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Math · Differential Equations

Chapter 5: Second-Order Linear Equations

Second-Order Linear Equations

Two derivatives, two constants, two basic solutions.

Lesson
1
Time
About 23 minutes
0 of 12 done

Step 1: Let's Learn

Read it, or press Listen and follow the words.

A constant-coefficient homogeneous equation looks like ay″ + by′ + cy = 0. Newton’s second law produces these constantly.

Superposition

If y₁ and y₂ both solve the equation, so does any combination of them. Linearity is what makes this true.

Exactly two are needed

The general solution is c₁y₁ + c₂y₂ for two independent solutions. A second-order equation needs exactly two.

They must be independent

If y₂ is a multiple of y₁ it adds nothing, and the combination cannot meet arbitrary initial conditions.

Two conditions

Both the starting position and the starting velocity are needed. One alone leaves a whole family of solutions.

Why two

A thrown ball needs both where it started and how fast. Position alone does not determine the flight.

Two derivatives, two constants

A second-order equation needs two initial conditions to pin down a solution, typically a position and a velocity. The number of constants always matches the order.

Superposition

For a linear homogeneous equation, any combination of solutions is another solution. That is what makes the solution set a vector space and links this course directly to linear algebra.

Two independent solutions suffice

Find two solutions that are not multiples of each other and every solution is a combination of them. The general solution is their span, which is why finding two is enough.

Why second order is so common

Newton's second law relates acceleration to force, and acceleration is a second derivative. Almost every mechanical and electrical system therefore produces a second-order equation.

Step 2: Try It Yourself

Tap and try it out.

y″ + y = 0 has sine and cosine as its two basic solutions. Every solution is a combination of these.
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y = 2 sin(1x) + 0
  • Point(1, 1.68)

Step 3: Watch an Example

One step at a time.

Watch Hamza Combine Two Solutions

Hamza knows sine and cosine both solve y″ + y = 0 and needs the solution with y = 3 and y′ = 0 at t = 0.

  1. Step 1

    He writes the general solution as y = A cos t + B sin t.

Step 4: Your Turn

Practice makes it stick.

The Count

Problem 1 of 2

A second-order equation. How many arbitrary constants does its general solution have?

The Conditions

Problem 2 of 2

How many initial conditions are needed to pin down a second-order solution?

Structure of the Solution

1 of 8

A third-order equation. How many arbitrary constants?

2 of 8

y = A cos t + B sin t with y = 5 and y′ = 0 at t = 0. What is A?

3 of 8

The same solution with y = 5 and y′ = 0 at t = 0. What is B?

4 of 8

Two solutions where one is triple the other. Are they independent? 1 yes, 0 no.

5 of 8

y₁ and y₂ both solve a linear homogeneous equation. Does y₁ + y₂? 1 yes, 0 no.

6 of 8

How many independent solutions does a second-order equation need?

7 of 8

Sort each statement by whether it holds for a linear homogeneous equation.

Tap something to move it.

  • Empty
  • Empty

8 of 8

Is y = 0 always a solution of a homogeneous linear equation? 1 yes, 0 no.

Step 5: Quick Check

Show what you know.

Question 1 of 2

A fourth-order linear equation. How many arbitrary constants does the general solution have?

Question 2 of 2

What does superposition say?

What You Learned

  • A second-order linear homogeneous equation has a general solution c₁y₁ + c₂y₂.
  • Superposition means any combination of solutions is again a solution.
  • Two initial conditions are needed, typically the starting position and velocity.