The general solution of the first-order differential equation $\sin y y ^ { \prime } + \sin x \cos y = \sin x$ is

A) $\cos y = 1 + C e ^ { - \cot x }$
B) $\sin y = 1 + C e ^ { - \sin x }$
C) $\cos y = 1 + C e ^ { - \sin x }$
D) $\sin y = 1 + C e ^ { - \cos x }$
E) $\cos y = 1 + C e ^ { - \cos x }$

Question

The general solution of the first-order differential equation $\sin y y ^ { \prime } + \sin x \cos y = \sin x$ is

A) $\cos y = 1 + C e ^ { - \cot x }$
B) $\sin y = 1 + C e ^ { - \sin x }$
C) $\cos y = 1 + C e ^ { - \sin x }$
D) $\sin y = 1 + C e ^ { - \cos x }$
E) $\cos y = 1 + C e ^ { - \cos x }$

Quizplus · Accepted Answer

The Answer of The general solution of the first-order differential equation $\sin y y ^ { \prime } + \sin x \cos y = \sin x$ is

A) $\cos y = 1 + C e ^ { - \cot x }$
B) $\sin y = 1 + C e ^ { - \sin x }$
C) $\cos y = 1 + C e ^ { - \sin x }$
D) $\sin y = 1 + C e ^ { - \cos x }$
E) $\cos y = 1 + C e ^ { - \cos x }$

The General Solution of the First-Order Differential Equation sin⁡yy′+sin⁡xcos⁡y=sin⁡x\sin y y ^ { \prime } + \sin x \cos y = \sin xsinyy′+sinxcosy=sinx

The General Solution of the First-Order Differential Equation $\sin y y ^ { \prime } + \sin x \cos y = \sin x$