What is the solution for (dy)/(dx)=-x^4-4x^3+5x^2 ?

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

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(dy)/(dx)=-x^4-4x^3+5x^2

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

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

Solution steps

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = - x^{4} - 4 x^{3} + 5 x^{2}

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

then

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

y = \int - x^{4} - 4 x^{3} + 5 x^{2} d x

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

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

\frac{d}{d x} (3 x^{2} + 8 x + 5)

\frac{\partial}{\partial y} (6 x y - 5 y)

- 10 x + 11, x^{2} - 5 x - 13, - 8, 3

x \to 0 lim (\frac{7}{x} - 7 cot (x))

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