MAT 276 • Distinct Real Roots
Algorithm Example Interactive Lab Summary
Constant Coefficients
Homogeneous Linear ODE
Step-by-Step

Solving DEs with Distinct Real Roots
Step-by-Step

This is the classic move for a second-order homogeneous linear differential equation with constant coefficients. You turn the differential equation into an algebra problem, solve for the roots, then rebuild the solution from exponential modes.

Big idea

Exponentials are the perfect probe because differentiating \(e^{rx}\) just keeps giving back the same shape times constants.

Decision test

If the characteristic equation has two different real roots, then the full solution is a sum of two exponentials.

Mental model

Each root gives one natural mode. The general solution is just both modes superposed together. Super clean.

Structured interrogation

Objective
Find the general solution, and if initial conditions are given, solve for the constants.

Variables
Independent variable: usually \(x\). Dependent variable: \(y(x)\).

Known information
The equation has the form \(ay''+by'+cy=0\), where \(a,b,c\) are constants.

Constraint
We are in the distinct real roots case, so \(b^2-4ac>0\).

Assumption
We try \(y=e^{rx}\) because exponentials survive differentiation.

The model

\[ay''+by'+cy=0\]

This equation is linear, second order, homogeneous, and has constant coefficients.

Geometrically, it ties together position, slope, and curvature at every point on the curve.

Lock-in

The whole trick is to find building blocks whose derivatives stay in the same family. Exponentials do that automatically.

Why the exponential guess works

Try

\[y=e^{rx}\]

Then

\[y'=re^{rx},\qquad y''=r^2e^{rx}\]

Substitute into the differential equation:

\[a(r^2e^{rx})+b(re^{rx})+c(e^{rx})=0\]
\[e^{rx}(ar^2+br+c)=0\]

Since \(e^{rx}\neq 0\), the only way this works is if

\[ar^2+br+c=0\]

That algebra equation is the characteristic equation or auxiliary equation.

The step-by-step algorithm

Step 1

Write the DE

\[ay''+by'+cy=0\]

Check that it is homogeneous and the coefficients are constants.

Step 2

Build the characteristic equation

\[ar^2+br+c=0\]

Replace the derivative structure by powers of \(r\).

Step 3

Solve for the roots

\[r_1,\; r_2\]

Distinct real roots means \(r_1\neq r_2\) and both are real.

Step 4

Write the solution

\[y=c_1e^{r_1x}+c_2e^{r_2x}\]

Then use initial conditions if the problem gives them.

Worked example from first principles

Solve

\[y''-5y'+6y=0\]
Objective
Find the general solution. No initial conditions yet, so we stop after finding the family of all solutions.
Translate to algebra
\[r^2-5r+6=0\]
Factor
\[(r-2)(r-3)=0\]
\[r_1=2,\qquad r_2=3\]
Build the solution
\[y=c_1e^{2x}+c_2e^{3x}\]
Why this is legit
Each exponential solves the DE by itself, and linearity lets us add valid modes together.

Geometry-first intuition

The curve is made from two exponential shapes. Depending on \(c_1\) and \(c_2\), one mode may dominate early and the other later.

Fast checklist

  • Homogeneous? Yes.
  • Constant coefficients? Yes.
  • Characteristic equation quadratic? Yes.
  • Discriminant positive? \(25-24=1>0\).
  • So this is the distinct-real-roots case.

Now add initial conditions

Solve

\[y''-5y'+6y=0,\qquad y(0)=4,\qquad y'(0)=9\]

From before,

\[y=c_1e^{2x}+c_2e^{3x}\]

Differentiate:

\[y'=2c_1e^{2x}+3c_2e^{3x}\]

Plug in \(x=0\):

\[c_1+c_2=4\]
\[2c_1+3c_2=9\]

Solve the system step-by-step:

\[c_1=4-c_2\]
\[2(4-c_2)+3c_2=9\]
\[8-2c_2+3c_2=9\]
\[c_2=1,\qquad c_1=3\]

So the particular solution is

\[y=3e^{2x}+e^{3x}\]
Lock-in

The roots determine the solution shape. The initial conditions determine the mix.

Interactive root lab

a
b
c
C₁
C₂
Discriminant
1
Case
distinct real roots
Characteristic equation
\[r^2-5r+6=0\]
\(r_1=3.000\), \(r_2=2.000\)

Nice. This is the exact case for this lesson.

Mode graph
mode 1
mode 2
sum = full solution
Change the coefficients and watch the roots change the geometry. Positive roots blow up. Negative roots decay. Opposite signs give one growing mode and one decaying mode.

Common mistakes

  • Using the characteristic-equation method on an equation that is not homogeneous.
  • Forgetting that the coefficients must be constants for this basic version.
  • Using the distinct-real-roots formula when the discriminant is zero or negative.
  • Mixing up root symbols \(r_1,r_2\) with solution constants \(c_1,c_2\).
  • Dropping signs when using the quadratic formula.

One-page summary

\[ay''+by'+cy=0\]

Assume \(y=e^{rx}\)

\[ar^2+br+c=0\]

If \(b^2-4ac>0\), roots are distinct and real.

\[y=c_1e^{r_1x}+c_2e^{r_2x}\]

Then use initial conditions to solve for \(c_1\) and \(c_2\).

Lock it in

Distinct real roots means the differential equation has two different exponential modes. The characteristic equation finds those modes. The general solution is their linear combination. That’s the whole game.