Review of Homogeneous and Particular Solutions
The method of undetermined coefficients

In E12 we will extend what you learned in E11, but it will be more general.  We will look at linear, time-invariant, continuous time, continuous variable, lumped parameter, dynamic physical systems.  We will develop models for mechanical, thermal, biological... systems and show that they all yield differential equations that are of the form:

 

where y(t) is the output of the system, and f(t) is some known function.

General Solutions method:  We solve for y(t) by assuming a solution of the form y(t)=yH(t)+yP(t).  The forms of the homogeneous and particular solution are given below.

Note: If f(t) is the sum of functions use superposition (superposition states that if input f1(t) yields output y1(t), and input f2(t) yields output y2(t) then the input f3(t)=f1(t)+f2(t) will yield output y3(t)=y1(t)+y2(t)).

Homogeneous Solution: First solve for form of homogenous equation:

by assuming yH(t)=Aest yields the characteristic equation and homogeneous solution

The roots of characteristic equation determine allowable values of s, which in turn determine the nature of the homogeneous response.  Special cases include repeated roots (If there are repeated values of s, the homogenous solution also includes a time-multiplied exponential.  For example, if the characterstic equation has repeated roots at s=-1, then yH(t)=A1e-t+A2te-t) and complex conjugate roots (damped sinusoids). 

Initial conditions and the particular solution determine the values of A (we can't find these coefficients until after we know the particular response). 

Particular Solution:  Assume yP(t) is of the same character as f(t) (and all of its derivatives).

f(t) yP(t)
A K
At K1t+K2
At2 K1t2+K2t+K3
Aeat Keat
Acos(ω·t) Csin(ω·t)+Dcos(ω·t) = Kcos(ω·t+φ)
Asin(ω·t) Csin(ω·t)+Dcos(ω·t) = Kcos(ω·t+φ)
Ateat K1teat+K2eat

Note: if a term in yH(t) also appears in yP(t), multiply the term in yP(t) by "t".  For example if the differential equation has yH(t)=Ae-2t and f(t)=Be-2t then yP(t)=Kte-2t.

Note: we can't solve for

Example 1:

γ(t)=unit step function, constant for t≥0.

Particular solution:

 

Homogeneous Solution:

 

Complete solution:

Example 2:

 

t=unit ramp.

Particular solution:

To solve for unknown coefficients equate like powers of t:

Complete solution:

The form of the homogeneous solution remains unchanged, so we can go right to finding the complete solution: