The solution of $${x^3}{{dy} \over {dx}} + 4{x^2}\tan y = {e^x}\sec y$$ satisfying y (1) = 0, is

The solution of the equation $${{dy} \over {dx}} + {1 \over x}\tan y = {1 \over {{x^2}}}\tan y\sin y$$ is

The solution of differential equation $$(x{y^5} + 2y)dx - xdy = 0$$, is

A curve passes through (2, 0) and the slope of the tangent at P(x, y) is equal to $${{{{(x + 1)}^2} + y - 3} \over {x + 1}}$$ then the equation of the curve is