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

The solution for (dy)/(dx)=(x^2-3x)^3 is y=(x^7)/7-(3x^6)/2+(27x^5)/5-(27x^4)/4+c_{1}

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(dy)/(dx)=(x^2-3x)^3

\frac{d y}{d x} = (x^{2} - 3 x)^{3}

y = \frac{x ^{7}}{7} - \frac{3 x ^{6}}{2} + \frac{27 x ^{5}}{5} - \frac{27 x ^{4}}{4} + c_{1}

Solution steps

\frac{d y}{d x} = (x^{2} - 3 x)^{3}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

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

then

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

y = \int (x^{2} - 3 x)^{3} d x

\int (x^{2} - 3 x)^{3} d x = \frac{x ^{7}}{7} - \frac{3 x ^{6}}{2} + \frac{27 x ^{5}}{5} - \frac{27 x ^{4}}{4} + c_{1}

y = \frac{x ^{7}}{7} - \frac{3 x ^{6}}{2} + \frac{27 x ^{5}}{5} - \frac{27 x ^{4}}{4} + c_{1}

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What is the solution for (dy)/(dx)=(x^2-3x)^3 ?
The solution for (dy)/(dx)=(x^2-3x)^3 is y=(x^7)/7-(3x^6)/2+(27x^5)/5-(27x^4)/4+c_{1}