Solution of Differential and Intego-Differential Equations

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Sources:

  1. B. P. Lathi & Roger Green. (2018). Chapter 4: The Laplace Transform. Signal Processing and Linear Systems (3rd ed., pp. 360-370). Oxford University Press.

The time-differentiation property of the Laplace transform has set the stage for solving linear differential (or integro-differential) equations with constant coefficients.

Because \(d^k y / d t^k \Longleftrightarrow\) \(s^k Y(s)\), the Laplace transform of a differential equation is an algebraic equation that can be readily solved for \(Y(s)\).

Next we take the inverse Laplace transform of \(Y(s)\) to find the desired solution \(y(t)\).

Introduction

The following examples demonstrate the Laplace transform procedure for solving linear differential equations with constant coefficients.

Some commonly used properties of Laplace transform in this article: \[ \begin{aligned} \frac{d y(t)}{d t} & \Longleftrightarrow s Y(s)-y\left(0^{-}\right) \\ \frac{d^2 y(t)}{d t^2} & \Longleftrightarrow s^2 Y(s)-s y\left(0^{-}\right)-\dot{y}\left(0^{-}\right) \end{aligned} \] Some commonly used properties of Laplace transform in this article: \[ \begin{aligned} \delta(t) & \Longleftrightarrow 1 \\ u(t) & \Longleftrightarrow \frac{1}{s} \\ t & \Longleftrightarrow \frac{1}{s^2} \\ t^2 & \Longleftrightarrow \frac{2}{s^3} \\ e^{-at} & \Longleftrightarrow \frac{1}{s+a} \\ te^{-at} & \Longleftrightarrow \frac{1}{(s+a)^2} \end{aligned} \] For more results, refer to Table of the Laplace Transform Pairs.

Example

Solve the second-order linear differential equation \[ \left(D^2+5 D+6\right) y(t)=(D+1) x(t) \] for the initial conditions \(y\left(0^{-}\right)=2\) and \(\dot{y}\left(0^{-}\right)=1\) and the input \(x(t)=e^{-4 t} u(t)\).

Solution:

The equation is \[ \frac{d^2 y(t)}{d t^2}+5 \frac{d y(t)}{d t}+6 y(t)=\frac{d x(t)}{d t}+x(t) \]

Let \[ y(t) \Longleftrightarrow Y(s) \]

Then from Eq. (4.15), \[ \frac{d y(t)}{d t} \Longleftrightarrow s Y(s)-y\left(0^{-}\right)=s Y(s)-2 \] and \[ \frac{d^2 y(t)}{d t^2} \Longleftrightarrow s^2 Y(s)-s y\left(0^{-}\right)-\dot{y}\left(0^{-}\right)=s^2 Y(s)-2 s-1 \]

Moreover, for \(x(t)=e^{-4 t} u(t)\), \[ X(s)=\frac{1}{s+4} \quad \text { and } \quad \frac{d x(t)}{d t} \Longleftrightarrow s X(s)-x\left(0^{-}\right)=\frac{s}{s+4}-0=\frac{s}{s+4} \]

Taking the Laplace transform of Eq. (4.22), we obtain \[ \left[s^2 Y(s)-2 s-1\right]+5[s Y(s)-2]+6 Y(s)=\frac{s}{s+4}+\frac{1}{s+4} \]

Collecting all the terms of \(Y(s)\) and the remaining terms separately on the left-hand side, we obtain \[ \left(s^2+5 s+6\right) Y(s)-(2 s+11)=\frac{s+1}{s+4} \]

Therefore, \[ \left(s^2+5 s+6\right) Y(s)=(2 s+11)+\frac{s+1}{s+4}=\frac{2 s^2+20 s+45}{s+4} \] and \[ Y(s)=\frac{2 s^2+20 s+45}{\left(s^2+5 s+6\right)(s+4)}=\frac{2 s^2+20 s+45}{(s+2)(s+3)(s+4)} \]

Expanding the right-hand side into partial fractions yields \[ Y(s)=\frac{13 / 2}{s+2}-\frac{3}{s+3}-\frac{3 / 2}{s+4} \]

The inverse Laplace transform of this equation yields \[ y(t)=\left(\frac{13}{2} e^{-2 t}-3 e^{-3 t}-\frac{3}{2} e^{-4 t}\right) u(t) \]

Comments on initial conditions at \(0^{-}\)and at \(0^{+}\)

# TODO

The initial conditions in previous example are \(y\left(0^{-}\right)=2\) and \(\dot{y}\left(0^{-}\right)=1\). If we let \(t=0\) in the total response in Eq. (4.24), we find \(y(0)=2\) and \(\dot{y}(0)=2\), which is at odds with the given initial conditions. Why? Because the initial conditions are given at \(t=0^{-}\)(just before the input is applied), when only the zero-input response is present. The zero-state response is the result of the input \(x(t)\) applied at \(t=0\). Hence, this component does not exist at \(t=0^{-}\). Consequently, the initial conditions at \(t=0^{-}\)are satisfied by the zero-input response, not by the total response. We can readily verify in this example that the zero-input response does indeed satisfy the given initial conditions at \(t=0^{-}\). It is the total response that satisfies the initial conditions at \(t=0^{+}\), which are generally different from the initial conditions at \(0^{-}\).

Zero-State response (or transfer function)

Consider an \(N\) th-order LTIC system specified by the equation \[ Q(D) y(t)=P(D) x(t) \] or \[ \left(D^N+a_1 D^{N-1}+\cdots+a_{N-1} D+a_N\right) y(t)=\left(b_0 D^N+b_1 D^{N-1}+\cdots+b_{N-1} D+b_N\right) x(t) \]

We shall now find the general expression for the zero-state response of an LTIC system. Zero-state response \(y(t)\), by definition, is the system response to an input when the system is initially relaxed (in zero state). Therefore, \(y(t)\) satisfies \[ \left(D^N+a_1 D^{N-1}+\cdots+a_{N-1} D+a_N\right) y(t)=\left(b_0 D^N+b_1 D^{N-1}+\cdots+b_{N-1} D+b_N\right) x(t) \] with zero initial conditions \[ y\left(0^{-}\right)=\dot{y}\left(0^{-}\right)=\ddot{y}\left(0^{-}\right)=\cdots=y^{(N-1)}\left(0^{-}\right)=0 \] Moreover, the input \(x(t)\) is causal so that (# TODO) \[ x\left(0^{-}\right)=\dot{x}\left(0^{-}\right)=\ddot{x}\left(0^{-}\right)=\cdots=x^{(N-1)}\left(0^{-}\right)=0 \]

Let \[ y(t) \Longleftrightarrow Y(s) \quad \text { and } \quad x(t) \Longleftrightarrow X(s) \]

Because of zero initial conditions, \[ \begin{aligned} & D^r y(t)=\frac{d^r}{d t^r} y(t) \Longleftrightarrow s^r Y(s) \\ & D^k x(t)=\frac{d^k}{d t^k} x(t) \Longleftrightarrow s^k X(s) \end{aligned} \]

Therefore, the Laplace transform of \[ \left(D^N+a_1 D^{N-1}+\cdots+a_{N-1} D+a_N\right) y(t)=\left(b_0 D^N+b_1 D^{N-1}+\cdots+b_{N-1} D+b_N\right) x(t) \] yields \[ \left(s^N+a_1 s^{N-1}+\cdots+a_{N-1} s+a_N\right) Y(s)=\left(b_0 s^N+b_1 s^{N-1}+\cdots+b_{N-1} s+b_N\right) X(s) \] or \[ Y(s)=\frac{b_0 s^N+b_1 s^{N-1}+\cdots+b_{N-1} s+b_N}{s^N+a_1 s^{N-1}+\cdots+a_{N-1} s+a_N} X(s)=\frac{P(s)}{Q(s)} X(s) \]

But we showed that that \(Y(s)=H(s) X(s)\) before. Consequently, \[ H(s)=\frac{P(s)}{Q(s)} \]

The same result has been derived earlier using an alternate (time-domain) approach.

Eexmple: Laplace transform to find system transfer functions

Show that the transfer function of: (a) an ideal delay of \(T\) seconds is \(e^{-s T}\) (b) an ideal differentiator is \(s\) (c) an ideal integrator is \(1 / s\)

(a) Ideal delay

For an ideal delay of \(T\) seconds, the input \(x(t)\) and output \(y(t)\) are related by \[ y(t)=x(t-T) \quad \text { and } \quad Y(s)=X(s) e^{-s T} \quad \text { [see Eq. (4.12)] } \] Therefore, \[ H(s)=\frac{Y(s)}{X(s)}=e^{-s T} \]

(b) Ideal differentiator

For an ideal differentiator, the input \(x(t)\) and the output \(y(t)\) are related by \[ y(t)=\frac{d x(t)}{d t} \]

The Laplace transform of this equation yields \[ Y(s)=s X(s) \quad\left[x\left(0^{-}\right)=0 \text { for a causal signal }\right] \] and \[ H(s)=\frac{Y(s)}{X(s)}=s \]

(c) Ideal integrator

For an ideal integrator with zero initial state, that is, \(y\left(0^{-}\right)=0\), \[ y(t)=\int_0^t x(\tau) d \tau \quad \text { and } \quad Y(s)=\frac{1}{s} X(s) \]

Therefore, \[ H(s)=\frac{1}{s} \]