Question

solve (1+2xy cos x^2 -2xy)dx+(sinx^2-x^2)d y=0

Answer 1

$\left(1+2xy\:cos\:x^2\:-2xy\right)dx+\left(sinx^2-x^2\right)d\:y=0$

$\left(1+2xy\cos \left(x^2\right)-2xy\right)dx+\left(\sin \left(x^2\right)-x^2\right)dy=0$

First order linear Ordinary Differential Equation

$\mathrm{A\:first\:order\:linear\:ODE\:has\:the\:form\:of\:}y'\left(x\right)+p\left(x\right)y=q\left(x\right)$

Let y be the dependent variable, Divide by dx:

$1+2xy\cos \left(x^2\right)-2xy+\left(\sin \left(x^2\right)-x^2\right)\frac{dy}{dx}=0$

$\mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'$

$1+2xy\cos \left(x^2\right)-2xy+\left(\sin \left(x^2\right)-x^2\right)y'\:=0$

We rewrite in the form of a first order linear ODE

$y'\:+\frac{2x\left(\cos \left(x^2\right)-1\right)}{-x^2+\sin \left(x^2\right)}y=-\frac{1}{\sin \left(x^2\right)-x^2}$

We find integration factor: $\mu = -x^2 + \sin \left(x^2\right)$

We put the equation in the from $(\mu(x)⋅y)' = \mu(x) ⋅ q(x); ((-x^2 + sin(x^2))y)' = -1$

We solve:

$((-x^2 + sin(x^2))y)' = -1;\, y=-\frac{x}{-x^2+\sin \left(x^2\right)}+\frac{c_1}{-x^2+\sin \left(x^2\right)}$

$y=-\frac{x}{-x^2+\sin \left(x^2\right)}+\frac{c_1}{-x^2+\sin \left(x^2\right)}$