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}

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