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}

Solution steps

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

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f (x) = x^{2} - 2 x + 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))