What is the solution for 3yy^'+2x=0 ?

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

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3yy^'+2x=0

3 y y^{^{'}} + 2 x = 0

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

Solution steps

3 y y^{'} (x) + 2 x = 0

Solve separable ODE:

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

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

\frac{d}{d x} \frac{5}{( 7 - x ^{2} ) ^{5}}

12 x^{5} - 8 x^{4} + 20 x^{3}

\frac{d}{d x} (\frac{sin ( 2 x + 1 )}{cos ( - 1 - 2 x )})

\frac{d ^{2}}{d x ^{2}} (cos (6 x))

x \to - 1 lim (x (0))

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