What is the solution for y^'=x(x-1)(x-5) ?

The solution for y^'=x(x-1)(x-5) is y=(x^4)/4-2x^3+(5x^2)/2+c_{1}

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y^'=x(x-1)(x-5)

y^{^{'}} = x (x - 1) (x - 5)

y = \frac{x ^{4}}{4} - 2 x^{3} + \frac{5 x ^{2}}{2} + c_{1}

Solution steps

y^{'} (x) = x (x - 1) (x - 5)

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int x (x - 1) (x - 5) d x

\int x (x - 1) (x - 5) d x = \frac{x ^{4}}{4} - 2 x^{3} + \frac{5 x ^{2}}{2} + c_{1}

y = \frac{x ^{4}}{4} - 2 x^{3} + \frac{5 x ^{2}}{2} + c_{1}

\int \frac{y}{1 + x ^{2} y ^{2}} d x

\int \frac{- cos ( n x )}{n} d x

t \frac{d y}{d t} = t^{3} + 24 t^{3} y

\frac{\partial}{\partial x} (7 y cos (x))

n = 2 \sum \infty \frac{5}{3 ^{n}}

What is the solution for y^'=x(x-1)(x-5) ?
The solution for y^'=x(x-1)(x-5) is y=(x^4)/4-2x^3+(5x^2)/2+c_{1}