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SSSSS [86.1K]
2 years ago
13

Help please i don’t know how to do it

Mathematics
1 answer:
Norma-Jean [14]2 years ago
4 0

Answer:

Step-by-step explanation:

\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

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