What is the solution for 2sin(x)y^'+ycos(x)=y^2cos(x) ?

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

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2sin(x)y^'+ycos(x)=y^2cos(x)

2 sin (x) y^{^{'}} + y cos (x) = y^{2} cos (x)

y = \frac{1 + e ^{c_{1}} sin ^{\frac{1}{2}} ( x )}{1 - e ^{2 c_{1}} sin ( x )}

Solution steps

2 sin (x) y^{'} (x) + y cos (x) = y^{2} cos (x)

Solve separable ODE:

y = \frac{1 + e ^{c_{1}} sin ^{\frac{1}{2}} ( x )}{1 - e ^{2 c_{1}} sin ( x )}

y = \frac{1 + e ^{c_{1}} sin ^{\frac{1}{2}} ( x )}{1 - e ^{2 c_{1}} sin ( x )}

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