微分方程求解

Solving differential equations

The procedure for solving differential equation is given

below:

Step 1: Determine the particular solution yp(t).

Note that the particular solution has the same form to the

input signal x(t). This means

Input signal x(t) Particular solution

e-atu(t) Ke-atu(t)

u(t) Ku(t)

cos(ω0t) K1cos(ω0t)+ K2sin(ω0t)

By substituting the particular solution in to the

differential equation, the coefficient(s) K (or K1 and K2) can be

determined.

Step 2: Determine the characteristic roots.

Write the characteristic equation from the differential

equation, then determine the roots of characteristic equation,

λ1, λ2, λ3,…….

Step 3: Determine the homogeneous solution

If no repeated roots, then the homogeneous solution is

given by

yh(t) = A1eλ1t + A2eλ2t + A3eλ3t +……..for t > 0.

Step 4: Determine the complete solution The complete solution is y(t) = yp(t) + yh(t). Substituting

the complete solution in to the differential equation, and

appling the initial conditions, the coefficients A1, A2,

A3,……can be determined.

Example

A causal LTI system is described by the differential

equation

)()(2)(3)(22txtydttdydttyd

Given the input and initial conditions :

x(t) = u(t), y(0) = - 0.5, 5.0)(0tdttdy.

Determine the complete solution of the system.

Solution:

First, we determine the particular solution. To this end,

we assume that the particular solution is yp(t) = Ku(t), where

K is a constant which will be determined. Substituting yp(t) =

Ku(t) and x(t) = u(t) in to the differential equation we have

12K for t > 0 (Note: for a causal LTI system, the

response y(t) starts always at t = 0+)

i.e., K=1/2 for t > 0, the particular solution is yp(t) =

0.5u(t).

Next, we determine the homogeneous solution.

The characteristic equation: 0232 The characteristic root: 2,121

The homogeneous solution is tthBeAety21)( for t > 0.

i.e., tthBeAety2)( for t > 0.

Thus the complete solution is

tthpBeAetytyty221)()()( for t > 0.

Making the first differentiation to y(t) we have

ttBeAedttdy22)(

Applying the initial conditions we get

2121)0(BAy

212)(0BAdttdyt

2123BA

The complete solution is )()212321()(2tueetytt

Discussion:

How to determine the impulse response of the system in

the above example?

First, we must understand that the impulse response h(t)

is the zero-state response to the unit impulse signal δ(t). And if

the input signal is δ(t), the input signal excite the system only

at t = 0. For t > 0, the input is 0, means there is no input signal

applied to the system.

So the above method for solving differential equations can

not be used to find the impulse response h(t). Why? An important reason is that the initial conditions must be

changed because of the input δ(t). So the initial conditions are

no longer zero initial conditions.

But the above method can be used to determine the unit

step response s(t). By the use of the relationship between the

unit impulse response and unit step response

dttdsth)()(

The impulse response h(t) can be easily determined.

Remember, the unit step response s(t) is the zero-state

response.

Again consider the system in the above example. Lets

determine the unit step response s(t) and unit impulse

response h(t).

The unit step response is also ttBeAets221)(

and ttBeAedttds22)(

Applying the zero-initial conditions y(0) = 0, 0)(0tdttds

[not y(0) = - 0.5, 5.0)(0tdttdy] we have the following equations

021)0(BAy

02)(0BAdttdyt

Solving the equations

211BA

The unit step response is )()2121()(2tueetstt The impulse response h(t) is the first differentiation of

s(t).

)()()()2121()()(22tueetueedtddttdsthtttt

You can verify this solution by the use of the Laplace

transform.