos y+\sin^2y)\,\mathrm dy=0$

The ODE is now exact, since

$\dfrac{\partial(\sin y)}{\partial y}=\cos y$

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

so there exists a solution of the form $\Psi(x,y)=C$. This solution satisfies

$\dfrac{\partial\Psi}{\partial x}=\sin y$

$\dfrac{\partial\Psi}{\partial y}=x\cos y+\sin^2y$

Integrating both sides of the first PDE wrt $x$ gives

$\Psi(x,y)=x\sin y+f(y)$

and differentiating wrt $y$ gives

$\dfrac{\partial\Psi}{\partial y}=x\cos y+\sin^2y=x\cos y+\dfrac{\mathrm df}{\mathrm dy}$

$\implies\dfrac{\mathrm df}{\mathrm dy}=\sin^2y=\dfrac{1-\cos2y}2$

$\implies f(y)=\dfrac y2-\dfrac{\sin2y}4+C$

So the ODE has solution

$\Psi(x,y)=x\sin y+\dfrac y2-\dfrac{\sin2y}4=C$

which can be rewritten and simplified as

$\Psi(x,y)=\boxed{\sin y(2x-\cos y)+y=C}$