Math101learn.math101.caCharacteristic Equation
A precise guide to characteristic roots for constant-coefficient linear differential equations.
Precise definition
For a homogeneous linear equation with constant coefficients $a_ny^{(n)}+\cdots+a_1y'+a_0y=0$, substituting $y=e^{rt}$ gives the characteristic polynomial $a_nr^n+\cdots+a_1r+a_0$. Its roots, with multiplicity, determine a fundamental set of exponential, polynomial-exponential, or sinusoidal solutions.
Notation and mathematical language
For a second-order equation $ay''+by'+cy=0$, write $ar^2+br+c=0$. Distinct real roots $r_1,r_2$ give $C_1e^{r_1t}+C_2e^{r_2t}$. A repeated root $r$ gives $(C_1+C_2t)e^{rt}$. Complex roots $\alpha\pm i\beta$ give $e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t)$.
Conceptual picture
Differentiation acts on $e^{rt}$ by multiplication by $r$, so a polynomial in the derivative operator becomes the same polynomial in $r$. Repeated roots require extra factors of $t$ because a second independent solution cannot be another constant multiple of $e^{rt}$.
Conditions and key results
The method applies directly to homogeneous linear equations whose coefficients are constant. It does not solve variable-coefficient or nonlinear equations merely by writing a polynomial. Complex roots occur in conjugate pairs for real coefficients, allowing an entirely real solution basis.
A reliable strategy
- Move all homogeneous terms to one side and verify constant coefficients.
- Form and factor or solve the characteristic polynomial, recording multiplicities.
- Translate each root type into the correct independent real solution terms.
- Apply initial or boundary conditions and verify by substitution into the differential equation.
Fully worked example
Interpretation and application
Characteristic roots encode dynamics: negative real parts decay, positive real parts grow, and nonzero imaginary parts oscillate. These statements describe the homogeneous linear model. A forcing term changes the particular response but not the roots that govern the complementary solution.
Common mistakes
Verification and reasonableness
- Substitute each basis function into the homogeneous equation.
- Compute the Wronskian or observe distinct root construction to confirm independence.
- Check initial conditions after solving for constants, including derivative product-rule terms.
Practice
- Solve the characteristic equation for $y''-5y'+6y=0$.
- What form corresponds to a double root $r=-1$?
- What real form corresponds to roots $-2\pm4i$?
Answers and brief solutions
- $r=2,3$.
- $(C_1+C_2t)e^{-t}$.
- $e^{-2t}(C_1\cos4t+C_2\sin4t)$.
Further deduction
Root multiplicity has an operator interpretation. If the differential operator factors as $(D-r)^m$, then $e^{rt},te^{rt},\ldots,t^{m-1}e^{rt}$ all lie in its null space. For an $n$th-order equation, counting these functions across all roots yields exactly $n$ independent homogeneous solutions, matching the number of initial values required by the existence theorem.
Initial stability can often be read without solving constants. For a real second-order polynomial $r^2+br+c$, both roots have negative real part exactly when $b>0$ and $c>0$. If $c<0$, their product is negative and a growing positive root exists. These coefficient tests are limited to the normalized second-order real case; higher-order stability needs stronger criteria.
Complex conjugate roots arise because real-coefficient polynomials conjugate their nonreal zeros. Combining $e^{(\alpha\pm i\beta)t}$ into real sine and cosine modes preserves a real solution basis and makes the roles of decay rate $\alpha$ and angular frequency $\beta$ explicit.
Related topics
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
What is the larger characteristic root of $r^2-5r+6=0$?
- $r^2-5r+6=(r-2)(r-3)$.
- The roots are 2 and 3, so the larger is 3.
End of lesson
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