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

The solution for y^'=-(5x^4-5)/(4y) is y=sqrt((-x^5+5x+c_{1))/2},y=-sqrt((-x^5+5x+c_{1))/2}

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y^'=-(5x^4-5)/(4y)

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

y = \frac{- x ^{5} + 5 x + c _{1}}{2}, y = - \frac{- x ^{5} + 5 x + c _{1}}{2}

Solution steps

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

Solve separable ODE:

y = \frac{- x ^{5} + 5 x + c _{1}}{2}, y = - \frac{- x ^{5} + 5 x + c _{1}}{2}

y = \frac{- x ^{5} + 5 x + c _{1}}{2}, y = - \frac{- x ^{5} + 5 x + c _{1}}{2}

(e^{4} - y) cos (x) \frac{d y}{d x} = e^{4} sin (x)

y^{^{'}} = y - 2 y^{2}, y (0) = t (t)

\frac{d y}{d x} = - \frac{4}{9} x^{6} y^{4}

y^{^{'}} = (cos (x)) y^{2}

\frac{d R}{d t} = - a R (t)

What is the solution for y^'=-(5x^4-5)/(4y) ?
The solution for y^'=-(5x^4-5)/(4y) is y=sqrt((-x^5+5x+c_{1))/2},y=-sqrt((-x^5+5x+c_{1))/2}