$\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\$

From the initial condition,

$\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C$

So we have that $\ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1$

4 0

2 years ago

Consider the line y=- 2x+7.

iVinArrow [24]

Answer:

the slope is rhe same -2

Step-by-step explanation:

the slope is the opposite +1/2

3 0

2 years ago

Read 2 more answers

Serena paid $194.99 for a game system and then bought two equally-priced games. If she spent a total of $284.97, what is the pri

Nataly_w [17]

$284.99 - 4194.99 = $89.98
$89.98 ÷ 2 = $ 44.99

p = $44.99 for each game.

3 0

3 years ago