A basic linear differential equation wrong solution: cosx * y' = cosx + 2sinx * y

[Ognjen]

The equation is the following:

$$cosx*y'=cosx + 2sinx*y$$
Here is how I tried to solve it (attached image of blank paper with the problem solved on it).

The official solution, however, is:

$$y=1/(cos^2x)*(x/2 + sin2x/4)$$
I don't understand why my solution is not correct, when I successfully identified P(x) and Q(x) (according to the differential equation solver website), and plugged them in the formula for y(x) for linear DE. For clarification, the last formula I was talking about was this one:

$$y(x) = e^(-\int P(x) \,dx) * (C + \int Q(x)*e^(\int P(x) \,dx))$$
I would appreciate any help given.

 

[nasi112]

$\displaystyle y(x) = \frac{1}{e^{\int P(x) \ dx}}\left(\int e^{\int P(x) \ dx} \ Q(x) \ dx + c\right)$

 

[topsquark]

The equation is the following:

$$cosx*y'=cosx + 2sinx*y$$
Here is how I tried to solve it (attached image of blank paper with the problem solved on it).

The official solution, however, is:

$$y=1/(cos^2x)*(x/2 + sin2x/4)$$
I don't understand why my solution is not correct, when I successfully identified P(x) and Q(x) (according to the differential equation solver website), and plugged them in the formula for y(x) for linear DE. For clarification, the last formula I was talking about was this one:

$$y(x) = e^(-\int P(x) \,dx) * (C + \int Q(x)*e^(\int P(x) \,dx))$$
I would appreciate any help given.

Can you show us how you derived your expression for y?

-Dan