# Lecture 12 Higher Order ODEs - Particular Solutions by etssetcf

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```									                     Department of Engineering Mathematics
EMAT10100 Eng Maths 1 - Calculus

Lecture 12: Higher Order ODEs - Particular Solutions
In the previous lecture we studied solutions of second order, linear, homogeneous ODEs with
constant coeﬃcients. Recall that equations of this type:

x    ˙
a¨ + bx + cx = 0,

can be solved by computing roots of:

am2 + bm + c = 0,

resulting in 2 roots, m1 and m2 . This results in 3 scenarios:

1. m1 = m2 ∈ R resulting in the general solution:

x(t) = Cem1 t + Dem2 t

2. m1 = m2 = m ∈ R resulting in the general solution:

x(t) = Cemt + Dtemt

3. m1,2 = α ± jβ - a complex conjugate pair, results in the general solution:

x(t) = eαt [C cos (βt) + D sin(βt)]

In this ﬁnal lecture, we consider how to solve non-homogenous equations. We again consider
only those with constant coeﬃcients, i.e. of the form:

d2 x    dx
a      2
+b    + cx = f(t),
dt      dt
where f(t) = 0. Noting that the ODE of interest may be rewritten as the LINEAR OPERATOR
L:
L[x] = f(t),
then it can be shown that the most general solution of the ODE is:

x(t) = x(t) + xp (t),

where

• x(t) is the solution of the associated homogeneous equation: L[x] = 0, which is termed
the COMPLEMENTARY FUNCTION.

• xp (t) is any solution you can ﬁnd of the full non-homogenous equation, which is termed
the PARTICULAR SOLUTION.

This result is easy to prove.

(i) x(t) is a solution of L[x] = 0
(ii) xp (t) is a solution of L[x] = f(t)

then:
L[x + xp ] = f(t)
From (i) we have L[x] = 0 and from (ii) L[xp ] = f(t) then:

L[x + xp ] = L[x] + L[xp ] = f(t).

Unfortunately, that is as good as it gets as currently we have no general methodology for deter-
mining the particular solution xp (t) for any given f(t) and we must resort to essentially trial and
error (with some educated guessing thrown in for good measure!)

Example:
Consider the equation:
x + x − 2x = 3e−t
¨ ˙
The solution is given by
x(t) = x(t) + xp (t)
To ﬁnd x we solve the associated homogeneous equation:

¨ ˙
x + x − 2x = 0

⇒ m2 + m − 2 = 0

⇒ m1 = −2 and m2 = 1

From this we can conclude that the complementary function is given by:

x(t) = Ae−2t + Bet

Now to ﬁnd the particular solution. Since f(t) = 3e−t , our trial function must be something that
when diﬀerentiated involves e−t , so we try a trial function xp (t) = Ce−t .
To check whether this trial function is a solution of the DE, we substitute xp (t) = Ce−t into the
DE by observing that:

xp = −Ce−t
˙                                           xp = Ce−t
¨

we substitute these values as follows:
¨
xp     ˙
+xp    −2xp = 3e−t ,
⇒ Ce  −t
−Ce −t
−2Ce−t = 3e−t ,
⇔ −2C = 3
3
⇒ C=− .
2
Therefore:
3
xp (t) = − e−t
2
From these we can write down the GENERAL SOLUTION of the ODE:
3
x(t) = x + xp = Ae−2t + Bet − e−t
2
Note that you should only make use of any initial conditions once you have written down the
general solution. If you attempt to ﬁnd the parameters A and B before you have evaluated the
particular solution, you will not ﬁnd the correct parameter values!

Example: (The Failure Case)
Suppose that we now consider a similar looking ODE:

x − 5x + 4x = et
¨    ˙

We ﬁrst ﬁnd the complementary function by evaluating the roots of the characteristic polynomial
associated with the ODE:

m2 − 5m + 4 = 0 ⇒ m1 = 4, m2 = 1.

Therefore the complementary function is given by:

x(t) = Ae4t + Bet .

Now to ﬁnd the particular solution, we would be tempted to try a trial function of the form:
xp (t) = Ce−t , as before since the forcing term involves the function et . However, e−t forms
part of the complementary function, which we know when diﬀerentiated will give a solution to
the homogeneous equation, which is inconsistent with the solution we need. What should we do
You may recall from our study of the homogeneous case, that when we had repeated roots of the
equation, we consider a solution that involved multiplying the root by a power of t. We shall try
the same trick again and assume that our trial solution takes the form xp (t) = Cte−t instead.
In this case:

xp = Ctet ⇒ x˙p = C(tet + et ),
⇒ xp = C(tet + 2et ),
¨

so that
C(tet + 2et ) − 5C(tet + et ) + 4Ctet = et
1
i.e. C = − 3 . Thus the General Solution in this case is given by:
1
x(t) = Ae4 t + Bet − tet .
3

Table 1: In general, the choice of xp (t) can be done using the following guidelines

Case                            f(t)                                     xp (t)
kt
(i)                            Ae                                       αekt
(ii) f(t) is an exponential (ekt ) which features in the CF             αtekt
A cos(ωt) or
(iii)                                                           C sin(ωt) + D cos(ωt)
A sin(ωt)
(iv)              f(t) is a polynomial of order n            polynomial in t of same order

As we mentioned before, ﬁnding particular solutions is in general not possible. Essentially if we
do not have a RHS that is of the above form then ﬁnding the particular solution isn’t going to
happen!
Example
Find the particular solution of:
¨    ˙
x − 5x + 6x = 24t
In this case we consider a particular solution of the form:

xp (t) = K1 t + K2

¨
In this case we see that x˙p = K1 and xp = 0. Substituting these into the ODE we have obtain:

−5K1 + 6(K1 t + K2 ) = 24t

From this we have two simultaneous equations:

6K1 = 24,
−5K1 + 6K2 = 0.
10
This tells us that K1 = 4 and K2 =      3
.

Example
Solve the ODE
d2 y    dy
2
+3    + 2y = 10 cos x,                        y(0) = 1, y (0) = 0
dx      dx
This problem is slightly diﬀerent in that we have some initial data, that we will utilise to evaluate
the coeﬃcients of the complementary function. The ﬁrst step though is to solve the equation
and we proceed in a similar manner to before.
We ﬁrst ﬁnd the complementary function by evaluating the roots of the characteristic polynomial:

m2 + 3m + 2 = 0 ⇒ m1 = −2, m1 = −1,

from which we write down the CF:

y = Ae−x + Be−2x .

Consulting table 1, we shall try a trial solution of the form:

yp (x) = λ cos x + µ sin x,
⇒ yp (x) = −λ sin x + µ cos x,
⇒ yp (x) = −λ cos x − µ sin x.

substituting these into the ODE we obtain:

(−λ cos x − µ sin x) + 3(−λ sin x + µ cos x) + 2(λ cos x + µ sin x) = 10 cos x.

Equating coeﬃcients of cos x and sin x we see that:

λ + 3µ = 10
⇒      λ = 1,     µ=3
3λ − µ = 0

This gives us the general solution:

y(x) = Ae−x + Be−2x + cos x + 3 sin x.
To ﬁnd the values of the parameters A and B, we make use of the initial data: y(0) = 1, y (0) =
0, which gives us another set of simultaneous equations to solve for A and B:

1=A+B+1
⇒      A = −3,      B = 3.
0 = −A − 2B + 3

y(x) = 3e−2x − 3e−x + cos x + 3 sin x.

Exercises:

1. Suppose that:
¨
x + x = t,
and that x(0) = 0 and x( π ) = 0. Determine the value of x when t = π .
2                                          4

2. Evaluate the general solution of the ODE:

d2 y    dy
2
+8    + 25y = 48 cos x − 16 sin x.
dx      dx
Find also the particular solution given y(0) = 8 and y (0) = −27.

3. Find the particular solution of the ODE:

x − 5x + 6x = (12t − 7)e−t
¨    ˙

HINT: You should try a trial solution, xp (t) of the form (At + B)e−t .

HOMEWORK: Read James pg. 723–734. Exercises 10.9.2 and 10.9.4
Well that about wraps it up for this part of the course. In the next part of the course we will be
considering probability theory.

John R. Terry    Weeks 11-16

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