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Revision:Differential Equations

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Differential Equations

A differential equation is an equation which contains a derivative in (such as dy/dx). When given a differential equation, you will often be asked to 'solve' the differential equation or find the 'general solution'. This basically means find an expression which does not contain any derivatives. To do this you will need to integrate.

Example

Here is a popular one that appears in exams quite frequently,

According to Newton's law of cooling, the rate at which the temperature of a body falls is proportional to the amount by which the temperature exceeds that of it's surroundings.

A room is at a constant temperature of 20^{\circ}C. An object has temperature 80^{\circ}C when it is brought into the room and 5 minutes later it's temperature is 65^{\circ}C. What will it's temperature be after a further interval of 5 minutes?

Solution:

Let T^{\circ}C be the temperature of the object at time t minutes after being brought into the room.

\frac{dT}{dt}\propto T-20

Let k be the positive constant of proportionality.


\frac{dT}{dt}=-k(T-20)

We have -k in the formula because the temperature is falling.

Rearranging to put T on the left hand side,

\frac{1}{T-20}\frac{dT}{dt}=-k

Integrating both sides with respect to t,

\displaystyle \int \frac{1}{T-20}\ dT=-\int k\ dt

\ln(T-20)=-kt+c

T-20=Ae^{-kt}, by letting e^c=A.

We know that when t=0,T=80.

We use this to find A.

80-20=A

A=60, so

T=20+60e^{-kt}

When t=5, T=65,so

45=60e^{-5k}

e^{-5k}=\frac{3}{4}

-5k=\ln(\frac{3}{4})

k=\frac{1}{5}\ln(\frac{4}{3})

So the differential equation becomes,

T-20=60e^{-(\frac{1}{5}\ln(\frac{4}{3}))t}

When t=10,

T-20=60e^{-(\frac{1}{5}\ln(\frac{4}{3}))\times 10}

T=20+60e^{-2\ln(\frac{4}{3})}

T=20+60 \times \frac{9}{16}

T=53.75

After a further 5 mintues the temperature of the object is 53.75^{\circ}C.

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