Find the solution of $$\frac{d y}{d x}=\frac{1}{\cos (x-y)}$$

The solution of the equation $$\frac{d y}{d x}+x(x+y)=x^3(x+y)^3-1$$ is

If $$p$$ and $$q$$ are order and degree of the question $$\left(\frac{d^2 y}{d x^2}\right)^4+4 \frac{\left(\frac{d^2 y}{d x^2}\right)^2}{\left(\frac{d^3 y}{d x^3}\right)^3}+\frac{d^3 y}{d x^3}=x^2-1$$, then

The solution of $$\frac{d y}{d x}=1+x+y+x y$$ is