inhomogeneous_second_order

Inhomogeneous Second-Order ODE

Definition

The inhomogeneous linear second-order ODE:

$$\ddot{x}+p(t)\dot{x}+q(t)x=g(t),x(t_0)=x_0,\dot{x}(t_0)=u_0,$$

where $g(t)\ne 0$ is the inhomogeneous (forcing) term.

Intuition

Linearity guarantees that the total response equals the free (homogeneous) response plus a forced (particular) response. The homogeneous part encodes the system’s natural behaviour; the particular part encodes how the system responds to the specific input $g(t)$. Discontinuous or impulsive forcing terms are most cleanly handled by converting to an algebraic problem via the Laplace transform.

Formal Description

Three-step solution method

1. Solve the associated homogeneous ODE $\ddot{x}+p\dot{x}+qx=0$ for two independent solutions $x_1(t)$, $x_2(t)$, and form

$$x_h(t)=c_1{x}_1(t)+c_2{x}_2(t).$$

2. Find a particular solution $x_p(t)$ satisfying the full inhomogeneous ODE. When $p,q$ are constants and $g$ is a polynomial, exponential, sine, or cosine, an undetermined-coefficients ansatz works.
3. Write the general solution $x(t)=x_h(t)+x_p(t)$ and apply the initial conditions to determine $c_1,c_2$.

Linearity guarantees $x_h+x_p$ solves the ODE:

$$({\ddot{x}}_h+p{\dot{x}}_h+q{x}_h)+({\ddot{x}}_p+p{\dot{x}}_p+q{x}_p)=0+g=g.$$

Discontinuous inhomogeneous term

For an inhomogeneous term involving the Heaviside step function, e.g.\ $g(t)=1-u_1(t)$, the Laplace transform method is most efficient.

Example. Solve $\ddot{x}+3\dot{x}+2x=1-u_1(t)$, $x(0)=\dot{x}(0)=0$.

Taking the Laplace transform:

$$(s^2+3s+2)X(s)=\frac{1-e^{-s}}{s},X(s)=\frac{1-e^{-s}}{s(s+1)(s+2)}.$$

Setting $F(s)=\frac{1}{s(s+1)(s+2)}=\frac{1}{2}\cdot \frac{1}{s}-\frac{1}{s+1}+\frac{1}{2}\cdot \frac{1}{s+2}$, the solution is

$$x(t)=f(t)-u_1(t)f(t-1),f(t)=\frac{1}{2}-e^{-t}+\frac{1}{2}{e}^{-2t}.$$

Impulsive inhomogeneous term

For a Dirac delta forcing $g(t)=\delta (t)$, the Laplace transform gives $\mathcal{L}\{\delta (t)\}=1$.

Example. Solve $\ddot{x}+3\dot{x}+2x=\delta (t)$, $x(0)=\dot{x}(0)=0$.

$$(s^2+3s+2)X(s)=1,X(s)=\frac{1}{(s+1)(s+2)}=\frac{1}{s+1}-\frac{1}{s+2}.$$ $$x(t)=e^{-t}-e^{-2t}.$$

Note: $x(t)$ is continuous at $t=0$ but $\dot{x}$ is not — impulsive forcing produces a velocity discontinuity.

Key Results

• The general solution is always a sum of the homogeneous and a particular solution.
• Laplace transforms are the preferred method for discontinuous or impulsive forcing terms.
• Impulsive forcing $\delta (t)$ causes a jump in the first derivative but not in $x$ itself.

Applications

• Driven mechanical oscillators (springs with periodic or impulsive forcing).
• Electrical circuits with switched or pulsed voltage sources.
• Any control-theory system with a specified input signal $g(t)$.

Trade-offs

• The undetermined-coefficients method is only systematic when $g(t)$ is a polynomial, exponential, sine, or cosine (or products thereof); variation of parameters is needed otherwise.
• For piecewise or impulsive $g(t)$, the Laplace method is far more efficient but requires facility with partial fractions and the transform table.
• Initial conditions are applied to the full solution $x_h+x_p$, not to $x_h$ alone — a common source of error.

Links

• Homogeneous Second-Order ODE
• Principle of Superposition
• Laplace Transform
• Heaviside and Dirac Functions