Characteristic Roots Classify Linear ODE Solutions

For $ay''+by'+cy=0$, try $y=e^{rt}$. Substitution produces

$ar^2+br+c=0,$

the characteristic equation.

Three root patterns

Distinct real roots give $y=C_1e^{r_1t}+C_2e^{r_2t}$. A repeated root gives $y=(C_1+C_2t)e^{rt}$. Complex roots $\alpha\pm i\beta$ give

$y=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t).$

The real part controls exponential growth or decay; the imaginary part controls oscillation frequency.

Why the trial solution is legitimate

The operator $L[y]=ay''+by'+cy$ sends $e^{rt}$ to $(ar^2+br+c)e^{rt}$. Because $e^{rt}$ is never zero, $L[e^{rt}]=0$ exactly when $r$ is a root of the characteristic polynomial. The trial is therefore not a guess pulled from nowhere: exponentials are eigenfunctions of constant-coefficient differentiation.

When roots are distinct, the two exponential solutions are linearly independent. A second-order equation needs two independent solutions because two initial conditions, usually $y(0)$ and $y'(0)$, must determine two constants.

Check your understanding

Without solving for constants, classify the behavior of $y''+6y'+9y=0$. Will generic solutions oscillate?

Show the reasoning

The polynomial is $(r+3)^2$, so the repeated root is $-3$. Solutions have the form $(C_1+C_2t)e^{-3t}$. They decay without sinusoidal oscillation; the factor $t$ comes from the repeated root.