Solve for $y = C\cdot F(x)$ from $y' = y \cos(x)$ I have this function 
$y^\prime = y \cos(x)$
I want to use integration to reach to the form $y = C\cdot f(x)$.
I can reach to this level $y = e^{\sin(x)+C}$.
How can I solve it? Please help.
 A: We have $$\frac{dy}{dx} = y \cos x$$
$$\int \frac{dy}{y} = \int \cos x \ dx$$
$$\log y = \sin x + C$$
$$\implies y = e^{\sin x} \cdot e^C = C’ e^{\sin x}$$
A: $y'=y\cos(x) \implies \frac{dy}{dx}=y\cos(x) \implies \frac{1}{y} \frac{dy}{dx}=\cos(x) \implies \frac{1}{y} dy = \cos(x)dx$
$\implies \int \frac{1}{y}dy = \int \cos(x)dx \implies \ln |y|= \sin(x)+k \implies |y|=e^{\sin(x)+k}=C_0e^{\sin(x)}$
for some $C_0 \in \mathbb{R}^+$
$\implies y= \pm C_0e^{\sin(x)} \implies y=Ce^{\sin(x)}$ for any non-zero $C \in \mathbb{R}$. It is easy to see that $C=0$ will also yield a solution. Hence
$y=Ce^{\sin(x)}$
for any $C \in \mathbb{R}$.
A: As you and others have shown, your solution is $$y(x) = e^{\cos x + c} = C e^{\cos x}$$
for some $C$.
The only way you can determine $C$ is with initial conditions $(x=0, f(0))$ or other constraints $(x_0, f(x_0))$ or $(x_0, f^\prime (x_0))$...
A: Mate ,You have got your answer already ,just take 
$$e^c=C$$, from the last line of your answer
$$y=e^{\sin x +c}= e^{\sin x}e^c=Ce^{\sin x}$$
 you are done..
