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Second-Order Equations & Linear Systems

Useful preparation: First-Order Equations & Initial Values

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A second-order equation tracks a state and its rate of change. Two initial numbers are needed because position alone does not determine motion. For constant coefficients, exponentials reproduce their own shape under differentiation, making them natural building blocks. The characteristic roots identify whether these building blocks grow, decay, or oscillate.

Second-order linear equation
The equation a y double prime+b y prime+c y=g (t)\displaystyle c\, y = g\, \left(t\right), with nonzero constant a, is linear. It is homogeneous when g is identically zero.
Characteristic polynomial
For the homogeneous constant-coefficient equation, substituting exp⁡ (rt)\displaystyle \operatorname{exp}\, \left(rt\right) gives a r squared+b r+c\displaystyle b\, r + c. Its roots determine the homogeneous solution modes.
State system
Writing u=y\displaystyle u = y and v=y\displaystyle v = y prime turns one second-order equation into two first-order equations, so the state records both the value and its velocity.

Conditions before calculation

  • The elementary characteristic-polynomial method applies directly to homogeneous equations with constant coefficients, not arbitrary variable-coefficient equations.
  • Distinct real roots give two exponentials; a repeated root requires exp⁡ (rt)\displaystyle \operatorname{exp}\, \left(rt\right) and t exp⁡ (rt)\displaystyle t\, \operatorname{exp}\, \left(rt\right); conjugate roots alpha plus or minus i beta give real sine and cosine modes times exp⁡\displaystyle \operatorname{exp}(alpha t).
  • For the homogeneous constant-coefficient system x′\displaystyle x'=Ax, all eigenvalues with strictly negative real parts imply asymptotic stability of zero. Zero real parts require further analysis, including possible repeated-root growth.

Why the method works

1. Choose a complete family

A second-order homogeneous equation needs two independent solutions. Writing the same exponential twice provides only one adjustable direction and cannot satisfy arbitrary initial data.

2. Fit both initial conditions

Differentiate the proposed general solution before using the initial velocity. The resulting two linear equations determine its two constants.

3. Connect to a system

A forcing term enters the velocity equation. For a nonhomogeneous problem, add one particular solution to the full homogeneous family.

(uv)′=(01−c/a−b/a)(uv)+(0g(t)/a)\begin{pmatrix}u\\v\end{pmatrix}'=\begin{pmatrix}0&1\\-c/a&-b/a\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}+\begin{pmatrix}0\\g(t)/a\end{pmatrix}

Worked examples

EXAMPLE 1

Solve y double prime+3y prime+2 y=0\displaystyle 2\, y = 0, with y (0)=1\displaystyle y\, \left(0\right) = 1 and y prime(0)=0.

  1. Factor the characteristic polynomial to find two distinct decay rates.

    r2+3r+2=(r+1)(r+2)r^2+3r+2=(r+1)(r+2)

  2. Write the general solution and its initial-value equations.

    y=C1e−t+C2e−2t,C1+C2=1,−C1−2C2=0y=C_1e^{-t}+C_2e^{-2t},\quad C_1+C_2=1,\quad -C_1-2C_2=0

  3. Solving gives C 1=2\displaystyle C\, 1 = 2 and C 2=−1\displaystyle C\, 2 = -1. Each exponential satisfies the differential equation, and the two initial conditions check.

Conclusion: The solution is 2 exp⁡ (−t)−exp⁡ (−2 t)\displaystyle 2\, \operatorname{exp}\, \left(-t\right) - \operatorname{exp}\, \left(-2\, t\right), which tends to zero as t increases.

EXAMPLE 2

Solve y double prime-2y prime+y=0\displaystyle y = 0, with y (0)=0\displaystyle y\, \left(0\right) = 0 and y prime(0)=1.

  1. The characteristic polynomial has a repeated root.

    r2−2r+1=(r−1)2r^2-2r+1=(r-1)^2

  2. Use an independent second solution; y (0)=0\displaystyle y\, \left(0\right) = 0 forces C 1=0\displaystyle C\, 1 = 0.

    y=(C1+C2t)ety=(C_1+C_2t)e^t

  3. Differentiate to obtain y prime=C 2 (1+t) exp⁡ (t)\displaystyle C\, 2\, \left(1 + t\right)\, \operatorname{exp}\, \left(t\right); the initial derivative gives C 2=1\displaystyle C\, 2 = 1. Direct substitution yields zero.

    y=tet,y′′−2y′+y=[(t+2)−2(t+1)+t]et=0y=te^t,\quad y''-2y'+y=[(t+2)-2(t+1)+t]e^t=0

Conclusion: The solution grows without bound; the repeated positive root produces an unstable mode.

Pitfalls & counterexamples

Tempting claim: Purely imaginary roots make every linear system asymptotically stable.

Correction: They can produce persistent oscillation rather than decay; repeated structures can also cause growth.

Test the claim: The equation y double prime+y=0\displaystyle y = 0 has the nondecaying solution cos⁡ (t)\displaystyle \operatorname{cos}\, \left(t\right).

Check your understanding

What extra factor is needed for the second solution at a repeated real root r?

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Compare your reasoning

A factor of t.

The functions exp⁡ (rt)\displaystyle \operatorname{exp}\, \left(rt\right) and t exp⁡ (rt)\displaystyle t\, \operatorname{exp}\, \left(rt\right) are independent, unlike two constant multiples of exp⁡ (rt)\displaystyle \operatorname{exp}\, \left(rt\right).

Connect this idea

Eigenvalues of the companion matrix equal the characteristic roots. Mechanical damping, circuits, and coupled linear systems use the same mode analysis; nonlinear stability needs additional hypotheses.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

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