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

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

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y^'=(x(x^2+1))/(4y^3),y(0)=-1/(sqrt(2))

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

y = - \frac{4 x ^{4} + 1 + 2 x ^{2}}{2}

Solution steps

y^{'} (x) = \frac{x ( x ^{2} + 1 )}{4 y ^{3}}

Solve separable ODE:

y = - \frac{4 x ^{4} + 1 + 2 x ^{2}}{2}

y = - \frac{4 x ^{4} + 1 + 2 x ^{2}}{2}

\frac{\partial}{\partial x} ((a^{2} x^{2} + b^{2} y^{2})^{4})

d er i v a t i v e f (x) = x^{2} - 2 x + 1

\frac{d}{d x} (\frac{x ^{3}}{3} + c)

x \to 5 lim (\frac{x ^{3} + 8}{x + 2})

x (x + 1) \frac{d y}{d x} + x y = 1, y (e) = 1

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