What is the solution for 9yy^'+4x=0 ?

The solution for 9yy^'+4x=0 is y=(sqrt(-4x^2+c_{1)})/3 ,y=-(sqrt(-4x^2+c_{1)})/3

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9yy^'+4x=0

9 y y^{^{'}} + 4 x = 0

y = \frac{- 4 x ^{2} + c _{1}}{3}, y = - \frac{- 4 x ^{2} + c _{1}}{3}

Solution steps

9 y y^{'} (x) + 4 x = 0

Solve separable ODE:

y = \frac{- 4 x ^{2} + c _{1}}{3}, y = - \frac{- 4 x ^{2} + c _{1}}{3}

y = \frac{- 4 x ^{2} + c _{1}}{3}, y = - \frac{- 4 x ^{2} + c _{1}}{3}

e^{c o s (x)}

\frac{d}{d x} (csc^{9} (x))

\frac{4 x}{x + 2} (2.2)

\frac{1}{10 x}, at x = 9

\frac{d}{d x} (\frac{x ^{3}}{x ^{2} - 9})

What is the solution for 9yy^'+4x=0 ?
The solution for 9yy^'+4x=0 is y=(sqrt(-4x^2+c_{1)})/3 ,y=-(sqrt(-4x^2+c_{1)})/3