A calf that weighs wo pounds at birth gains weight at the rate shown below, wher

Question

A calf that weighs wo pounds at birth gains weight at the rate shown below, where w is weight in pounds and t is time in years. Dw/dt = 1500 - w Solve the differential equation. (Use w_O for w0 as necessary.) The rate of change in the number of miles s of road cleared per hour by a snowplow is inversely proportional to the depth h of snow. That is, Ds/dh = k/h Find s as a function of h given that s = 26 miles when h = 2 inches and s = 9 miles when h = 6 inches (2 less than equal to h less than equal to 15). s(h) =

Explanation / Answer

dw/dt + w = 1500

dw/dt + Pw = Q

P = 1 and Q = 1500

Integrating factor = e^(integral of 1) --> e^t

Multiply the given DE by e^t on both sides :

e^t * [dw/dt + w = 1500]

e^t*dw/dt + we^t = 1500e^t

d/dt(we^t) = 1500e^t

Integrating :

we^t = 1500e^t + C

Divide all over by e^t :

w = 1500 + Ce^(-t)

At t = 0, w = w0 :

w0 = 1500 + C
C = w0 - 1500

So, w = 1500 + (w0 - 1500)e^(-t)

w = w0*e^(-t) + 1500(1 - e^(-t)) ----> ANSWER

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ds/dh = k/h
ds = k*dh/h
Integrating :
s = k*ln|h| + C

26 = k*ln(2) + C
9 = k*ln(6) + C

Subtracting :

17 = k*(ln(2) - ln(6))

17 = -k * ln(3)

k = -17/ln(3)

9 = k*ln(6) + C
9 = -17ln(6)/ln(3) + C
C = 9 + 17ln(6)/ln(3)
C = (9ln(3) + 17ln(6))/ln(3)

So, solution becomes :

s = k*ln|h| + C

Plug in for k and C :

s(h) = (-17/ln(3))*ln|h| + (9ln(3) + 17ln(6))/ln(3) ---> ANSWER

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