Math 555: Differential Equations

1.2 Solutions To Some Differential Equations

---

1.2.1 Field Mice and Owls

Population models do not derive their equations from Newton's physical laws but are still based on common sense about the system we are modeling. A first principal of population models is that the growth rate of the population is directly proportional to the current population. This kind of assumption is called a simplifying assumption because it ignores a lot of very specific details of population growth and uses the growth rate, the proportionality constant, to approximate all of these interactions of these specific details into a easy to write equation,

$$\dfrac{dp}{dt} = \kappa p.$$
Here $\kappa$ is the proportionality constant or growth rate and $p(t)$ is the population at time $t$, while $\frac{dp}{dt}$ is the instantaneous rate of change of the population.

We can solve this differential equation.

1.2.2 Initial Conditions

If our proportionality constant is $\kappa = \frac{1}{2}$ then our differential equation becomes

$$\dfrac{dp}{dt} = \dfrac{p}{2}.$$
If the population at some point in time is $p(t_0) = 0$ then the population neither grows nor shrinks; it always stays zero. We know this from our differential equation

$$\dfrac{dp}{dt} = \dfrac{p(t)}{2} = \dfrac{0}{2} = 0.$$
So it appears we have an equilibrium solution when $p(t_0) = 0$. What about $t_0$? We can arrange our horizontal axis so that the time we are interested in, $t_0$, is the center of our axis. That is $t_0=0$ so our known or measured population is given by $p(0)=0$. It is so common to arrange our graph so that the known point of our differential equation is at $t_0=0$. We call this known value an initial condition . This is because we start from a time that we know the size of the population and use our model to predict the population for future times.

1.2.3 Initial Value Problems

A mathematical model that consists of a differential equation and initial condition(s) is called an Initial Value Problem . In our course we will often abbreviate an Ordinary Differential Equation with ODE , and an Initial Value Problem with IVP . In our example we have

$$\dfrac{dp}{dt} = \dfrac{p}{2}\text{,}\qquad p(0) = 0.$$
We have noted that the solution to our differential equation is $p(t) = 0$. This is due to the fact that

$$\dfrac{d}{dt}\,p(t) = \dfrac{d}{dt}\,0 = 0.$$
So

$$\dfrac{dp}{dt}= 0 = \dfrac{0}{2}$$
and

$$p(0) = 0.$$
Thus the function $p(t)=0$ makes the differential equation a true statement for all real numbers $t$ and satisfies the initial condition $p(0)=0$. That is to say, the function $p(t)=0$ solves the differential equation and satisfies the initial condition. Hence we say the function $p(t)=0$ solves the initial value problem.

1.2.4 Separable Differential Equations

Now our example of Field Mice is given by the initial value problem

$$\dfrac{dp}{dt} = \dfrac{p}{2},\qquad p(0) = p_0$$
where $p_0$ is the population we measure at time $t=0$. Let's solve this initial value problem. We are going to be using the chain rule and differentials so it may appear that we don't pay a lot of attention to the difference between the independent variable and the dependent variable when we are using algebra to manipulate our equations. We will also need to keep track of special cases that would invalidate our algebraic steps. For example we already noticed that if $p(0)=0$, then the solution to the differential equation is given by $p(t)=0$, a constant function.

If we want to consider other solutions then we can continue remembering that from this point in our computations $p(t)\neq 0$. We divide both sides of our differential equation by our non-zero $p$ to get

$$\dfrac{1}{p}\,\dfrac{dp}{dt} = \dfrac{1}{2}.$$
This differential equation is called separable because we can arrange for the independent variable to appear on one side of the equation and the dependent variable to appear on the other, $t^0 = 1$, so the independent variable is still on the right-hand side of the equation.

One significant difference between this class and many of your previous math classes is that the vocabulary we introduce here is an integral part of the material you are to learn. Engineers and Scientists use this vocabulary to discuss their mathematical models and the solutions that are used to interpret the physical world and predict its future behavior. When you see terms such as the ones in this page emphasized in bold memorize them and learn what they mean. We will continue to use these vocabulary terms throughout the course without re-defining them. This is to help you become familiar with them so that you can communicate with your fellow scientists and engineers.

If we can write a differential equation as

$$g(y)\,\dfrac{dy}{dt} = f(t)$$
so that only the dependent variable multiplied by its derivative appears on one side of the equation and only the independent variable appears on the other, then we call this differential equation a separable differential equation .

1.2.5 Solving our Separable Differential Equation

Let us take a look at the form of our differential equation so far.

$$\dfrac{1}{p}\,\dfrac{dp}{dt} = \dfrac{1}{2}.$$
We integrate both sides with respect to our independent variable $t$ to get

$$\displaystyle\int\dfrac{1}{p}\,\dfrac{dp}{dt}\,dt = \displaystyle\int\dfrac{1}{2}\,dt.$$
According to the chain rule $\frac{dp}{dt}\,dt = dp$ so we can re-write the last line

$$\displaystyle\int\dfrac{1}{p}\,dp = \displaystyle\int\dfrac{1}{2}\,dt.$$
Both are indefinite integrals so computing the anti-derivative of both sides gives us

$$\log(p) + c_1 = \dfrac{t}{2} + c_2.$$
Subtracting $c_1$ from both sides we obtain

$$\log(p) = \dfrac{t}{2} + c_2-c_1.$$
Here $c_1$ and $c_2$ are arbitrary constants; they are called arbitrary because their value does not affect whether the statement about the anti-derivative is true. Notice we have changed to the more modern notation for natural logarithm. If we want to compute a common logarithm or logarithm base 10 we will write $\text{log}_{10}(x)$. A logarithm without a subscript to denote its base is logarithm base $e$. Also the difference of two arbitrary constants is also an arbitrary constant so we will always combine them in one step and write the anti-derivative of both sides $$\log(p) = \dfrac{t}{2} + c.$$

Since the natural logarithm function and the exponential function are inverses of each other for all value greater than or equal to zero, we can write

$$\exp\left(\log(p)\right) = \exp\left(\dfrac{t}{2}+c\right)$$ or $$e^{\log(p)} = e^{\frac{t}{2}+c}.$$
In either case $\exp(\log(p)) = p$ so we have

$$p(t) = e^{\frac{t}{2}+c} = e^{\frac{t}{2}}\cdot e^{c} = C\,e^{\frac{t}{2}}.$$
Here $e$ raised to the power of an arbitrary constant is an arbitrary constant $C = e^c$. This gives us a general solution to the differential equation. Notice that the equation doesn't provide a single solution to our differential equation. It gives us infinitely many solutions since every real number $C$ will give us a solution. We can prove to ourselves that any of these functions are solutions by plugging them into the differential equation:

$$ \dfrac{dp}{dt} = p'(t) = \dfrac{1}{2}\,C\,e^{\frac{t}{2}} = \dfrac{p}{2}.\,{\huge \color{#307fe2}{\checkmark}}$$
We call this infinite set of solutions to our differential equation

$$p(t) = C\,e^{\frac{t}{2}}$$
a family of solutions or general solution .

1.2.6 Solving our Initial Value Problem

We solved our first differential equation and now we want to solve our first initial value problem.

$$\dfrac{dp}{dt} = \dfrac{p}{2},\qquad p(0) = p_0.$$
To solve the initial value problem we will "plug" our initial condition into the differential equation

$$p_0 = p(0) = C\,e^{\frac{0}{2}} = C\,e^0 = C.$$
Now our constant isn't arbitrary anymore. The value must be equal to $p_0$ from our calculus. We now have a unique solution (or integral curve) $$p(t) = p_0\,e^{\frac{t}{2}}.$$
This is the unique solution to our initial value problem.

Integral Curves and Solutions to Initial Value Problems

When we graph the direction field we see that there are infinitely many integral curves, each corresponding to an integral curve.

However none of these integral curves intersect; they never touch or cross each other. This is because if they were to touch they would share a value in common. If we have two integral curves $y = p_1(t)$ and $y = p_2(t)$ and they have a point in common at $t = t_{*}$, $$p_1(t_{*}) = p_2(t_{*})$$
However both of these solutions solve the differential equation so we have that

$$p_1'(t_{*}) = \dfrac{dp_1}{dt}(t_{*}) = \dfrac{p_1(t_{*})}{2} = \dfrac{p_2(t_{*})}{2} = \dfrac{dp_2}{dt}(t_{*}) = p_2'(t_{*}).$$
Moreover we can calculate the second derivative of our solutions using the differential equation

$$\dfrac{d^2p}{dt^2} = \dfrac{d}{dt}p'(t) = \dfrac{d}{dt}\left(\dfrac{dp}{dt}\right) = \dfrac{d}{dt}\left(\dfrac{p}{2}\right) = \dfrac{1}{2}\dfrac{dp}{dt} = \dfrac{p}{4}.$$
We can evaluate the second derivative of each solution at $t_{*}$ to get

$$p_1''(t_{*}) = \dfrac{d^2p_1}{dt^2}(t_{*}) = \dfrac{p_1(t_{*})}{4} = \dfrac{p_2(t_{*})}{4} = \dfrac{d^2p_2}{dt^2}(t_{*}) = p_2''(t_{*}).$$
We can compute as many derivatives as we want and get that the n$^{th}$ derivative of $p_1(t_{*})$ and $p_2(t_{*})$ are equal.

$$p_1^{(n)}(t_{*}) = \dfrac{d^np_1}{dt^n}(t_{*}) = \dfrac{p_1(t_{*})}{2^{n+1}} = \dfrac{p_2(t_{*})}{2^{n+1}} = \dfrac{d^np_2}{dt^n}(t_{*}) = p_2^{(n)}(t_{*}).$$
If we recall our calculus we have from Taylor's theorem that we can represent an analytic function using their derivatives if they have them as
$$f(t) = \displaystyle\sum_{n=0}^{\infty}\dfrac{f^{(n)}(x_{*})}{n!}(x - x_{*})^n.$$ Thus we have that $$p_1(t) = \displaystyle\sum_{n=0}^{\infty}\dfrac{p_1^{(n)}(x_{*})}{n!}(x - x_{*})^n = \displaystyle\sum_{n=0}^{\infty}\dfrac{\frac{p_1(t_{*})}{2^{n+1}}}{n!}(x-x_{*})^n = \dfrac{p_1(x_{*})}{2}\displaystyle\sum_{n=0}^{\infty}\dfrac{(x-x_{*})^n}{2^n\cdot n!}$$
and that
$$p_2(t) = \dfrac{p_2(x_{*})}{2}\displaystyle\sum_{n=0}^{\infty}\dfrac{(x-x_{*})^n}{2^n\cdot n!}$$
Since $p_1(t_{*}) = p_2(t_{*})$ we have $p_1(t) = p_2(t)$, they are the same solution. This should prove to us that the integral curves we see on our graph never touch at all. It also tells us that there is only one integral curve that passes through the point $(0, p(0))$ on our graph of the direction field. This unique solution we found to be $$p(t) = p_0\,e^{\frac{t}{2}}.$$ We could take any point on the graph $(t_{*}, p_{*})$ and there would be a unique integral curve, unique solution, to our initial value problem given by solving for our arbitrary constant. $$p_{*} = p(t_{*}) = C\,e^{\frac{t_{*}}{2}}$$ so $$C = p_{*}\,e^{-\frac{t_{*}}{2}}$$ Our arithmetic is easier if we pick $t_*=0$ but that is not necessary.

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

Creative Commons Attribution-NonCommercial-ShareAlike 4.0


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.


Noncommercial
You may not use the material for commercial purposes.


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.