Examples from Second-order constant-coefficient equations

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

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

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.

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.