What is the solution for x^3(dy)/(dx)+4y^2=0,y(1)=1 ?

The solution for x^3(dy)/(dx)+4y^2=0,y(1)=1 is y=-(x^2)/(2-3x^2)

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x^3(dy)/(dx)+4y^2=0,y(1)=1

x^{3} \frac{d y}{d x} + 4 y^{2} = 0, y (1) = 1

y = - \frac{x ^{2}}{2 - 3 x ^{2}}

Solution steps

x^{3} \frac{d y}{d x} + 4 y^{2} = 0

Solve separable ODE:

y = - \frac{x ^{2}}{2 - 3 x ^{2}}

y = - \frac{x ^{2}}{2 - 3 x ^{2}}

2 x y^{^{'}} ln (x) = - 2 y

y^{^{'}} sec (t) = \frac{- y ^{23}}{22}

\frac{d y}{d x} 6 cos (4 x) sin (2 y) = 1

x \cdot \frac{d y}{d x} + x^{3} y = 0

\frac{d y}{d x} = y^{2} - 13 y + 42

What is the solution for x^3(dy)/(dx)+4y^2=0,y(1)=1 ?
The solution for x^3(dy)/(dx)+4y^2=0,y(1)=1 is y=-(x^2)/(2-3x^2)