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Chapter 6: Problem 40

Weight Gain A calf that weighs \(w_{0}\) pounds at birth gains weight at the rate \(d w / d t=1200-w,\) where \(w\) is weight in pounds and \(t\) is time in years. Solve the differential equation.

Short Answer

Expert verified
The solution to the differential equation is \(w = 1200 + C e^{-t}\).

Step by step solution

01

Identify the type of differential equation

The equation given \(dw / dt = 1200 - w \) is a first-order linear differential equation.
02

Rewrite the differential equation

Rearrange the equation to standard form \(dw / dt + w = 1200\). This standard form shows that the equation is linear and will enable us to find an integrating factor.
03

Determine the integrating factor

The integrating factor for a first-order linear differential equation is \(e^{鈭玴(t) dt}\) where \(p(t)\) is the coefficient of \(y\) in the standard form. Here, \(p(t)\) is 1, so the integrating factor is \(e^{鈭�1 dt} = e^t\).
04

Multiply through by the integrating factor

Multiplying both sides of the equation by the integrating factor \(e^t\) we get: \(e^t dw/dt + e^t w = 1200 e^t\).
05

Simplify the left side of the equation

The left side of this equation is the derivative of \(e^t w\) with respect to \(t\), so we can rewrite the equation as: \((e^t w)' = 1200 e^t\).
06

Integrate both sides of the equation

Integrating, we get: \(e^t w = 1200 e^t + C\), where C is the constant of integration.
07

Solve for \(w\)

Finally, solve for \(w\): \(w = 1200 + C e^{-t}\).

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