What is the solution for sin(3x)dx+2ycos^2(3x)dy=0 ?

The solution for sin(3x)dx+2ycos^2(3x)dy=0 is y=sqrt((-sec(3x)+c_{1))/3},y=-sqrt((-sec(3x)+c_{1))/3}

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sin(3x)dx+2ycos^2(3x)dy=0

sin (3 x) d x + 2 y cos^{2} (3 x) d y = 0

y = \frac{- sec ( 3 x ) + c _{1}}{3}, y = - \frac{- sec ( 3 x ) + c _{1}}{3}

Solution steps

sin (3 x) d x + 2 y cos^{2} (3 x) d y = 0

Solve separable ODE:

y = \frac{- sec ( 3 x ) + c _{1}}{3}, y = - \frac{- sec ( 3 x ) + c _{1}}{3}

y = \frac{- sec ( 3 x ) + c _{1}}{3}, y = - \frac{- sec ( 3 x ) + c _{1}}{3}

y^{^{'}} = e^{- 4 y}

\frac{d y}{d x} + x y = 3 x

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

y^{^{'}} = 1 - y - x y + x

3 y^{^{'}} = y cos (3 x), y (0) = 1

What is the solution for sin(3x)dx+2ycos^2(3x)dy=0 ?
The solution for sin(3x)dx+2ycos^2(3x)dy=0 is y=sqrt((-sec(3x)+c_{1))/3},y=-sqrt((-sec(3x)+c_{1))/3}