微分方程求解:y' - ysinx - y^2 + cosx = 0

微分方程求解:y' - ysinx - y^2 + cosx = 0

原方程化為(y+sin x)'=y(y+sin x),令z=y+sin x,
z'=z(z-sin x),即z'+zsin x=z^2這是貝努利方程,就可求解了.