Solve ODE with exactness solve the following ODE with exactness.
$$ e^{2x}(2\cos(y)-\sin(y)y')=0 \;, y(0)=0 $$
working through this question I got the answer $e^{2x}\cos(y)=0$ which doesn't seem correct. Assistance in finding the actual answer would be gratefully appreciated thank you
 A: Just integrate ...since $(e^{2x}\ne 0)$
$$e^{2x}(2\cos(y)-\sin(y)y')=0 \;, y(0)=0$$
$$2\cos(y)-\sin(y)y'=0$$
$$\int \frac{\sin(y)} {\cos(y)}dy=2x+K$$
$$\ln|\cos(y)|=-2x+K$$
$$y(0)=0 \to K=0$$
$$x=-\frac {\ln|\cos(y)|} 2$$ 

Edit 
For exactness
$$e^{2x}2\cos(y)dx-e^{2x}\sin(y)dy=0$$
$$df=P(x,y)dx+Q(x,y)dy$$
$$\partial_yP=\partial_ye^{2x}2\cos(y)=-2e^{2x}\sin(y)$$
$$\partial_xQ=-\partial_xe^{2x}\sin(y)=-2e^{2x}\sin(y)$$
The differential is exact, after integration we get
$$e^{2x}\cos(y)=K$$
using the cauchy condition:
 $$y(0)=0 \to K=1 \to e^{2x}\cos(y)=1$$
it's the same answer as in the first part of my answer
$$e^{2x}\cos(y)=1 \to \ln{(e^{2x}\cos(y))}=0 \to \ln|\cos(y)|=-2x$$
A: As the OP stated the equation to be correct, we can get rid of the exponential and solve:
$$2\cos(y)-\sin(y)y'=0$$
$$2\cos(y)dx=\sin(y)dy$$
$$2x=\int \tan(y)dy =\int \frac{ \sin y }{ \cos y} dy =- \int \frac{d \cos y }{ \cos y} =-\ln \cos y + C$$

Edit:
Since the OP actually needs to use the exact equation methods, let's get the exponential back:
$$2e^{2x}\cos(y)dx-e^{2x}\sin(y)dy=0$$
Checking for exactness:
$$\partial_y \left(2e^{2x}\cos(y) \right)=-2e^{2x}\sin(y)$$
$$\partial_x \left(-e^{2x}\sin(y) \right)=-2e^{2x}\sin(y)$$
The equation is exact.
The implicit solution will be:
$$\int 2e^{2x}\cos(y) dx=e^{2x} \cos y =C_1$$
Which is equivalent to the solution obtained by separation of variables (the constants are different).
