use the differential equation given by dy/dx = xy/3 , y > 0. (help needed)

Question

For Questions 1–3, use the differential equation given by  , y > 0.

1. Complete the table of values

2.) On the axes below, sketch a slope field for the given differential equation using the nine points on the table that you found.

3.find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 4.

Use the following information for Questions 4 and 5.

Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. So the rate of cooling for a bottle of lemonade at a room temperature of 75°F which is placed into a refrigerator with temperature of 38°F can be modeled by where T(t) is the temperature of the lemonade after t minutes and T(0) = 75. After 30 minutes the lemonade has cooled to 60°F, so T(30) = 60.

4.) To the nearest degree, what is the temperature of the lemonade after an additional 30 minutes?

5.) To the nearest minute, how long does it take for the lemonade to cool to 55°F?

Answer 1

1) Here you are just plugging in values for x, y to get xy/3 = dy/dx.

2) Here you are plotting those values. For instance, x = 1, y = -1 ⇒ dy/dx = -1/3. Plot this as a line segment with slope of -1/3.

3) dy/dx = xy/3 ⇒ ∫dy/y = ∫xdx/3 ⇒ ln(y) = x²/6 + C. y(0) = 4 ⇒ C = 4.

4) We can express "rate of temperature change is proportional to difference in temperature between object and surroundings" as dT/dt = A(T-T_env).

Thus, ∫ dt/(T-T_env) = ∫Adt ⇒ ln(T-T_env) = At + ln(C) ⇒ T = T_env + Ce^At. T(0) = 75 ⇒ 75 = 38 + C ⇒ C = 37. T(30) = 60 ⇒ 60 = 38 + 37e^A(30) ⇒ ln(22) = ln(37) + A(30) ⇒ A = 1/30 ln(22/37) ⇒ T = 38 + 37 e^{ln(22/37)t/30} = 38 + 37 (22/37)^t/30

This gives T(60) = 38 + 37 (22/37)² = 38 + 37 (0.3535) ≈ 51 F°.

5) 55 = 38 + 37 (22/37)^t/30 ⇒ ln(17) = ln(37) + t/30 ln(22/37) ⇒ 30 ln(17/37)/ln(22/37) = t = 44.88 ≈ 45 mins.