What is the solution for y^'-x/(5y^4)=0,y(0)=2 ?

The solution for y^'-x/(5y^4)=0,y(0)=2 is y=\sqrt[5]{(x^2)/2+32}

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y^'-x/(5y^4)=0,y(0)=2

y^{^{'}} - \frac{x}{5 y ^{4}} = 0, y (0) = 2

Solution steps

y^{'} (x) - \frac{x}{5 y ^{4}} = 0

Solve separable ODE:

\frac{d y}{d x} = y - a

3 t^{5} \frac{d x}{d t} + 6 x - 5 = 0

\frac{d y}{d x} = 3 - 0.04 y

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

\frac{d}{d θ} (y (θ)) + y (θ) = 0

What is the solution for y^'-x/(5y^4)=0,y(0)=2 ?
The solution for y^'-x/(5y^4)=0,y(0)=2 is y=\sqrt[5]{(x^2)/2+32}