Differential equations

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Differential equations show you relationships between rates of functions.

A differential equation is simply an equation with a derivative in it. Here is an example:

\[ a\cdot f''(x) + b\cdot f'(x) + c\cdot f(x) = g(x). \]

What is a differential equation?

An equation that you take the derivative of. An equation that relates the rate of a function to other values. It is a formula for the slope of a tangent line at a given point.

When a mathematician solves a differential equation, they are finding functions satisfying the equation.

Which of the following functions solve the differential equation

\[ f^{(4)}(x) = f(x)? \]

Remember, $f^{(4)}$ is the fourth derivative of $f$.

$f(x) = \sin (x)$ $f(x) = x^2$ $f(x) = e^x$ $f(x) = e^{-x}$ $f(x) = \tan (x)$

It turns out that the complete solution to this differential equation is $c_1\sin (x)+c_2\cos (x)+c_3e^x+c_4e^{-x}$. In other words, every solution to this differential equation can be written in this form. You should check that these are all solutions (for example $f(x) = \sin (x)+3\cos (x)-7e^x+\pi e^{-x}$ is a solution). Proving that these are all of the solutions is beyond the scope of this course.

The differential equation above is an example of a fourth order differential equation, because the highest derivative in the equation is a “fourth” derivative. In general the highest derivative in a differential equation is the order.

Differential equations are one of the most practical objects of mathematical study. They appear constantly in every field of science and engineering. They are a powerful way to model many diverse situations.

1 Modeling with differential equations

Setting up differential equations is a skill to be acquired. However, you can try your hand with our next question.

Imagine that a glass of water has initial temperature $5^\circ \mathrm {C}$, and that the ambient temperature is $22^\circ \mathrm {C}$. The water will warm up over time. Assume that the rate of change in the temperature of the water is directly proportional to the difference between the current water temperature and the ambient temperature. Which of the following differential equations must be satisfied by the function $H(t)$ which measures the temperature of the water with respect to time?

$y' = 5+\frac {y}{22}$ $y' = k(22-y)$ for some $k>0$ $y' = k(y-22)$ for some $k>0$ $y' = k(5-y)$ for some $k>0$ $y' = k(y-5)$ for some $k>0$

This is just a straight translation job. “The rate of change in the temperature of the water” is $y'$. “Directly proportional to” means that it is equal to some constant (say $k$) times whatever it is proportional to. “The difference between the current water temperature and the ambient temperature” is either $22-y$ or $y-22$, since $y$ is the temperature of the water and $22$ is the ambient temperature. Think about which we should choose before looking at the next hint. Will it be $y'=k(22-y)$ or $y'=k(y-22)$ where $k>0$?

Since the temperature of the water is increasing over time, we want $y'>0$. Since the temperature will be increasing (and it is reasonable to assume it never surpasses the ambient temperature!) $22-y$ is positive. So we can conclude that $y' = k(22-y)$ for some $k>0$.

The differential equation does not involve the number $5$. If we wanted to incorporate that piece of data into our model we could ask “Which solution(s) to this differential equation satisfy $H(0) = 5$?” This is known as an initial value problem.

Sometimes the rates in question are constant.

One can approximate the force of gravity as constant near the Earth. So the acceleration of a falling object is a constant $g>0$. If $h(t)$ is the height of an object at time $t$, which differential equation must $h$ satisfy?

$y' = g$ $y=-g$ $y''=-g$ $y'' = g$ $y'' = -gy$

The acceleration of an object is the second derivative of its position, so the differential equation should say the second derivative, $y''$, is constant. Should it be a positive of negative constant?

A falling object will fall quicker and quicker, so the second derivative of its height should be negative. Thus $y''=-g$ is the correct answer.

2 Initial value problems

We have already seen, and solved, a particular kind of differential equation in this course. Namely a solution $F$ to the differential equation $y' = f(x)$ is just an antiderivative of $f$! We know the “general solution” of this differential equation is just $F(x)+C$, as long as the domain of $f$ is an interval. We can use this idea to solve differential equations of the form $y^{(n)} = f(x)$, by just repeatedly integrating and solving for “$C$” when we can.

Find the general solution to the differential equation $y''(x) = x$.

Since $y''(x) = x$, we know that \begin{align*} y'(x) &= \int x dx\\ &= \answer [given]{\frac {x^2}{2}} + C_1. \end{align*}

This further implies that

\[ y(x) = \int \frac {x^2}{2} + C_1 dx, \]

so we must have that

\[ y(x) = \answer [given]{\frac {x^3}{6}}+C_1x+C_2, \]

for some constant $C_2$. This is the general solution of the differential equation, in fact, every solution of this differential equation is of the form $y(x) = \frac {1}{6}x^3+C_1x+C_2$.

We can use the general solution to give specific solutions.

Find the solution to the differential equation $y''(x) = x$ that passes through the points $(0,1)$ and $(1,2)$.

From our work above we know that

\[ y(x) = \frac {x^3}{6}+C_1x+C_2. \]

To find the particular solution we are interested in, we solve the system of equations \begin{align*} \answer [given]{1}= y(\answer [given]{0}) &= \frac {1}{6}\cdot 0^3+C_1\cdot 0+C_2,\\ \answer [given]{2}= y(\answer [given]{1}) &= \frac {1}{6}\cdot 1^3+C_1\cdot 1+C_2. \end{align*}

The first line tells us that $C_2 = \answer [given]{1}$, and now the second line is

\[ 2= \frac {1}{6}+C_1+1. \]

and so $C_1 = \answer [given]{\frac {5}{6}}$. Our solution is now

\[ y(x) = \answer [given]{\frac {x^3}{6}+\frac {5 x}{6}+1}. \]

2025-01-06 20:08:11