What is the solution for 3x^2+4y(y^')=0 ?

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

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3x^2+4y(y^')=0

3 x^{2} + 4 y (y^{^{'}}) = 0

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

Solution steps

3 x^{2} + 4 y (y^{'} (x)) = 0

Solve separable ODE:

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

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

\frac{d}{d x} (x^{4} - 17 x^{2} + 16)

\int \frac{a ^{x}}{ln ( a )} d x

f (x) = \frac{x}{x + 6}

\int (ln (x) + 1) e^{x l n (x)} d x

\int x^{- 2} e^{- \frac{1}{x}} d x

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