What is the solution for y^'=y(xy^5+2) ?

The solution for y^'=y(xy^5+2) is y=\sqrt[5]{(20e^{10x})/(-10xe^{10x)+e^{10x}+c_{1}}}

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y^'=y(xy^5+2)

y^{^{'}} = y (x y^{5} + 2)

y = 5 \frac{20 e ^{10 x}}{- 10 x e ^{10 x} + e ^{10 x} + c _{1}}

Solution steps

y^{'} (x) = y (x y^{5} + 2)

Solve Bernoulli ODE:

y = 5 \frac{20 e ^{10 x}}{- 10 x e ^{10 x} + e ^{10 x} + c _{1}}

y = 5 \frac{20 e ^{10 x}}{- 10 x e ^{10 x} + e ^{10 x} + c _{1}}

f (t) = t cos (t)

\frac{\partial}{\partial y} (sin (x))

f (x) = - 3 x^{3} + 4, at x = - 1

\frac{d}{d x} (nπ x)

\frac{d}{d x} (\frac{2}{1 + x})

What is the solution for y^'=y(xy^5+2) ?
The solution for y^'=y(xy^5+2) is y=\sqrt[5]{(20e^{10x})/(-10xe^{10x)+e^{10x}+c_{1}}}