What is the solution for e^{-y}sin(x)-y^'cos^2(x)=0 ?

The solution for e^{-y}sin(x)-y^'cos^2(x)=0 is y=ln(sec(x)+c_{1})

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e^{-y}sin(x)-y^'cos^2(x)=0

e^{- y} sin (x) - y^{^{'}} cos^{2} (x) = 0

y = ln (sec (x) + c_{1})

Solution steps

e^{- y} sin (x) - y^{'} (x) cos^{2} (x) = 0

Solve separable ODE:

y = ln (sec (x) + c_{1})

y = ln (sec (x) + c_{1})

\frac{d}{d x} (- 0.8 sin (x) + cos (2 x))

y^{^{'}} = y (x y^{5} + 3)

x \to 8 lim (500 x - 6 x^{2})

x \to 4 - lim (\frac{x}{x - 4})

x \to - 3 lim (x^{3} - 2 x^{2} + x + 7)

What is the solution for e^{-y}sin(x)-y^'cos^2(x)=0 ?
The solution for e^{-y}sin(x)-y^'cos^2(x)=0 is y=ln(sec(x)+c_{1})