MATH 2410 — Differential Equations
Lecture Notes. This is NOT meant to be a substitute for coming to lecture or consulting the textbook directly, but rather a helpful resource to review for exams/ review what was discussed in lecture.
Lecture 8: Section 3.1 - Mathematical Modeling (Linear) ↑ Back to top
In this lecture, we will use some of the methods discussed in Lecture 3 on how to construct models for certain real-world situations/systems and now solve those differential equations.
Example: Radioactive Decay
Suppose that we have a substance that decays at a rate proportional to the amount present at time $t$, and that has a half-life of $3.3$ hours. If $1$g of the substance is present initially, how long will it take for $90\%$ of the substance to decay?
First, we define the variables:
\[ \begin{aligned} A(t) &= \text{amount of substance present at time } t \\ t &= \text{time in hours} \\ A'(t) &= \text{rate of decay} \end{aligned} \]The statement “decays at a rate proportional to the amount present” gives
\[ \frac{dA}{dt} = -kA, \quad k>0. \]The half-life of $3.3$ hours means
\[ A(3.3) = \frac{1}{2}A(0). \]Since initially $A(0)=1$, we obtain
\[ A(3.3) = \frac{1}{2}. \]If $90\%$ decays, then $10\%$ remains, so we want
\[ A(t) = \frac{1}{10}. \]The differential equation is separable, giving the general solution
\[ A(t) = Ce^{-kt}. \]Using the initial condition:
\[ 1 = A(0) = C. \]Using the half-life:
\[ \frac{1}{2} = e^{-k(3.3)} \quad \Rightarrow \quad k = \frac{-1}{3.3}\ln\!\left(\frac{1}{2}\right). \]Finally:
\[ e^{-kt} = \frac{1}{10} \quad \Rightarrow \quad t = \frac{-1}{k}\ln\!\left(\frac{1}{10}\right) \approx 10.96 \text{ hours}. \]This type of model is used for exponential growth and decay processes, including population growth.
Example: SI Model for Infectious Disease
Suppose that we have a fixed isolated population of $n$ people and we introduce one infected person into the population. The rate of spread of the disease is proportional to the number of interactions between susceptible and infected individuals.
Define the variables:
\[ \begin{aligned} S(t) &= \text{number of susceptible individuals at time } t \\ I(t) &= \text{number of infected individuals at time } t \\ t &= \text{time} \\ I'(t) &= \text{rate of spread of the disease} \end{aligned} \]The statement that the disease spreads proportional to encounters between susceptible and infected individuals gives:
\[ \frac{dI}{dt} = kSI. \]Since the population is fixed and starts with one infected person:
\[ S + I = n + 1 \] \[ I(0)=1, \quad S(0)=n. \]Substituting $S = n+1-I$ gives the model:
\[ I'(t) = kI(n+1-I), \qquad I(0)=1. \]This is a logistic-type differential equation that models the spread of infection over time.
Lecture 9: Section 3.2 - Mathematical Modeling (Nonlinear) ↑ Back to top
So far, many models we have studied were linear differential equations. However, real systems often depend on multiple interacting factors, leading to nonlinear models.
General Nonlinear Growth Model
Suppose a growth rate depends both on the amount present and another factor that itself depends on the amount present. Then we may have a model of the form
This is nonlinear because the dependent variable appears multiplied by a function of itself.
Carrying Capacity and Logistic Equation
Suppose an environment has a carrying capacity for a population. Then we model
where
and \(K\) is the carrying capacity.
A common choice is a linear function:Solving the Logistic Equation
Separate variables:Example — Disease Spread
A student carrying a flu virus returns to a school of 1000 students. The rate of spread is proportional to both infected students and not infected students.
Model:Adjusting Logistic Models
Suppose a constant number of individuals is removed (harvesting):- Separable
- Autonomous
Lecture 10: Section 3.3 - Mathematical Modeling (Systems of Linear Equations) ↑ Back to top
Definition of a System
A system of first-order differential equations has the formWhy Systems?
Systems model interactions between multiple quantities, such as:- Predator–prey populations
- Mixing tanks
- Multi-step radioactive decay
- Economic models
Example — Radioactive Decay Chain
Suppose substances decay in sequence:X → Y → Z
Model:Example — Mixing Tanks
Two tanks exchange fluid. Let:- \(x_1(t)\) = salt in tank A
- \(x_2(t)\) = salt in tank B
Key Takeaways
- Nonlinear models often arise when growth depends on interactions.
- The logistic equation models limited growth with carrying capacity.
- Systems of ODEs model multiple interacting variables.
- Linear systems have special structure that simplifies solving.
Lecture 11: Section 4.1 - Higher-Order Differential Equations ↑ Back to top
In the previous lectures, we have looked at methods to solve certain types of first-order differential equations. Now, we will look at methods of solving higher-order linear differential equations. Let’s begin with a definition.
A linear \(n\)th-order differential equation of the form \[ a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdots + a_1(x)y'+a_0(x)y = 0 \] is said to be homogeneous. Otherwise, we have \[ a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\cdots + a_1(x)y'+a_0(x)y = g(x) \] where \(g(x) \neq 0\), which is said to be nonhomogeneous.
Here we insist that the functions \(a_i(x)\) depend only on \(x\) and are continuous on the interval we are considering.
All homogeneous differential equations have a trivial constant solution \(y=0\). So we are more interested in finding non-constant solutions. From now on, when we refer to “solutions,” we mean a nontrivial solution.
When we find one solution to a higher-order linear differential equation, we actually find infinitely many. Suppose that \(y\) is a solution to \[ a_n(x)y^{(n)}+\cdots + a_1(x)y'+a_0(x)y =0 \] and \(c\) is a constant. Then \[ a_n(x)(cy)^{(n)}+\cdots+a_0(x)(cy) = c\left(a_n(x)y^{(n)}+\cdots+a_0(x)y\right)=0. \]
This reveals that for any constant \(c\), the curve \(cy\) is also a solution. This follows from the linearity of derivatives: \[ (cy)' = cy', \qquad (y_1+y_2)' = y_1'+y_2'. \]
Similarly, if \(y_1\) and \(y_2\) are solutions, then \[ a_n(x)(y_1+y_2)^{(n)}+\cdots+a_0(x)(y_1+y_2)=0, \] so \(y_1+y_2\) is also a solution.
Let \(y_1, \dots, y_n\) be solutions to a homogeneous differential equation. Then for any constants \(c_1, \dots , c_n\), \[ y = c_1y_1 + \cdots + c_ny_n \] is also a solution.
The question remains: how do we find all the solutions?
A set of functions \(f_1(x), \dots, f_n(x)\) are said to be linearly dependent on an interval \(I\) if there exist constants \(c_1, \dots, c_n\) (not all zero) such that \[ c_1f_1(x) + \cdots + c_nf_n(x) = 0 \] for all \(x\) in \(I\). Otherwise, the set is said to be linearly independent.
To find all solutions, we need a set of linearly independent solutions. Any solution can then be written as a linear combination of these.
Show that \(f_1(x)=x\) and \(f_2(x)=x^2\) are linearly independent.
We set \[ c_1x+c_2x^2=0. \] Factoring gives \[ x(c_1+c_2x)=0. \] This would require \(c_2=-c_1/x\), which is not constant. Thus the only constants that work are \(c_1=c_2=0\), so the functions are linearly independent.
An alternative way to check linear independence is using the Wronskian.
The functions \(f_1,\dots,f_n\) are linearly independent if \[ W(f_1,\dots,f_n)= \det \begin{bmatrix} f_1 & \cdots & f_n\\ \vdots & \ddots & \vdots\\ f_1^{(n-1)} & \cdots & f_n^{(n-1)} \end{bmatrix} \neq 0. \]
For the previous example, \[ W(x,x^2)= \det \begin{bmatrix} x & x^2\\ 1 & 2x \end{bmatrix} = x(2x)-x^2(1)=x^2 \neq 0. \] Thus the functions are linearly independent.
For two functions, remember \[ \det \begin{bmatrix} a & b\\ c & d \end{bmatrix} = ad-bc. \]
Any set of \(n\) linearly independent solutions to an \(n\)th-order homogeneous differential equation is called a fundamental set of solutions.
Given a fundamental set of solutions, the general solution is \[ y = c_1f_1 + \cdots + c_nf_n \] where \(c_1,\dots,c_n\) are constants.
The differential equation \[ y''+y=0 \] has general solution \[ y=c_1\sin(x)+c_2\cos(x). \]
We check independence using the Wronskian: \[ \det \begin{bmatrix} \sin(x) & \cos(x)\\ \cos(x) & -\sin(x) \end{bmatrix} = \sin^2(x)+\cos^2(x)=1\neq 0. \]
Since the equation is second order, we only need two linearly independent solutions.
Lecture 12: Section 4.2 - Reduction of Order ↑ Back to top
In this lecture, we discuss a method called reduction of order, which allows us to find a second linearly independent solution of a second-order linear homogeneous differential equation when one solution is already known.
Idea of the Method
We assume the second solution has the form \[ y_2 = v(x)y_1(x), \] where \(v(x)\) is an unknown nonconstant function to be determined.
Then via product rule \[ y_2' = v'y_1 + vy_1', \quad y_2'' = v''y_1 + 2v'y_1' + vy_1''. \]
Substituting into the differential equation and using the fact that \(y_1\) satisfies the equation simplifies the expression and eventually leads to a first-order equation in \(v'\).
Example 1
Here \(P(x)=0\), so the formula becomes \[ y_2 = y_1 \int \frac{1}{y_1^2}\,dx. \]
Thus, \[ y_2 = \cos x \int \frac{1}{\cos^2 x}\,dx = \cos x \int \sec^2 x\,dx = \cos x \tan x = \sin x. \]
So the general solution is \[ y = c_1\cos x + c_2\sin x. \]
Example 2
Again \(P(x)=0\), so \[ y_2 = y_1 \int \frac{1}{y_1^2}\,dx = e^x \int e^{-2x}\,dx. \]
Compute the integral: \[ \int e^{-2x}\,dx = -\frac{1}{2}e^{-2x}. \]
Therefore, \[ y_2 = e^x\left(-\frac{1}{2}e^{-2x}\right) = -\frac{1}{2}e^{-x}. \]
Since constants do not matter for linear independence, we take \[ y_2 = e^{-x}. \]
Hence the general solution is \[ y = c_1 e^x + c_2 e^{-x}. \]
Summary
- Reduction of order is used when one solution is already known.
- Assume \(y_2 = v(x)y_1(x)\).
- This converts the problem into a first-order equation.
- The formula provides a direct shortcut.
Lecture 13: Section 4.3 - Auxillary Equations ↑ Back to top
Now that we have seen the method of reduction of order, we know how to find more linearly independent solutions to a higher-order differential equation once we have one solution already.
The missing piece in solving homogeneous differential equations is finding the first solution.
We now focus on equations of the form
\[ y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y' + a_0y = 0 \]where each \(a_i\) is a constant.
Choosing a Candidate Solution
We want a function whose derivatives are constant multiples of itself. The most natural choice is
\[ y_1 = e^{mx}. \]Substituting into the differential equation gives:
\[ e^{mx}(m^n + a_{n-1}m^{n-1} + \cdots + a_1m + a_0). \]The polynomial \[ f(m) = m^n + a_{n-1}m^{n-1} + \cdots + a_1m + a_0 \] is called the auxiliary equation (or characteristic equation).
Roots of the auxiliary equation determine solutions of the differential equation.
Second-Order Case
Consider \[ y'' + a_1 y' + a_2 y = 0. \] The auxiliary equation is \[ m^2 + a_1 m + a_2 = 0. \] There are three possible cases.Case 1: Two Distinct Real Roots
If \[ f(m) = (m-a)(m-b), \quad a \neq b, \] then the solutions are \[ y = c_1 e^{ax} + c_2 e^{bx}. \] This is the simplest case.Case 2: Repeated Root
If \[ f(m) = (m-a)^2, \] then we only get one solution \(y_1 = e^{ax}\). We must use reduction of order.Solve \[ y'' - 2ay' + a^2 y = 0. \]
Auxiliary equation: \[ (m-a)^2 = 0. \]
First solution: \[ y_1 = e^{ax}. \] Let \(y_2 = u e^{ax}\). Substitution simplifies to: \[ u'' e^{ax} = 0. \] So \[ u'' = 0. \] Integrating twice: \[ u = c_1 x + c_2. \] To maintain independence, take \(u = x\). Thus \[ y_2 = x e^{ax}. \] General solution: \[ y = c_1 e^{ax} + c_2 x e^{ax}. \]Case 3: Complex Roots
If the auxiliary equation has complex roots \[ m = \alpha \pm i\beta, \] we use Euler's identity \[ e^{i\theta} = \cos(\theta) + i\sin(\theta). \] The real-valued solution becomes \[ y = e^{\alpha x}\left( c_1 \cos(\beta x) + c_2 \sin(\beta x) \right). \] Special case: \[ y'' + k^2 y = 0 \] has general solution \[ y = c_1 \cos(kx) + c_2 \sin(kx). \]Worked Examples
Solve \[ y'' + 4y' + 3y = 0. \] Auxiliary equation: \[ m^2 + 4m + 3 = 0. \] Factor: \[ (m+1)(m+3)=0. \] Roots: \[ m=-1, \quad m=-3. \] General solution: \[ y = c_1 e^{-x} + c_2 e^{-3x}. \]
Solve \[ y'' + 2y' + y = 0. \] Auxiliary equation: \[ m^2 + 2m + 1 = 0. \] Factor: \[ (m+1)^2=0. \] Repeated root \(m=-1\). General solution: \[ y = c_1 e^{-x} + c_2 x e^{-x}. \]
Summary
- Write the auxiliary equation.
- Solve for its roots.
- Three cases: distinct real, repeated real, complex conjugate.
- Each case gives a different form of the general solution.
Lecture 14: Section 4.4 - Method of Undetermined Coefficients ↑ Back to top
We now study equations of the form \[ y'' + a_1 y' + a_2 y = g(x), \] where \(g(x) \neq 0\). The general solution has the form \[ y = y_h + y_p, \] where:- \(y_h\) solves the associated homogeneous equation
- \(y_p\) is a particular solution
Idea of the Method
The method of undetermined coefficients works when \(g(x)\) is one of:- Polynomials
- Exponentials \(e^{ax}\)
- Sines and cosines
- Sums,Differences, or Products of the above
Step-by-Step Procedure
- Solve the homogeneous equation to find \(y_h\).
- Guess the form of \(y_p\) based on \(g(x)\).
- If guess duplicates a homogeneous term, multiply by \(x\).
- Plug into the equation and solve for coefficients.
Table of Guesses
| \(g(x)\) | Guess for \(y_p\) |
|---|---|
| Polynomial degree n | General polynomial degree n |
| \(e^{ax}\) | \(Ae^{ax}\) |
| \(\sin kx\), \(\cos kx\) | \(A\cos kx + B\sin kx\) |
| Sum/Difference/Product or functions above | Product of guesses |
Solve \[ y''+5y'-6y = -6x^2-2x+1. \] Step 1: Homogeneous Solution Auxiliary equation: \[ m^2+5m-6=0. \] Factor: \[ (m+6)(m-1)=0. \] \[ m=-6, \quad m=1. \] \[ y_h = c_1 e^{-6x}+c_2 e^{x}. \] Step 2: Guess Particular Solution Since the right-hand side is a quadratic polynomial, guess \[ y_p=Ax^2+Bx+C. \] Step 3: Compute Derivatives \[ y_p'=2Ax+B \] \[ y_p''=2A \] Step 4: Substitute \[ y''+5y'-6y \] Substitute: \[ 2A + 5(2Ax+B) -6(Ax^2+Bx+C). \] Simplify: \[ -6Ax^2 + (10A-6B)x + (2A+5B-6C). \] Match coefficients with \[ -6x^2-2x+1. \] System: \[ -6A=-6 \] \[ 10A-6B=-2 \] \[ 2A+5B-6C=1 \] Solve: \[ A=1 \] \[ 10-6B=-2 \Rightarrow B=2 \] \[ 2+10-6C=1 \Rightarrow C=\frac{11}{6}. \] Final Answer \[ y=c_1 e^{-6x}+c_2 e^{x}+x^2+2x+\frac{11}{6}. \]
Solve \[ y''-5y'+4y=\sin(2x). \] Step 1: Homogeneous Solution Auxiliary equation: \[ m^2-5m+4=0. \] Factor: \[ (m-1)(m-4)=0. \] \[ y_h=c_1 e^{x}+c_2 e^{4x}. \] Step 2: Guess Particular Solution Since the right-hand side is \(\sin(2x)\), guess \[ y_p=A\cos(2x)+B\sin(2x). \] Step 3: Compute Derivatives \[ y_p'=-2A\sin(2x)+2B\cos(2x) \] \[ y_p''=-4A\cos(2x)-4B\sin(2x) \] Step 4: Substitute Substitute into \[ y''-5y'+4y. \] After simplifying: \[ (10A-10B)\cos(2x) + (10A+10B)\sin(2x). \] Match coefficients with \(\sin(2x)\): \[ 10A-10B=0 \] \[ 10A+10B=1 \] Solve: \[ A=B \] \[ 20A=1 \Rightarrow A=\frac{1}{20}, \quad B=\frac{1}{20}. \] Final Answer \[ y=c_1 e^{x}+c_2 e^{4x} +\frac{1}{20}\cos(2x) +\frac{1}{20}\sin(2x). \]
Solve \[ y''+y=10x+e^{5x}, \quad y(0)=1. \] Step 1: Homogeneous Solution Auxiliary equation: \[ m^2+1=0. \] \[ y_h=c_1\cos x+c_2\sin x. \] Step 2: Guess Particular Solution Split RHS into two parts: Polynomial \(10x\) → guess \(Ax+B\) Exponential \(e^{5x}\) → guess \(Ce^{5x}\) So, \[ y_p=Ax+B+Ce^{5x}. \] Step 3: Compute Derivatives \[ y_p'=A+5Ce^{5x} \] \[ y_p''=25Ce^{5x} \] Step 4: Substitute \[ y''+y \] Substitute: \[ 25Ce^{5x} + Ax+B+Ce^{5x}. \] \[ =Ax+B+26Ce^{5x}. \] Match coefficients with \[ 10x+e^{5x}. \] System: \[ A=10 \] \[ B=0 \] \[ 26C=1 \Rightarrow C=\frac{1}{26}. \] Thus \[ y=c_1\cos x+c_2\sin x+10x+\frac{1}{26}e^{5x}. \] Apply initial condition: \[ y(0)=c_1+\frac{1}{26}=1. \] \[ c_1=\frac{25}{26}. \] Final Answer \[ y=\frac{25}{26}\cos x +c_2\sin x +10x +\frac{1}{26}e^{5x}. \]
- \(g(x)\) is not of special type (like \(\ln x\), \(\tan x\), etc.)
- Coefficients are not constant
Summary
- General solution = homogeneous + particular
- Guess form based on RHS
- Multiply by \(x\) if duplication occurs
- Plug in and solve for constants