Chapter 5: Applications of Integration
Section 5.6: Differential Equations
Example 5.6.6
If y>0, solve the initial-value problem consisting of the differential equation y′t=k yt and the initial condition y0=y0.
Solution
Mathematical Solution
Separation of variables leads to
dyy
=k dt
∫dyy
= ∫k dt
ln(y)
=k t+c
y
=ek t+c
=ek t ec
=A ek t
The fourth line is obtained by exponentiating both sides of the equation in the third line. In the fifth line, the absolute value bars are dropped because y>0. If the initial condition y0=y0 is imposed, A=y0 and the solution to the IVP becomes y=y0 ek t.
The given differential equation is used to model radioactive decay and exponential (Malthusian) population growth. In these models, it is often stated that the equation y′=k y can be interpreted to mean that the rate of change is proportional to the amount of substance present.
If the substance present is uranium, then this interpretation implies a kind of anthropomorphism wherein an individual atom knows how big a body in which it resides, and adjusts its tendency to decay accordingly. It is much more appealing to interpret the differential equation as y′/y=k, and to say that the tendency to decay, per unit substance, is universally constant.
Likewise for population growth. To say that the rate of change of the population is proportional to the size of the population is less appealing than to say that the rate of change per unit of population is constant.
Maple Solution
Solution via Context Panel
Write the differential and the initial condition.
Context Panel: Solve DE≻yt
y′t=k yt,y0=y0→solve DEyt=y0ⅇkt
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