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Lecture Notes in Differential Equations - Bruce E. Shapiro

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49<br />

This is identical to<br />

with<br />

Therefore the voltage is given by<br />

dV<br />

dt = V ∞ − V<br />

τ<br />

(6.30)<br />

V ∞ = V batt (6.31)<br />

τ = RC (6.32)<br />

V = V batt + (V 0 − V batt )e −t/RC (6.33)<br />

where V 0 is the voltage at t = 0. Regardless of the <strong>in</strong>itial voltage, the<br />

voltage always tends towards the battery voltage. The current through the<br />

circuit is<br />

i = C dV<br />

dt = V 0 − V batt<br />

e −t/RC (6.34)<br />

R<br />

so the current decays exponentially.<br />

Example 6.3. Newton’s Law of Heat<strong>in</strong>g (or Cool<strong>in</strong>g) says that the rate<br />

of change of the temperature T of an object (e.g., a potato) is proportional<br />

to the difference <strong>in</strong> temperatures between the object and its environment<br />

(e.g., an oven), i.e.,<br />

dT<br />

dt = k(T oven − T ) (6.35)<br />

Suppose that the oven is set to 350 F and a potato at room temperature<br />

(70 F) is put <strong>in</strong> the oven at t = 0, with a bak<strong>in</strong>g thermometer <strong>in</strong>serted <strong>in</strong>to<br />

the potato. After three m<strong>in</strong>utes you observe that the temperature of the<br />

potato is 150. How long will it take the potato to reach a temperature of<br />

350?<br />

The <strong>in</strong>itial value problem we have to solve is<br />

Divid<strong>in</strong>g and <strong>in</strong>tegrat<strong>in</strong>g,<br />

From the <strong>in</strong>itial condition<br />

}<br />

T ′ = k(400 − T )<br />

T (0) = 70<br />

∫<br />

∫<br />

dT<br />

400 − T =<br />

(6.36)<br />

kdt (6.37)<br />

− ln(400 − T ) = kt + C (6.38)<br />

C = − ln(400 − 70) = − ln 330 (6.39)

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