Question

SOLVE DE

(cosx+Iny) + dy/dx= -( x/y +e^y)dy/dx

Answer 1

Question #7500 Solve $(\cos x + \log y) + dy/dx = -(x/y + e^y)dy/dx$.

Solution. Denote $M(x,y) \coloneqq \cos + \log y$, $N(x,y) \coloneqq x/y + e^y + 1$. Our equation is equivalent to $(\cos x + \log y)dx + (x/y + e^y + 1)dy = 0$. Due to $\frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x} = \frac{1}{y}$ our DE is exact. Denote by $U(x,y)$ such function that $U(x,y) = 0$ defines the general solution. $U$ satisfies $\frac{\partial U}{\partial x} = \cos x + \log y$, thus $U(x,y) = \sin x + (\log y)x + \varphi(y)$. One can get $\frac{\partial U}{\partial y} = x/y + \varphi'(y) = N(x,y) = x/y + e^y + 1$, hence $\varphi'(y) = e^y + 1$, so $\varphi(y) = e^y + y + C$. $U(x,y) = \sin x + \log y \cdot x + e^y + y + C$, and the general solution is obtained from $U(x,y) = 0$.

Answer. The general solution $\sin x + \log y \cdot x + e^y + y + C = 0$