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

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

What is the coordinates of {2, 1} and {2, 4}?

blondinia [14]

Xm=(2+2)/2=2 Ym=(1+4)/2=2.5 (2,2.5)

7 0

3 years ago

Darrien read 97 pages last week.Evan read pages last week.How many 8 my 6. pages did, the boys read?

const2013 [10]

I’m really confused about this

7 0

3 years ago

A farmer sells 8.9 kilograms of pears and apples at the farmer's market.

Vlad1618 [11]

Answer:

2.225 kg

Step-by-step explanation:

We know that the total weight the farmer sold was 8.9 kilograms.

- 3/4 of this weight is apples, and the rest is pears.

Therefore, to calculate the amount of kilograms of pears she sold at the farmer's market (which you can call x), you need to apply the following procedure.

7 0

3 years ago