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Vlada [557]
3 years ago
7

A function y(t) satisfies the differential equation dy dt = y4 − 6y3 + 5y2.

Mathematics
1 answer:
mojhsa [17]3 years ago
8 0

\dfrac{\mathrm dy}{\mathrm dt}=y^4-6y^3+5y^2=y^2(y-1)(y-5)

a. If y(t) is constant, then the derivative 0, which means we would have

y=0\text{ or }y=1\text{ or }y=5

as constant solutions.

Next, we have 4 possible intervals to consider where the derivative doesn't vanish:

  • for t, we have \frac{\mathrm dy}{\mathrm dt}>0 (consider the sign of the derivative for, say, y=-1);
  • for 0, we have \frac{\mathrm dy}{\mathrm dt}>0;
  • for 1, we have \frac{\mathrm dy}{\mathrm dt};
  • and for t>5, we have \frac{\mathrm dy}{\mathrm dt}>0

Taking all these facts together, we see that ...

b. y is increasing on the interval (-\infty,0)\cup(0,1)\cup(5,\infty), and

c. y is decreasing on the interval (1,5).

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