Second-order constant-coefficient equations

Constant-coefficient equation

The same idea extends to an $n$th-order equation: each derivative has a constant coefficient and the corresponding characteristic polynomial has degree $n$.

Superposition of homogeneous solutions

The two arbitrary constants are what allow two initial conditions to be imposed. If the proposed solutions are scalar multiples, they provide only one mode and cannot represent all possible initial data.

Characteristic equation

The trial is useful because

so substitution gives

Since $e^{rx}\neq0$, the differential equation reduces to an algebraic equation. For an $n$th-order constant-coefficient equation, replace $y^{(j)}$ by $r^j$ to obtain the corresponding degree-$n$ polynomial.

Read the characteristic roots

For a repeated root, $e^{rx}$ and $xe^{rx}$ are independent; the factor $x$ is required. The extra mode comes from writing $y=e^{rx}v(x)$, where $r$ is the repeated root. Since $b=-2ar$ and $c=ar^2$, substitution reduces the ODE to $ae^{rx}v''=0$, so $v=A+Bx$.

For a complex pair, $\alpha$ controls exponential growth or decay and $\beta$ controls oscillation, as with the modulus and argument in polar form. For real coefficients, a non-real root always occurs with its complex conjugate. Since $e^{(\alpha+i\beta)x}=e^{\alpha x}(\cos\beta x+i\sin\beta x)$, its real and imaginary parts give the two real modes in the last row and the final solution needs no complex constants.

Two distinct real roots

For

the characteristic polynomial factors as

The roots are $5$ and $-3$, so

A repeated real root

For

the characteristic polynomial is $(r-3)^2$. The homogeneous solution is

Using only $Ae^{3x}$ would lose one arbitrary constant because the two proposed modes would be scalar multiples.

Initial data for distinct modes

For

the roots are $5$ and $-1$, so

The initial conditions give

Thus $A=2$, $B=0$ and

Homogeneous part plus particular part

The homogeneous part contains the arbitrary constants. The particular part accounts for the forcing and contains no new arbitrary constants.

Undetermined coefficients

The trial families work because differentiating a polynomial, exponential or sinusoidal expression keeps it within the same finite-dimensional family, so substitution reduces the problem to algebraic coefficient matching. The correction accounts for resonance: a trial that already lies in the homogeneous solution space cannot produce the forcing after applying $L$. Oscillation by itself is not resonance; resonance requires a forcing term whose trial shape overlaps a natural homogeneous mode.

Constant forcing

For

the homogeneous solution is $Ae^x+Be^{2x}$. A constant trial $y_p=C$ gives $2C=4$, so $C=2$. Hence

Linear forcing

For

the homogeneous roots are $2$ and $3$. Use $y_p=ax+b$. Substitution gives

Thus $a=1$ and $b=5/6$, so

Forcing at a repeated root

For

the homogeneous equation has the repeated root $-2$ and homogeneous solution $(A+Bx)e^{-2x}$. The trial $Ce^{-2x}$ and the corrected trial $Cxe^{-2x}$ both overlap with this solution space, so use

Here the operator factors as $(D+2)^2$, where $D=d/dx$. For $y_p=e^{-2x}v(x)$, applying this operator gives $e^{-2x}v''(x)$. With $v=Cx^2$, the left side is $2Ce^{-2x}$; matching the forcing gives $C=1/2$. Therefore

Trigonometric resonance

For

the homogeneous modes are $\cos x$ and $\sin x$, so the forcing overlaps with the homogeneous solution. Use

Since

we have $C=1/2$. With $y(0)=3$ and $y'(0)=0$,

Polynomial forcing

For

take $y_p=ax^2+bx+c$. Matching coefficients gives $a=1$, $b=1$ and $c=-2$. Thus

Exponential-polynomial forcing

For

the homogeneous roots are $1$ and $2$, so there is no overlap with $e^{3x}$. Set

Writing $v=Ax^2+Bx+C$, direct differentiation gives

Matching this with $x^2e^{3x}$ gives $2A=1$, $6A+2B=0$ and $2A+3B+2C=0$. Hence

The general solution is

Recover an IVP from an integral equation

A Volterra integral equation

For

setting $t=0$ gives $x(0)=0$. Differentiate using the Leibniz rule:

so $x'(0)=0$. Differentiating again gives

The integral equation therefore implies the IVP

The unique solution is

The initial conditions are essential. Solving only the differentiated equation would introduce nonzero sine and cosine solutions that need not satisfy the original integral equation.

Spring-mass-damper model

With an external force, the right side becomes a forcing term. The solution then has a transient homogeneous response and a particular forced response. In the usual mechanical units, $m$ is mass, $c$ has mass/time units and $k$ has mass/time$^2$ units.

Damped oscillation

For $m=1$, $c=0.2$ and $k=1.01$,

has characteristic roots

Therefore

With $x(0)=2$ and $\dot{x}(0)=-0.2$, we obtain $A=2$ and $B=0$, so

The cosine describes the oscillation, while $e^{-0.1t}$ is the decaying amplitude. Hence $x(t)\to0$.

Periodic forcing

If an external periodic force is applied,

the solution has a transient part, which decays and a steady periodic part at frequency $\omega$. Take $\omega=2$ and the same initial conditions. For the particular response, set

Substitution gives

so $a\approx-0.4137$ and $b\approx-0.6574$. Adding the homogeneous part $e^{-0.1t}(C\cos t+D\sin t)$ and imposing $x(0)=2$, $x'(0)=-0.2$ gives $C\approx2.657$ and $D\approx0.893$.

Therefore the response is approximately

As $t$ increases, only the forced periodic terms remain. The choice $\omega=2$ shows a forced response away from the peak. The response is largest when the forcing frequency is near the undamped natural frequency,

This is the near-resonant peak of the damped frequency response. Damping prevents the ideal undamped amplitude from growing without bound, but still produces a pronounced finite peak near that frequency.