What is the solution for y^'=x^5+2x^4-x-2 ?

The solution for y^'=x^5+2x^4-x-2 is y=(x^6)/6+(2x^5)/5-(x^2)/2-2x+c_{1}

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y^'=x^5+2x^4-x-2

y^{^{'}} = x^{5} + 2 x^{4} - x - 2

y = \frac{x ^{6}}{6} + \frac{2 x ^{5}}{5} - \frac{x ^{2}}{2} - 2 x + c_{1}

Solution steps

y^{'} (x) = x^{5} + 2 x^{4} - x - 2

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

then

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

y = \int x^{5} + 2 x^{4} - x - 2 d x

\int x^{5} + 2 x^{4} - x - 2 d x = \frac{x ^{6}}{6} + \frac{2 x ^{5}}{5} - \frac{x ^{2}}{2} - 2 x + c_{1}

y = \frac{x ^{6}}{6} + \frac{2 x ^{5}}{5} - \frac{x ^{2}}{2} - 2 x + c_{1}

What is the solution for y^'=x^5+2x^4-x-2 ?
The solution for y^'=x^5+2x^4-x-2 is y=(x^6)/6+(2x^5)/5-(x^2)/2-2x+c_{1}