Calculus II Lesson 12: Differential Equations

1. Presentations
2. Warm up
3. Upcoming
4. Differential Equations
   1. Example
   2. Solution
5. Overview
   1. Solutions to differential equations
   2. Order
   3. Example
   4. Solution
   5. Wrap-up
6. Separable Equations
   1. Example and check
   2. Mass Example
   3. Textbook Section
7. Linear First Order Differential Equations
   1. Examples from the book

Presentations

Warm up

Hand in (classwork):

1. Section 2.3 #145
2. Set up (don’t solve, just set it up) the integral to find the length of the curve $y = \frac{e^x + e^{-x}}{2}$ from x = -1 to x = 1.

Upcoming

• MyOpenMath HW due tonight (questions?)
• Quiz Thursday
• Exam 2 next week.

Differential Equations

• Scientific experiments: can be easier to observe changes to data
• Might not know the actual formula describing your data
• How a quantity changes may be related to the quantity itself
  • Ex: the more money you have, the more it earns in interest / investment

Sometimes:

• Can be easier to find the relationship between the change and the quantity
• Set up a differential equation

Example

• Suppose we are growing a bacteria culture.
• The larger the culture, the bigger the growth of the culture at any given time.
• Let $m(t)$ be the mass of the culture at time $t$. (Can easily measure its mass.)
• We notice: at any given $t$ (in seconds?), the culture grows at a rate $.05m(t)$
  • Idea: take lots of measurements, compare mass to previous mass. Look for a pattern.
• Initial mass: 2 grams. Mass in 10 minutes?

We have a differential equation:

\[m^\prime(t) = .05 m(t)\]

What functions are proportional to their own derivatives? Exponential functions! Guess $m(t) = Ae^{kt}$, and use $m^\prime$ and $m(0)$ to find these constants $A$ and $k$.

Solution

\[\begin{align} m(t) &= Ae^{kt} \\ m^\prime(t) &= Ake^{kt} \\ &= .05 m(t) \\ &= .05 Ae^{kt} \\ k &= .05 \end{align}\]

So we have:

• $m(t) = Ae^{.05t}$
• $m(0) = 2$

Find $A$?

\[\begin{align} Ae^{0} = 2 \\ A = 2 \end{align}\]

Solution:

• $m(t) = 2e^{.05t}$
• $m(10) = 2e^{.5} \approx 3.3$ grams.

Overview

• Any equation involving the derivative of a quantity is known as a differential equation.
• Example: $y^\prime = x$.
• A function $y = f(x)$ satisfying that equation is called a solution to the differential equation.
• In general: differential equations do not have unique solutions:
  • $y = \frac{x^2}{2}$ is a solution to $y^\prime = x$.
  • $y = \frac{x^2}{2} + 1$ is another solution.

Solutions to differential equations

• We look for a general form for a solution:
  • $y = \frac{x^2}{2} + C$.
• If we have an initial condition, we can find an exact solution.
  • Suppose $y^\prime = x$ and $y(0) = -1$.
  • Then $y = \frac{x^2}{2} + C$.
  • $-1 = \frac{0^2}{2} + C$
  • $C = -1$.
  • So: $y = \frac{x^2}{2} - 1$.

Order

The order of a differential equation is the highest derivative that appears in the equation. Exercise: Identify the orders of the following:

1. $y^\prime = y + x$
2. $(y^\prime)^2 = (y - x)^2$
3. $y^{\prime\prime} = -y$
4. $y^{\prime\prime\prime} + (y^\prime)^5 = y$

We will mostly focus on first-order differential equations, but some simple higher order differential equations can be solved.

Example

A ball is thrown straight into the air with an initial velocity of 5 meters per second. It is acted on by a constant, downward force of gravity, causing an acceleration of $-9.8$ $m/s^2$.

1. Find a formula $v(t)$ for the velocity at time $t$ seconds after the ball is thrown.
2. How many seconds after the ball is thrown does it reach its maximum height?
3. If the ball is thrown at a height of 2 meters above the ground, what is the maximum height reached?

Solution

Since $v^\prime(t) = a(t)$, and $a(t) = -9.8$, we know:

• $v^\prime(t) = -9.8$
• $v(0) = 5$.

So: $v(t) = -9.8t + C$. Since $v(0) = 5$, we can compute $5 = -9.8(0) + C$, so $C = 5$.

So our formula for the velocity is $v(t) = -9.8t + 5$. Going back to Calc I:

• the ball reaches its maximum height when its upward velocity is 0
• $v(t) = -9.8t + 5 = 0$
• Solve for $t$: $t = \frac{-5}{-9.8} \approx .51$ seconds.
• Derivative of height: velocity.
• $h^\prime(t) = -9.8t + 5$.
• So: $h(t) = -4.9t^2 + 5t + C$.
• $h(0) = 2$ means $-4.9(0) + 5(0) + C = 2$
• So $C = 2$

Therefore $h(t) = -4.9t^2 + 5t + 2$. Then $h(\frac{5}{9.8}) \approx 3.28$ meters. This is basically the exam 1 extra credit problem.

Wrap-up

• We really solved a second-order differential equation:
• $h^{\prime\prime}(t) = -9.8$
• Initial condition: $h(0) = 2$ and $h^\prime(0) = 5$.
• We did this by splitting it up into two steps:
  • First find $h^\prime(t)$
  • Then find $h(t)$.

Separable Equations

A differential equation of the form $y^\prime = f(x)g(y)$ is called a separable equation. We solve it by separating varaibles:

\[\frac{y^\prime}{g(y)} = f(x)\]

Then we integrate both sides: the left side with respect to $y$, and the right side with respect to $x$:

\[\int \frac{dy}{g(y)} = \int f(x) dx\]

Let’s do a simple example: $y^\prime = y$, with $y(0) = 1$. We separate variables and set up the integrals:

\[\int \frac{dy}{y} = \int 1 dx\]

We end up with:

\[\ln(y) = x + C\]

Solve this for $y$. Since the inverse of $\ln(y) = x$ is $e^x = y$, we have:

\[y = e^{x + C}\]

Since $e^{x+C} = e^x \cdot e^C$, and $e^C$ is just another constant, we re-write our solution as $y = A e^x$, where $A$ is a constant.

Now we plug in our initial conditions: $y(0) = 1$ means $1 = A e^0$, or $A = 1$. Therefore our solution is $y = e^x$.

As a side note: there are many equivalent ways of defining the number $e$. One way is the classic limit definition using compound interest: $e = \lim\limits_{n \rightarrow \infty} (1 + \frac{1}{n})^n$. Another way is to define $e$ using differential equations: here we used our knowledge of $e$ and $\ln$ to solve the above differential equation, but you can start with the differential equation $f^\prime(x) = f(x)$ and $f(0) = 1$, and define $e$ to be $f(1)$.

Each of these definitions can be used as a starting point, and then we can prove, mathematically, that the other properties of $e$ hold. That is, for example, we could start with $e$ using the differential equations definition, and then we can prove that $f(1)$ is exactly equal to that limit definition.

There are still other definitions of $e$ (the textbook provides another one in section 2.7). Why do mathematicians define $e$ in so many ways? Really because mathematicians tried, for a long time, to make mathematics rigorous. They wanted to find the exact axioms and definitions that we can start with, so that, making those assumptions, we can prove everything else there is to know about mathematics. Calculus was especially tricky to make rigorous; the history of mathematics often involves terms (like limits and continuity) being redefined to try to make the intuitive results actually provable.

Example and check

In this video I go through another example of a separable differential equation:

\[\begin{align} y^\prime &= \frac{x}{y} \\ y(0) &= 2 \end{align}\]

We solve this by separating variables and integrating:

\[\int y dy = \int x dx\]

Integrating:

\[\frac{y^2}{2} = \frac{x^2}{2} + C\]

Or $y^2 = x^2 + C$. Taking square roots, either $y = \sqrt{x^2 + C}$ or $y = -\sqrt{x^2 + C}$. Since $y(0) = 2 > 0$, we know that the second option is not possible, so it must be that $y = \sqrt{x^2 + C}$.

Now we plug in our initial condition: $y(0) = 2$ means that $\sqrt{C} = 2$. Therefore $C = 4$, and so the solution to the original equation is $y = \sqrt{x^2 + 4}$.

We can check that this solution works by plugging it back into the original differential equation. First, it’s clear that this solution satisfies the initial condition, $y(0) = 2$. Next, we show that it satisfies $y^\prime = \frac{x}{y}$.

Since $y = (x^2 + 4)^{\frac{1}{2}}$, we take the derivative. Using the power rule and chain rule: $y^\prime = \frac{1}{2} (x^2 + 4)^{-\frac{1}{2}} \cdot (2x)$, or $y^\prime = \frac{x}{\sqrt{x^2+4}}$. But this is exactly $\frac{x}{y}$, which is what we wanted to show!

Mass Example

Problem: A culture of bacteria grows at a rate proportional to its mass. If the culture initially weighs 10g, and in one hour, it weighs 15g, find a formula $m(t)$ for its mass (in grams) $t$ hours after that initial measurement.

I solve this problem in the above video. First, we set up a differential equation: $m^\prime(t) = km(t)$. We don’t know what $k$ is, but we know it must be a constant, since the problem states that $m^\prime$ is proportional to $m$. Then we have two initial conditions: $m(0) = 10$ and $m(1) = 15$.

We solve the equation by separating variables:

\[\begin{align} \int \frac{dm}{m} = \int k dt \\ \ln(m) = kt + C \\ m = e^{kt + C} = e^{kt} e^C \end{align}\]

Since $e^C$ is a constant, we can refer to this as another constant $A$, and our solution is $m(t) = Ae^{kt}$.

Notice that we have two unknown constants here: $A$ and $k$. We can use the two initial conditions to find it: $m(0) = 10$ means taht $Ae^{0} = 10$. Therefore $A = 10$, so $m(t) = 10 e^{kt}$.

$m(1) = 15$ means $10e^{k} = 15$, or $e^k = 1.5$. Therefore $k = \ln(1.5) \approx 0.41$. The solution, then, is $y = 10 e^{\ln(1.5) t}$. Using rules for exponents, $e^{\ln(1.5) t} = (e^{\ln(1.5)})^t$, and since $e^{\ln(x)} = x$, we can further simplify our solution to $m(t) = 10 \cdot 1.5^t$.

Textbook Section

Material on separable equations is covered in Section 4.3 in the textbook. There are several other worked examples there, I strongly encourage going through those as well as the exercises in that section.

Linear First Order Differential Equations

A differential equation of the form $a(x)y^\prime + b(x) y = c(x)$ is called a linear first-order differential equation. It’s first-order because, as we saw last time, the highest derivative that shows up is $y^\prime$ (first derivative). It’s linear because the powers of $y$ and $y^\prime$ are both 1. The standard form for a linear first order differential equation is $y^\prime + p(x) y = q(x)$. That is, when the coefficient of $y^\prime$ is just 1.

To solve a linear first order differential equation, we first notice that if we multiply the expression $y^\prime + p(x) y$ by an expression $\mu(x)$, we end up with $\mu(x) y^\prime + (\mu(x) p(x)) y$. Moreover, notice that using the product rule for derivatives $(\mu(x) y)^\prime = \mu(x) y^\prime + \mu^\prime(x) y$. That is, $\mu(x) (y^\prime + p(x) y) = (\mu(x) y)^\prime$ if $\mu^\prime(x) = \mu(x) p(x)$. In other words, if we can solve the differential equation $\mu^\prime = \mu p(x)$, and then multiply both sides by this factor we can simplify the differential equation to just $(\mu(x) y)^\prime = \mu(x) q(x)$. We then can solve that by integrating and then solving for $y$. Note that there is a formula to find $\mu(x)$, which we will see.

This is known as the integrating factor method. So the steps to solving a linear first order differential equation are:

1. Put the equation in standard form $y^\prime + p(x) y = q(x)$.
2. Find an integrating factor, $\mu(x) = e^{\int p(x) dx}$. Notice that this solves the differential equation $\mu^\prime(x) = \mu(x) p(x)$.
3. Multiply both sides by $\mu(x)$ so the equation becomes $(\mu(x) y)^\prime = \mu(x) q(x)$.
4. Integrate both sides: $\mu(x) y = \int \mu(x) q(x) dx + C$.
5. Divide by $\mu(x)$.

Let’s work through the example of $y^\prime + 2y = 1$, with $y(0) = 0$. Since it’s already in standard form, we start by finding $\mu(x)$. Since $p(x) = 2$, the integrating factor is $\mu(x) = e^{\int 2 dx}$, which is just $e^{2x}$. Notice that any antiderivative of $p(x)$ will work here.

Then we multiply both sides by $e^{2x}$: $(e^{2x} y)^\prime = e^{2x}$. Integrate and the equation is $e^{2x} y = \frac{1}{2} e^{2x} + C$. We can solve this now:

$y = \frac{e^2x + C}{2e^{2x}} = \frac{1}{2} + \frac{C}{e^{2x}}$, or $y = \frac{1}{2} + Ce^{-2x}$. Now we can use $y(0) = 0$ to find the solution: $y(0) = 0$ means that $\frac{1}{2} + C = 0$. So $C = -\frac{1}{2}$.

Therefore our final answer is $y = \frac{1}{2} - \frac{1}{2} e^{-2x}$.

Examples from the book

This particular method can be challenging. I encourage you to also follow along the examples in the textbook. Try to use these 5 steps before you see the textbook do the problem, and then follow along as the textbook solves these problems as well.