$\text { A calf that weighs } 70 \text { pounds at birth gains weight at the rate } \frac { d w } { d t } = k ( 1200 - w ) \text {, }$ where $w$ is weight in pounds and $t$ is time in years. Use a computer algebra system to solve the differential equation for $k = 0.9$.
A) $w = 70 e ^ { - 0.9 t }$
B) $w = 1130 e ^ { - 0.9 t }$
C) $w = 1200 - 1130 e ^ { - 0.9 t }$
D) $w = 1130 - 0.9 e ^ { - t }$
E) $w = 1200 e ^ { t } - 1130 e ^ { - 0.9 t }$

Question

Quizplus · Accepted Answer

The differential equation $\frac{dw}{dt} = 0.9(1200 - w)$ is a first-order linear differential equation. Solving it using separation of variables or an integrating factor method, and applying the initial condition $w(0) = 70$, leads to the solution $w = 1200 - 1130e^{-0.9t}$. This represents the weight of the calf approaching 1200 pounds as time goes on, with a rapid initial weight gain that slows as the calf's weight approaches 1200 pounds.

A calf that weighs 70 pounds at birth gains weight at the rate dwdt=k(1200−w), \text { A calf that weighs } 70 \text { pounds at birth gains weight at the rate } \frac { d w } { d t } = k ( 1200 - w ) \text {, } A calf that weighs 70 pounds at birth gains weight at the rate dtdw​=k(1200−w),

$\text { A calf that weighs } 70 \text { pounds at birth gains weight at the rate } \frac { d w } { d t } = k ( 1200 - w ) \text {, }$