Solve ODE with exactness

Question

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

Answer 1

Just integrate ...since $(e^{2x}\ne 0)$

$$e^{2x}(2\cos(y)-\sin(y)y')=0 \;, y(0)=0$$ $$2\cos(y)-\sin(y)y'=0$$ $$\int \frac{\sin(y)} {\cos(y)}dy=2x+K$$ $$\ln|\cos(y)|=-2x+K$$ $$y(0)=0 \to K=0$$ $$x=-\frac {\ln|\cos(y)|} 2$$

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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 $$e^{2x}\cos(y)=1 \to \ln{(e^{2x}\cos(y))}=0 \to \ln|\cos(y)|=-2x$$

Answer 2

As the OP stated the equation to be correct, we can get rid of the exponential and solve:

$$2\cos(y)-\sin(y)y'=0$$

$$2\cos(y)dx=\sin(y)dy$$

$$2x=\int \tan(y)dy =\int \frac{ \sin y }{ \cos y} dy =- \int \frac{d \cos y }{ \cos y} =-\ln \cos y + C$$

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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).