What is the solution for sin(3x)dx+4ycos^2(3x)dy=0 ?

The solution for sin(3x)dx+4ycos^2(3x)dy=0 is y=sqrt((-sec(3x)+c_{1))/6},y=-sqrt((-sec(3x)+c_{1))/6}

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sin(3x)dx+4ycos^2(3x)dy=0

sin (3 x) d x + 4 y cos^{2} (3 x) d y = 0

y = \frac{- sec ( 3 x ) + c _{1}}{6}, y = - \frac{- sec ( 3 x ) + c _{1}}{6}

Solution steps

sin (3 x) d x + 4 y cos^{2} (3 x) d y = 0

Solve separable ODE:

y = \frac{- sec ( 3 x ) + c _{1}}{6}, y = - \frac{- sec ( 3 x ) + c _{1}}{6}

y = \frac{- sec ( 3 x ) + c _{1}}{6}, y = - \frac{- sec ( 3 x ) + c _{1}}{6}

\frac{d y}{d x} = \frac{2 x}{- 8 y - 6}, y (0) = - 8

(1 + x^{2}) y^{^{'}} + 2 x y = - x

y^{^{'}} = - 2 t y

\frac{d P}{d t} = Pt, P (1) = 3

x^{^{'}} = 3 x

What is the solution for sin(3x)dx+4ycos^2(3x)dy=0 ?
The solution for sin(3x)dx+4ycos^2(3x)dy=0 is y=sqrt((-sec(3x)+c_{1))/6},y=-sqrt((-sec(3x)+c_{1))/6}