What is the solution for 2y*sin(x)dx=cos(x)dy-2dx ?

The solution for 2y*sin(x)dx=cos(x)dy-2dx is y=(2sin(x))/(cos^2(x))+(c_{1})/(cos^2(x))

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2y*sin(x)dx=cos(x)dy-2dx

2 y \cdot sin (x) d x = cos (x) d y - 2 d x

y = \frac{2 sin ( x )}{cos ^{2} ( x )} + \frac{c _{1}}{cos ^{2} ( x )}

Solution steps

2 y \cdot sin (x) d x = cos (x) d y - 2 d x

Solve linear ODE:

y = \frac{2 sin ( x )}{cos ^{2} ( x )} + \frac{c _{1}}{cos ^{2} ( x )}

y = \frac{2 sin ( x )}{cos ^{2} ( x )} + \frac{c _{1}}{cos ^{2} ( x )}

\frac{d}{d x} (cosh^{5} (x))

\frac{\partial}{\partial y} (ln (3 x^{2} - y^{3} - 3))

\frac{d}{d x} (16 I n (x^{2}) - x^{2})

\int sin (2 x) cos^{2} (x) d x

\frac{d}{d x} (x^{3} + 2 x^{2} - 1)

What is the solution for 2y*sin(x)dx=cos(x)dy-2dx ?
The solution for 2y*sin(x)dx=cos(x)dy-2dx is y=(2sin(x))/(cos^2(x))+(c_{1})/(cos^2(x))