1 Answer

Pradeep M · Answer 1 · 2023-04-18T06:53:52+0000

Separate the variables.

$(dy)/(dx) = x (x - 3) \implies dy = x (x - 3) \, dx$

Expand the right side.

$dy = (x^2 - 3x) \, dx$

Integrate both sides. On the right side, use the power rule on each term.

$\displaystyle \int dy = \int (x^2 - 3x) \, dx \implies y = \frac{x^3}3 - \frac{3x^2}2 + C$

Given that
y=3 when
x=2 , solve for
.

$3 = \frac{2^3}3 - \frac{3\cdot2^2}2 + C \implies C = \frac{19}3$

Then the particular solution is

$\boxed{y = \frac{x^3}3 - \frac{3x^2}2 + \frac{19}3}$

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