Solve the given initial-value problem

Question

Answer 1

(y² + ysinx)dx + (2xy - cosx - 1/(1 + y²)) = 0;

This is exact equation,becouse ∂/∂y(y² + ysinx) = ∂/∂x((2xy - cosx - 1/(1 + y²))) = 2y + sinx;

Then ∂F/∂x = y² + ysinx; F(x, y) = ∫ (y² + ysinx)dx = y²x - ycosx + f(y).

Then 2yx - cosx +f'(y) = 2xy - cosx - 1/(1 + y²); f'(y) = - 1(1 + y²); f(y) = - tan^-1y + C

F(x, y) = y²x - ycosx - tan^-1y = C; Because y(0) = 1. we have - 1 - tan^-11 = C, C = -1 - π/4;

y²x - ycosx - tan^-1y + 1 + π/4 = 0 is solution of this IVP.