Wichita State University Logo

Math 555: Differential Equations

1.2 More Solutions To Some Differential Equations

1.2.7 Field Mice and Owls

I hope it occurred to many of us that we do not normally observe unbounded exponential growth of a population of field mice in the wild. Let's add a predator to our model. Let us suppose that the local owl population consumes 15 mice per day to survive. Notice that this is a very simplistic model for the owls because normally the number of mice consumed would depend on the size of the population of owls. We will add this to our mathematical model in a future chapter. For now let's keep the number of mice per day constant at 15; that is 450 mice per month. Since the earlier model was a population growth per month we need to use the same units for our differential equation. This gives us the differential equation

$$\dfrac{dp}{dt} = \dfrac{p}{2} - 450$$
and the initial value problem

$$\dfrac{dp}{dt} = \dfrac{p}{2} - 450,\qquad p(0) = p_0.$$

1.2.8 First Solve the Differential Equation

Let's solve the initial value problem. To factor the right-hand side so that we can show that the differential equation is separable we will combine the terms into one fraction

$$\dfrac{dp}{dt} = \dfrac{p-900}{2}$$
This allows us to consider when the numerator is zero and possible divide both sides by the number $p-900$. Since $p-900=0$ only when $p=900$ we suspect we will see the equilibrium solution at $p=900$. Dividing both sides by the factor $p-900$ gives us

$$\dfrac{1}{p-900}\,\dfrac{dp}{dt} = \dfrac{1}{2},\qquad p\neq 900$$
Last time we didn't consider the possibility that the denominator might be negative and only searched for a solution starting from $t=0$ and continuing forward in time. However our direction field indicated that the integral curves were graphs of solutions for all real number $t$, even negative ones. Keeping this in mind let's integrate appropriately and obtain

$$\log|p-900| = \displaystyle\int\dfrac{1}{p-900}\,\dfrac{dp}{dt}\,dt = \displaystyle\int \dfrac{1}{2}\,dt = \dfrac{t}{2} + c.$$
Using the exponential function we obtain

$$\left| p - 900\right| = \exp(\log|p-900|) = \exp\left(\frac{t}{2} + c\right) = Ce^{\frac{t}{2}}.$$
Remember that we combine the integration constants into one big arbitrary constant $C$. Now we deal with absolute value. The absolute value function is defined by

$$ |x| = \left\{\begin{array}{lcr} \ \ \,x & \text{ when } & x\ge 0 \\ -x & \text{ when } & x < 0 \end{array}\right. $$
Combing our solution with the definition of absolute value gives us

$$ |p - 900| = \left\{\begin{array}{lcr} \ \ \,p - 900 & \text{ when } & p > 900 \\ -(p-900) & \text{ when } & p < 900 \end{array}\right. = \left\{ \begin{array}{lcr} C\,e^{\frac{t}{2}} & \text{ when } & p > 900 \\ C\,e^{\frac{t}{2}} & \text{ when } & p < 900 \end{array} \right. .$$
or

$$ p - 900 = \left\{ \begin{array}{lcr} \ \ \,C\,e^{\frac{t}{2}} & \text{ when } & p > 900 \\ -C\,e^{\frac{t}{2}} & \text{ when } & p < 900 \end{array} \right. .$$
Notice that $p-900 > 0$ corresponds to the case when the arbitrary constant is positive, $C > 0$, and $p-900 < 0$ corresponds to the case that the arbitrary constant is negative, $C < 0$. Since our arbitrary constant is a real number and can be either negative or positive, we can write our solution more succinctly

$$ p - 900 = C\,e^{\frac{t}{2}},\qquad p\neq 900.$$
Adding $900$ to both sides give us a general solution to our differential equation.

$$ p(t) = 900 + C\,e^{\frac{t}{2}},\qquad p\neq 900.$$
Notice that when $C=0$ we get the equilibrium solution $p(t)=900$, so this general solution applies to every case of a solution to our differential equation and describes all of the integral curves in our direction field. Thus our general solution to the differential equation is given by

$$ p(t) = 900 + C\,e^{\frac{t}{2}}.$$

1.2.9 Solve the Initial Value Problem

To solve the initial value problem we can evaluate our general solution at $t_0$ and solve for our constant $C$

$$p_0 = p(0) = 900 + Ce^0 = 900 + C.$$
Thus

$$C = p_0 - 900.$$
So

$$p(t) = 900 + (p_0 - 900)e^{\frac{t}{2}}.$$
Notice that if $p_0=900$ then the coefficient $p_0-900=0$ and we have the equilibrium solution

$$p(t) = 900.$$
If $p_0 > 900$ then the coefficient $p_0-900>0$ and the solution will grow without bound as $t$ approaches infinity giving us unbounded growth.

If $0 < p_0 < 900$ then the coefficient $p_0-900 < 0$ and the solution will grow without bound in the negative direction. This does not make sense to use because if we have zero mice then we don't create a "black hole" of mice that consumes all mice who might wander into the area. Instead let us ask, "When will the population of mice be extinct?" That is when will $p(t) = 0? $

Since we know that $-900 < p_0-900 < 0$ we will use $0 < 900-p_0 < 900$ instead. Setting $p(t) = 0$ we have

$$p(t) = 900 - (900 - p_0)e^{\frac{t}{2}} = 0,$$
or

$$(900 - p_0)e^{\frac{t}{2}} = 900.$$
Dividing both sides by $(900 - p_0)$ gives us

$$ e^{\frac{t}{2}} = \dfrac{900}{900-p_0} > 1$$
since $0 < 900-p_0 < 900$. Computing the logarithm of both sides gives us

$$ \dfrac{t}{2} = \log\left(\dfrac{900}{900-p_0}\right),$$
or

$$ t = 2\cdot\log\left(\dfrac{900}{900-p_0}\right) = \log\left(\dfrac{900}{900-p_0}\right)^2.$$

1.2.10 A Falling Object

Recall that our mathematical model for a falling object consisted of the differential equation


$$\dfrac{dv}{dt} = 9.8 - \dfrac{v}{5}$$
If we "drop" the falling object at time $t=0$ then we mean that don't impart an initial velocity on the object; we simply let go of it. Thus our initial value is $v(0) = 0$ and out initial value problem is given by

$$\dfrac{dv}{dt} = 9.8 - \dfrac{v}{5},\qquad v(0)=0.$$
To solve the separable differential equation we need to combine the terms on the right-hand side into one expression

$$\dfrac{dv}{dt} = \dfrac{49 - v}{5}$$
or

$$\dfrac{1}{49-v}\,\dfrac{dv}{dt} = \dfrac{1}{5}$$
Integrating both sides with respect to $t$ and using substitution we obtain

$$-\log\left|49 - v\right| = \displaystyle\int\dfrac{1}{49-v}\,\dfrac{dv}{dt}\,dt = \displaystyle\int\dfrac{1}{49-v}\,dp = \displaystyle\int\dfrac{1}{5}\,dt = \dfrac{t}{5} + c.$$
Multiplying both sides by $(-1)$ gives us

$$\log\left|49 - v\right| = \dfrac{t}{5} + c$$
because the negative of an arbitrary constant is an arbitrary constant. Using the exponential function gives us

$$ \left|49 - v\right| = Ce^{-\frac{t}{5}}.$$
Again since the arbitrary constant may be positive, negative or zero we get the general solution, or family of solutions

$$v(t) = 49 - Ce^{-\frac{t}{5}}.$$

1.2.11 Solving the Initial Value Problem

Using our initial condition $v(0)=0$ we get

$$ 0 = v(0) = 49 - Ce^0 = 49 - C$$
or

$$ C = 49 $$
Hence our unique solution to our initial value problem is given by

$$ v(t) = 49\left(1 - e^{-\frac{t}{5}}\right).$$
If we compute the limit as $t$ gets large without bound we have

$$ \lim_{t\rightarrow\infty} 49\left(1 - e^{-\frac{t}{5}}\right) = 49(1 - 0) = 49.$$
Thus every integral curve approaches the equilibrium solution $v(t) = 49$ as $t$ gets large without bound. Therefore we call this equilibrium solution stable , or a stable equilibrium solution. In physics we refer to this velocity as the terminal velocity of the object.

Creative Commons Logo - White


Your use of this self-initiated mediated course material is subject to our Creative Commons License .


Creative Commons Attribution-NonCommercial-ShareAlike 4.0

Creative Commons Logo - Black
Attribution
You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.

Creative Commons Logo - Black
Noncommercial
You may not use the material for commercial purposes.

Creative Commons Logo - Black
Share Alike
You are free to share, copy and redistribute the material in any medium or format. If you adapt, remix, transform, or build upon the material, you must distribute your contributions under the same license as the original.