微分方程求解

Solving differential equations
The procedure for solving differential equation is given below:
Step 1: Determine the particular solution y p(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-at u(t) Ke-at u(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
y h(t) = A1eλ1t + A2eλ2t + A3eλ3t +……..for t > 0. Step 4: Determine the complete solution
The complete solution is y(t) = y p (t) + y h (t). Substituting the complete solution in to the differential equation, and appling the initial conditions, the coefficients A 1, A 2, A 3,……can be determined.
Example
A causal LTI system is described by the differential equation
)()(2)(3)(2
2t x t y dt t dy dt t y d =++ Given the input and initial conditions : x(t) = u(t), y(0) = - 0.5, 5.0)
(0==t dt t dy .
Determine the complete solution of the system.
Solution:
First, we determine the particular solution. To this end, we assume that the particular solution is y p (t) = Ku(t), where K is a constant which will be determined. Substituting y p (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 y p (t) = 0.5u(t).
Next, we determine the homogeneous solution.
The characteristic equation: 0232=++λλ
The characteristic root:
2,121-=-=λλ The homogeneous solution is
t t h Be Ae t y 21)(λλ+= for t > 0.
i.e., t t h Be Ae t y 2)(--+= for t > 0.
Thus the complete solution is t t h p Be Ae t y t y t y 221)()()(--++=+= for t > 0.
Making the first differentiation to y(t) we have
t t Be Ae dt
t dy 22)(----= Applying the initial conditions we get
2121)0(-=++=B A y
2
12)(0=--==B A dt t dy t 2
123=-=∴B A The complete solution is )()2
12321()(2t u e e t y t t --+-= 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
dt
t ds t h )()(= 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
t t Be Ae t s 221)(--++= and
t t Be Ae dt t ds 22)(----= Applying the zero-initial conditions y(0) = 0, 0)(0==t dt t ds [not y(0) = - 0.5, 5.0)
(0==t dt t dy ] we have the following equations
02
1)0(=++=B A y
02)(0=--==B A dt t dy t Solving the equations 2
11=-=B A The unit step response is )()2
121()(2t u e e t s t t --+-=
The impulse response h(t) is the first differentiation of s(t).
)()()()2
121()()(22t u e e t u e e dt d dt t ds t h t t t t -----=+-== You can verify this solution by the use of the Laplace transform.。