Try y = e^(rt). Each derivative brings down a factor of r, so the whole equation collapses into a polynomial in r.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The characteristic equation
For ay″ + by′ + cy = 0 the polynomial is ar² + br + c = 0. The calculus problem has become a quadratic.
Two distinct real roots
The solution is c₁e^(r₁t) + c₂e^(r₂t). Both terms decay if both roots are negative.
A repeated root
One root gives only one solution, so a factor of t supplies the second: c₁e^(rt) + c₂te^(rt).
Complex roots
Roots α ± βi give e^(αt) times a combination of cos βt and sin βt. Oscillation appears whenever the roots are complex.
Reading the behaviour
The real part decides growth or decay; the imaginary part decides oscillation. Both are visible in the roots.
Guess an exponential
Substituting y = e^(rt) into a constant-coefficient linear equation turns every derivative into a power of r. The calculus becomes a polynomial equation in r, which is the characteristic equation.
Three cases
Distinct real roots give two exponentials; a repeated root gives an exponential and t times it; complex roots give oscillation with an exponential envelope. Each case has a different form of solution.
Complex roots mean oscillation
Euler's formula converts complex exponentials into sines and cosines. The real part of the root controls growth or decay, the imaginary part the frequency, and both are physically meaningful.
The repeated root needs a second solution
A double root supplies only one exponential, so a second independent solution is needed. Multiplying by t provides it, which can be verified directly by substitution.
Step 2: Try It Yourself
Tap and try it out.
- Point(1, 0.10)
Step 3: Watch an Example
One step at a time.
Watch Aiko Solve a Second-Order Equation
Aiko solves y″ + 5y′ + 6y = 0.
- Step 1
She writes the characteristic equation r² + 5r + 6 = 0.
Step 4: Your Turn
Practice makes it stick.
The Roots
Problem 1 of 2
y″ + 7y′ + 12y = 0. The roots are −3 and what other value?
The Long Run
Problem 2 of 2
Both characteristic roots are negative. What value does the solution approach as t grows large?
Solve the Quadratic
1 of 8
y″ + 3y′ + 2y = 0. The roots are −1 and what other value?
2 of 8
y″ − 4y = 0. The roots are 2 and what other value?
3 of 8
y″ + y = 0. Are the roots real? 1 yes, 0 no.
4 of 8
A repeated root r. What extra factor multiplies the second solution? Enter 1 for t, 2 for t squared.
5 of 8
y″ + 6y′ + 9y = 0. What is the repeated root?
6 of 8
Complex roots with real part 0. Does the solution grow, decay or oscillate steadily? Enter 1 grow, 2 decay, 3 steady oscillation.
7 of 8
Match each root situation to the shape of the solution.
Tap a card on the left to start.
8 of 8
y″ − 9y = 0. The roots are −3 and what other value?
Step 5: Quick Check
Show what you know.
Question 1 of 2
y″ + 5y′ + 4y = 0. The roots are −1 and what other value?
Question 2 of 2
What do complex characteristic roots produce?
What You Learned
- Guessing y = e^(rt) turns a constant-coefficient equation into a polynomial in r.
- Distinct real roots give two exponentials; a repeated root needs an extra factor of t.
- Complex roots give oscillation, with the real part setting growth or decay.