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

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

Helllp im doing my exam !

Tcecarenko [31]

log x ( 5 × 12 ) = 7

x^7 = 60

..................................

7 0

3 years ago

Write the equation for the graph

melamori03 [73]

Answer:

i think its (4, -4) hope that helps u :0

Step-by-step explanation:

6 0

1 year ago

Read 2 more answers

Geometry help!! Hurry please

omeli [17]

I think is c I'm sorry if is wrong

5 0

2 years ago

Why is this function nonlinear?​

Why is this function nonlinear?