Problem with solving Differential Equation (Help needed)

[Vuki]

Hi! I can't solve one Differential Equation and I would be very grateful if someone could help me This is the equation I need to solve:

y' = 1/(x*cosy + sin2y)

 

[stapel]

Hi! I can't solve one Differential Equation and I would be very grateful if someone could help me This is the equation I need to solve:

y' = 1/(x*cosy + sin2y)

According to Wolfram Alpha, the solution isn't very pretty:

. . . . .\(\displaystyle y(x) \,= \,\sin^{-1}\left(\dfrac{1}{2}\, \left(-2 W\left(e^{-\frac{x}{2}-1} c_1\right)\,-\,x\,-\,2\right)\right)\)

...where "W" is the "product log function".

 

[Deleted member 4993]

Hi! I can't solve one Differential Equation and I would be very grateful if someone could help me This is the equation I need to solve:

y' = 1/(x*cosy + sin2y)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{1}{x*cos(y) + sin(2y)}\)

\(\displaystyle \dfrac{dy}{dx} * cos(y) = \dfrac{1}{x + 2sin(y)}\)

substitute u = sin(y)

\(\displaystyle \dfrac{du}{dx} = \dfrac{1}{x + 2u}\)

\(\displaystyle \dfrac{dx}{du} = \ x + 2u \)

x = C₁e^u - 2u - 2

x = C₁e^sin(y) - 2sin(y) - 2

Check whether this expression satisfies your original DE.

 

[Vuki]

Thank you very much!
This equation seems to be above my level.
I can only solve it for x and get x = C₁e^u - 2u - 2. I hope it will be enough for my professor