Lesson 1: What Is a Differential Equation?
Estimated time: 30-40 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Define a differential equation and distinguish ODEs from PDEs
- Classify a DE by its order and linearity
- Identify initial conditions and initial value problems (IVPs)
- Verify that a given function is a solution to a DE
- Distinguish between general solutions and particular solutions
What Is a Differential Equation?
In calculus you learned to find derivatives and integrals of functions. A differential equation flips the problem around: you are given a relationship involving an unknown function and its derivatives, and your job is to find the function.
Differential Equation (DE): An equation that contains an unknown function and one or more of its derivatives. The unknown function is what we want to find.
Example 1: A Simple DE
dy/dx = 2x
This says: "Find a function y(x) whose derivative equals 2x." You can solve this by integration: y = x² + C, where C is an arbitrary constant.
Example 2: A More Interesting DE
dy/dx = 3y
This says: "Find a function whose derivative is three times the function itself." The answer is y = Ce3x. This equation models exponential growth and decay -- one of the most important applications in science.
ODE vs PDE
Ordinary Differential Equation (ODE): A DE in which the unknown function depends on a single independent variable. All derivatives are ordinary derivatives (d/dx).
Partial Differential Equation (PDE): A DE in which the unknown function depends on two or more independent variables. The equation involves partial derivatives.
Example 3: ODE vs PDE
ODE: y'' + 5y' + 6y = 0. Here y depends only on one variable (say x).
PDE: ∂u/∂t = k(∂²u/∂x²). Here u depends on both x and t. This is the heat equation.
In this course, we study ODEs only.
Order of a DE
Order: The order of a DE is the highest derivative that appears in the equation.
Example 4: Identifying Order
| Equation | Highest Derivative | Order |
|---|---|---|
| y' + 2y = x | y' | 1 |
| y'' + 3y' + 2y = 0 | y'' | 2 |
| y''' - xy = sin x | y''' | 3 |
| (y')³ + y = 0 | y' | 1 |
Common mistake: In (y')³ + y = 0, the exponent 3 is the power, not the order. The highest derivative is y', so the order is 1.
Linearity
Linear ODE: An ODE of the form an(x)y(n) + an-1(x)y(n-1) + ... + a1(x)y' + a0(x)y = g(x). The unknown y and all its derivatives appear to the first power only, with no products of y and its derivatives. The coefficient functions ai(x) may be any functions of x.
Example 5: Linear vs Nonlinear
| Equation | Linear? | Reason |
|---|---|---|
| y'' + 5y = sin x | Yes | y terms to first power, coefficients depend only on x |
| xy' + exy = x² | Yes | Coefficients x and ex depend only on x |
| y' + y² = 0 | No | y² -- y is squared |
| (y')² + y = x | No | (y')² -- derivative is squared |
| yy'' + y' = 0 | No | yy'' is a product of y and y'' |
Solutions and Initial Value Problems
Solution: A function y = f(x) is a solution to a DE on an interval I if, when substituted into the equation, it produces a true statement for all x in I.
General Solution: A family of solutions containing arbitrary constants. For an nth-order ODE, the general solution typically has n arbitrary constants.
Initial Value Problem (IVP): A DE together with initial conditions that specify the value of the solution (and possibly its derivatives) at a particular point. The initial conditions determine the arbitrary constants, giving a unique particular solution.
Example 6: Verifying a Solution
Verify that y = e-2x is a solution to y' + 2y = 0.
Step 1: Compute y'. y' = -2e-2x.
Step 2: Substitute into the DE. y' + 2y = -2e-2x + 2e-2x = 0. True!
So y = e-2x is indeed a solution.
Example 7: Solving an IVP
The general solution to y' = 3y is y = Ce3x. Solve the IVP with y(0) = 5.
Apply the initial condition: y(0) = Ce0 = C = 5.
Particular solution: y = 5e3x.
Example 8: Second-Order IVP
The general solution to y'' + 4y = 0 is y = c1cos 2x + c2sin 2x. Solve the IVP y(0) = 3, y'(0) = -2.
y(0) = c1 = 3.
y' = -2c1sin 2x + 2c2cos 2x. y'(0) = 2c2 = -2, so c2 = -1.
Particular solution: y = 3 cos 2x - sin 2x.
Check Your Understanding
1. Classify by order and linearity: y''' + xy' - y = ex.
2. Is (y')² + 3y = x linear or nonlinear?
3. Verify that y = 2ex + 3e-x is a solution to y'' - y = 0.
4. The general solution to y' + y = 0 is y = Ce-x. Find the particular solution satisfying y(0) = 7.
5. What is the difference between a general solution and a particular solution?
Key Takeaways
- A differential equation relates an unknown function to its derivatives.
- ODEs involve one independent variable; PDEs involve two or more.
- The order is the highest derivative present (power does not equal order).
- A linear ODE has y and its derivatives to the first power only, with no products of y-terms.
- A general solution contains arbitrary constants; an IVP pins down those constants to give a unique particular solution.
- Always verify solutions by substituting back into the original equation.
Ready for More?
Next Lesson
In Lesson 2, you will learn to visualize differential equations using direction fields and solution curves -- even without solving them.
Start Lesson 2