Solutions of Ordinary Differential Equations (ODE)

5.3 Solutions of Ordinary Differential Equations (ODE)

A function y = fx () is a solution of a differential equation if the latter is satisfied when and its y derivatives are replaced throughout by fx () and its corresponding derivatives. Also, the initial conditions must be satisfied.

For example a solution of the differential equation

-------- y 2 + y = 0

since and its second derivative satisfy the given differential equation. y Any linear, time-invariant system can be described by an ODE which has the form

n – d 1 --------- y

d ---------------- y

dy

a n n + a n – 1 n – 1 + …a + 1 ------ + a 0 y

b m ---------- m + b m – 1 ----------------- n – 1 + …b + 1 ------ + b 0 x

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩ Excitation Forcing ( ) Function x t ()

NON – HOMOGENEOUS DIFFERENTIAL EQUATION

If the excitation in (B12) is not zero, that is, if xt ()0 ≠ , the ODE is called a non-homogeneous ODE . If xt () = 0 , it reduces to:

n – d 1 --------- y

---------------- d y

dy a n n + a n – 1 n – 1 + …a + 1 ------ + a

HOMOGENEOUS DIFFERENTIAL EQUATION

The differential equation of (5.13) above is called a homogeneous ODE and has different linearly n

independent solutions denoted as y 1 ()y t , 2 ()y t , 3 ()…y t ,, n () t .

We will now prove that the most general solution of (5.13) is:

(5.14) where the subscript on the left side is used to emphasize that this is the form of the solution of H

y H () t = k 1 y 1 ()k t + 2 y 2 ()k t + 3 y 3 ()…k t + + n y n () t

the homogeneous ODE and k 1 ,,,, k 2 k 3 …k n are arbitrary constants.

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Solutions of Ordinary Differential Equations (ODE)

Proof :

Let us assume that y 1 () t is a solution of (5.13); then by substitution,

1 d y 1 dy 1

a n ----------- n + a n – 1 ------------------ n – 1 + …a + 1 -------- + a y 1 = 0 (5.15)

A solution of the form k 1 y 1 () t will also satisfy (5.13) since

a n ------- k n ( 1 y 1 )a + n – 1 ------------- k n – 1 ( 1 y 1 )…a + + 1 ----- k ( 1 y 1 )a + 0 ( k 1 y 1 )

= k 1 ⎜ a n ----------- n + a n – 1 ------------------ n

+ …a + 1 -------- + a 0 y 1 ⎟ = 0

If y = y 1 () t and y = y 2 () t are any two solutions, then y = y 1 ()y t + 2 () t will also be a solution since

1 d y 1 dy 1

a n ----------- n + a n – 1 ------------------ n – 1 + …a + 1 -------- + a 0 y 1 = 0

a n ----------- n + a n – 1 ------------------ n + …a + 1 ---------- – + 1 a 0 y 2 = 0

a n ------- y n ( 1 + y 2 )a + n – 1 ------------- y n – 1 ( 1 + y 2 )…a + + 1 ----- y ( 1 + y 2 )a + 0 ( y 1 + y 2 )

= a n ------- y n 1 + a n – 1 ------------- y n – 1 1 + …a + 1 ----- y 1 + a y

+ a n ------- y n 2 + a n – 1 ------------- y n – 1 2 + …a + 1 ----- y 2 + a 0 y = 0

In general, if

y = k 1 y 1 ()k t , 2 y 1 ()k t , 3 y 3 ()…k t ,, n y n () t

are the solutions of the homogeneous ODE of (5.13), the linear combination n

y = k 1 y 1 ()k t + 2 y 1 ()k t + 3 y 3 ()…k t + + n y n () t

is also a solution. In our subsequent discussion, the solution of the homogeneous ODE, i.e., the complementary

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Chapter 5 Differential Equations, State Variables, and State Equations

solution, will be referred to as the natural response, and will be denoted as y N () t or simply y N . The particular solution of a non-homogeneous ODE will be referred to as the forced response, and will

be denoted as y F () t or simply y F . Accordingly, we express the total solution of the non-homoge- neous ODE of (5.12) as:

yt () = y Natural + y Forced = y N + y F (5.18)

Response

Response

The natural response y N contains arbitrary constants and these can be evaluated from the given initial conditions. The forced response y F , however, contains no arbitrary constants. It is impera-

tive to remember that the arbitrary constants of the natural response must be evaluated from the total response.