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

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

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

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

Solution steps

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

Solve separable ODE:

\frac{d y}{d x} = y^{2} - 25

(1 + x^{2}) d y - 1 - y^{2} d x = 0

\frac{d y}{d x} = 14 x sec (y)

y^{^{'}} = - \frac{x - 3}{36 y - 108}, y (3) = 4

\frac{d S}{d t} = 0.04 - \frac{S}{( 1140 )}

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