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

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

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y^'=(x^4)/y

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{'}} - \frac{1}{2} y = 2 cos (x)

x \to 1 lim (\frac{4}{2 x - 2})

\frac{\partial}{\partial x} (6 x^{2} y^{2} + 5 x + 8 y^{3})

\int \frac{6 x ^{3} - 5}{x} d x

x \to 1 lim (\frac{x ^{15} - 1}{x ^{5} - 1})

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