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Second Order
Differential Equations
Introduction
In this Section we start to learn how to solve second order differential equations of a particular type: those that are linear and have constant coefficients. Such equations are used widely in the modelling of physical phenomena, for example, in the analysis of vibrating systems and the analysis of electrical circuits.
The solution of these equations is achieved in stages. The first stage is to find what is called a ‘com- plementary function’. The second stage is to find a ‘particular integral’. Finally, the complementary function and the particular integral are combined to form the general solution.
Prerequisites
Before starting this Section you should...
- understand what is meant by a differential equation
- understand complex numbers ( 10) '
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%
Learning Outcomes
On completion you should be able to...
- recognise a linear, constant coefficient equation
- understand what is meant by the terms ‘auxiliary equation’ and ‘complementary function’
- find the complementary function when the auxiliary equation has real, equal or complex roots
30 HELM (2008): Workbook 19: Differential Equations
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1. Constant coefficient second order linear ODEs
We now proceed to study those second order linear equations which have constant coefficients. The general form of such an equation is:
a d^2 y dx^2
where a, b, c are constants. The homogeneous form of (3) is the case when f (x) ≡ 0 :
a d^2 y dx^2
To find the general solution of (3), it is first necessary to solve (4). The general solution of (4) is called the complementary function and will always contain two arbitrary constants. We will denote this solution by ycf.
The technique for finding the complementary function is described in this Section.
Task State which of the following are constant coefficient equations. State which are homogeneous.
(a) d^2 y dx^2
dy dx
- 3y = e−^2 x^ (b) x d^2 y dx^2
- 2y = 0
(c) d^2 x dt^2
dx dt
dy dx
Your solution (a) (b) (c) (d)
Answer (a) is constant coefficient and is not homogeneous.
(b) is homogeneous but not constant coefficient as the coefficient of d^2 y dx^2 is x, a variable.
(c) is constant coefficient and homogeneous. In this example the dependent variable is x. (d) is constant coefficient and homogeneous.
Note: A complementary function is the general solution of a homogeneous, linear differential equation.
HELM (2008): Section 19.3: Second Order Differential Equations
®
Example 6
Find values of k so that y = ekx^ is a solution of: d^2 y dx^2
dy dx
− 6 y = 0
Hence state the general solution.
Solution As suggested we try a solution of the form y = ekx. Differentiating we find dy dx = kekx^ and d^2 y dx^2 = k^2 ekx.
Substitution into the given equation yields: k^2 ekx^ − kekx^ − 6 ekx^ = 0 that is (k^2 − k − 6)ekx^ = 0 The only way this equation can be satisfied for all values of x is if k^2 − k − 6 = 0 that is, (k − 3)(k + 2) = 0 so that k = 3 or k = − 2. That is to say, if y = ekx^ is to be a solution of the differential equation, k must be either 3 or − 2. We therefore have found two solutions: y 1 (x) = e^3 x^ and y 2 (x) = e−^2 x These are linearly independent and therefore the general solution is ycf(x) = Ae^3 x^ + Be−^2 x The equation k^2 − k − 6 = 0 for determining k is called the auxiliary equation.
Task By substituting y = ekx, find values of k so that y is a solution of d^2 y dx^2
dy dx
Hence, write down two solutions, and the general solution of this equation.
First find the auxiliary equation:
Your solution
Answer k^2 − 3 k + 2 = 0
HELM (2008): Section 19.3: Second Order Differential Equations
Now solve the auxiliary equation and write down the general solution:
Your solution
Answer The auxiliary equation can be factorised as (k − 1)(k − 2) = 0 and so the required values of k are 1 and 2. The two solutions are y = ex^ and y = e^2 x. The general solution is ycf(x) = Aex^ + Be^2 x
Example 7
Find the auxiliary equation of the differential equation:
a d^2 y dx^2
Solution We try a solution of the form y = ekx^ so that dy dx = kekx^ and d^2 y dx^2 = k^2 ekx. Substitution into the given differential equation yields: ak^2 ekx^ + bkekx^ + cekx^ = 0 that is (ak^2 + bk + c)ekx^ = 0 Since this equation is to be satisfied for all values of x, then ak^2 + bk + c = 0 is the required auxiliary equation.
Key Point 5
The auxiliary equation of a d^2 y dx^2
34 HELM (2008):
Workbook 19: Differential Equations
Example 8
Find the general solution of: d^2 y dx^2
dy dx
− 10 y = 0
Solution
By letting y = ekx, so that
dy dx = kekx^ and
d^2 y dx^2 = k^2 ekx
the auxiliary equation is found to be: k^2 + 3k − 10 = 0 and so (k − 2)(k + 5) = 0 so that k = 2 and k = − 5. Thus there exist two solutions: y 1 = e^2 x^ and y 2 = e−^5 x. We can write the general solution as: y = Ae^2 x^ + Be−^5 x
Example 9
Find the general solution of: d^2 y dx^2
Solution
As before, let y = ekx^ so that dy dx = kekx^ and d^2 y dx^2 = k^2 ekx.
The auxiliary equation is easily found to be: k^2 + 4 = 0 that is, k^2 = − 4 so that k = ± 2 i, that is, we have complex roots. The two independent solutions of the equation are thus y 1 (x) = e^2 ix^ y 2 (x) = e−^2 ix so that the general solution can be written in the form y(x) = Ae^2 ix^ + Be−^2 ix. However, in cases such as this, it is usual to rewrite the solution in the following way. Recall that Euler’s relations give: e^2 ix^ = cos 2x + i sin 2x and e−^2 ix^ = cos 2x − i sin 2x so that y(x) = A(cos 2x + i sin 2x) + B(cos 2x − i sin 2x). If we now relabel the constants such that A + B = C and Ai − Bi = D we can write the general solution in the form: y(x) = C cos 2x + D sin 2x
Note: In Example 8 we have expressed the solution as y =... whereas in Example 9 we have expressed it as y(x) =.... Either will do.
36 HELM (2008):
Workbook 19: Differential Equations
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Example 10
Given ay′′^ + by′^ + cy = 0, write down the auxiliary equation. If the roots of the auxiliary equation are complex (one root will always be the complex conjugate of the other) and are denoted by k 1 = α + βi and k 2 = α − βi show that the general solution is: y(x) = eαx(A cos βx + B sin βx)
Solution Substitution of y = ekx^ into the differential equation yields (ak^2 +bk +c)ekx^ = 0 and so the auxiliary equation is: ak^2 + bk + c = 0 If k 1 = α + βi, k 2 = α − βi then the general solution is y = Ce(α+βi)x^ + De(α−βi)x where C and D are arbitrary constants. Using the laws of indices this is rewritten as: y = Ceαxeβix^ + Deαxe−βix^ = eαx(Ceβix^ + De−βix) Then, using Euler’s relations, we obtain:
y = eαx(C cos βx + Ci sin βx + D cos βx − Di sin βx) = eαx{(C + D) cos βx + (Ci − Di) sin βx}
Writing A = C + D and B = Ci − Di, we find the required solution: y = eαx(A cos βx + B sin βx)
Key Point 7
If the auxiliary equation has complex roots, α + βi and α − βi, then the complementary function is: ycf = eαx(A cos βx + B sin βx)
HELM (2008): Section 19.3: Second Order Differential Equations
®
Example 11
The auxiliary equation of ay′′^ + by′^ + cy = 0 is ak^2 + bk + c = 0. Suppose this equation has equal roots k = k 1 and k = k 1. Verify that y = xek^1 x^ is a solution of the differential equation.
Solution We have: y = xek^1 x^ y′^ = ek^1 x(1 + k 1 x) y′′^ = ek^1 x(k^21 x + 2k 1 ) Substitution into the left-hand side of the differential equation yields: ek^1 x{a(k^21 x + 2k 1 ) + b(1 + k 1 x) + cx} = ek^1 x{(ak 12 + bk 1 + c)x + 2ak 1 + b} But ak^21 + bk 1 + c = 0 since k 1 satisfies the auxiliary equation. Also,
k 1 = −b ±
b^2 − 4 ac 2 a but since the roots are equal, then b^2 − 4 ac = 0 hence k 1 = −b/ 2 a. So 2 ak 1 + b = 0. Hence ek^1 x{(ak^21 + bk 1 + c)x + 2ak 1 + b} = ek^1 x{(0)x + 0} = 0. We conclude that y = xek^1 x^ is a solution of ay′′^ + by′^ + cy = 0 when the roots of the auxiliary equation are equal. This illustrates Key Point
Example 12
Obtain the general solution of the equation:
d^2 y dx^2
dy dx
Solution As before, a trial solution of the form y = ekx^ yields an auxiliary equation k^2 + 8k + 16 = 0. This equation factorizes so that (k + 4)(k + 4) = 0 and we obtain equal roots, that is, k = − 4 (twice). If we proceed as before, writing y 1 (x) = e−^4 x^ y 2 (x) = e−^4 x, it is clear that the two solutions are not independent. We need to find a second independent solution. Using the result summarised in Key Point 8, we conclude that the second independent solution is y 2 = xe−^4 x. The general solution is then: y(x) = (A + Bx)e−^4 x
HELM (2008): Section 19.3: Second Order Differential Equations
Exercises
- Obtain the general solutions, that is, the complementary functions, of the following equations:
(a) d^2 y dx^2
dy dx
dy dx
dx dt
(d) d^2 y dt^2
dy dt
dy dx
dy dt
(g) d^2 y dx^2
dy dx
dy dt
dy dx − 2 y = 0
(j) d^2 y dx^2
dy dx = 0 (l) d^2 x dt^2 − 16 x = 0
- Find the auxiliary equation for the differential equation L d^2 i dt^2
+ R
di dt
C
i = 0
Hence write down the complementary function.
- Find the complementary function of the equation d^2 y dx^2
dy dx
Answers
- (a) y = Aex^ + Be^2 x (b) y = Ae−x^ + Be−^6 x (c) x = Ae−^2 t^ + Be−^3 t (d) y = Ae−t^ + Bte−t (e) y = Ae^2 x^ + Bxe^2 x (f) y = e−^0.^5 t(A cos 2. 78 t + B sin 2. 78 t) (g) y = Aex^ + Bxex (h) x = e−^0.^5 t(A cos 2. 18 t + B sin 2. 18 t) (i) y = Ae−^2 x^ + Bex (j) y = A cos 3x + B sin 3x (k) y = A + Be^2 x (l) x = Ae^4 t^ + Be−^4 t
- Lk^2 + Rk +
C
= 0 i(t) = Aek^1 t^ + Bek^2 t^ k 1 , k 2 =
2 L
−R ±
R^2 C − 4 L
C
- e−x/^2
A cos
√ 3 2 x^ +^ B^ sin^
√ 3 2 x
40 HELM (2008):
Workbook 19: Differential Equations
Task State what is meant by a particular integral.
Your solution
Answer A particular integral is any solution of a differential equation.
4. Finding a particular integral
In the previous subsection we explained what is meant by a particular integral. Now we look at a simple method to find a particular integral. In fact our method is rather crude. It involves trial and error and educated guesswork. We try solutions which are of the same general form as the f (x) on the right-hand side.
Example 13
Find a particular integral of the equation d^2 y dx^2
dy dx − 6 y = e^2 x
Solution We shall attempt to find a solution of the inhomogeneous problem by trying a function of the same form as that on the right-hand side of the ODE. In particular, let us try y(x) = Ae^2 x, where A is a constant that we shall now determine. If y(x) = Ae^2 x^ then dy dx = 2Ae^2 x^ and d^2 y dx^2 = 4Ae^2 x.
Substitution in the ODE gives: 4 Ae^2 x^ − 2 Ae^2 x^ − 6 Ae^2 x^ = e^2 x that is, − 4 Ae^2 x^ = e^2 x To ensure that y is a solution, we require − 4 A = 1, that is, A = −^14. Therefore the particular integral is yp(x) = −^14 e^2 x.
In Example 13 we chose a trial solution Ae^2 x^ of the same form as the ODE’s right-hand side. Table 2 provides a summary of the trial solutions which should be tried for various forms of the right-hand side.
42 HELM (2008): Workbook 19: Differential Equations
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Table 2: Trial solutions to find the particular integral f (x) Trial solution (1) constant term c constant term k
(2) linear, ax + b Ax + B
(3) polynomial in x polynomial in x of degree r: of degree r: axr^ + · · · + bx + c Axr^ + · · · + Bx + k
(4) a cos kx A cos kx + B sin kx
(5) a sin kx A cos kx + B sin kx
(6) aekx^ Aekx
(7) ae−kx^ Ae−kx
Task By trying a solution of the form y = αe−x^ find a particular integral of the equation d^2 y dx^2
dy dx
− 2 y = 3e−x
Substitute y = αe−x^ into the given equation to find α, and hence find the particular integral:
Your solution
Answer α = −^32 ; yp(x) = −^32 e−x
HELM (2008): Section 19.3: Second Order Differential Equations
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Task Find a particular integral for the equation: d^2 y dx^2
dy dx
First decide on an appropriate form for the trial solution, referring to Table 2 (page 43) if necessary:
Your solution
Answer From Table 2, y = A cos x + B sin x, A and B constants.
Now find dy dx and d^2 y dx^2 and substitute into the differential equation:
Your solution
Answer Differentiating, we find: dy dx
= −A sin x + B cos x d^2 y dx^2
= −A cos x − B sin x
Substitution into the differential equation gives: (−A cos x − B sin x) − 6(−A sin x + B cos x) + 8(A cos x + B sin x) = 3 cos x
HELM (2008): Section 19.3: Second Order Differential Equations
Equate coefficients of cos x:
Your solution
Answer 7 A − 6 B = 3
Also, equate coefficients of sin x:
Your solution
Answer 7 B + 6A = 0
Solve these two equations in A and B simultaneously to find values for A and B, and hence obtain the particular integral:
Your solution
Answer A = 2185 , B = −^1885 , yp(x) = 2185 cos x − 1885 sin x
46 HELM (2008):
Workbook 19: Differential Equations
Engineering Example 2
An LC circuit with sinusoidal input
The differential equation governing the flow of current in a series LC circuit when subject to an
applied voltage v(t) = V 0 sin ωt is L d^2 i dt^2
C
i = ωV 0 cos ωt
L C
i
v
Figure 3 Obtain its general solution.
Solution
The homogeneous equation is L d^2 icf dt^2
icf C
Letting icf = ekt^ we find the auxiliary equation is Lk^2 + (^) C^1 = 0 so that k = ±i/
LC. Therefore, the complementary function is:
icf = A cos t √ LC
where A and B arbitrary constants.
To find a particular integral try ip = E cos ωt + F sin ωt, where E, F are constants. We find: dip dt = −ωE sin ωt + ωF cos ωt d^2 ip dt^2 = −ω^2 E cos ωt − ω^2 F sin ωt
Substitution into the inhomogeneous equation yields:
L(−ω^2 E cos ωt − ω^2 F sin ωt) +
C
(E cos ωt + F sin ωt) = ωV 0 cos ωt
Equating coefficients of sin ωt gives: −ω^2 LF + (F/C) = 0. Equating coefficients of cos ωt gives: −ω^2 LE + (E/C) = ωV 0. Therefore F = 0 and E = CV 0 ω/(1 − ω^2 LC). Hence the particular integral is
ip = CV 0 ω 1 − ω^2 LC cos ωt.
Finally, the general solution is:
i = icf + ip = A cos √t LC
CV 0 ω 1 − ω^2 LC cos ωt
48 HELM (2008):
Workbook 19: Differential Equations
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6. Inhomogeneous term in the complementary function
Occasionally you will come across a differential equation a d^2 y dx^2
- b dy dx
- cy = f (x) for which the
inhomogeneous term, f (x), forms part of the complementary function. One such example is the equation
d^2 y dx^2
dy dx − 6 y = e^3 x
It is straightforward to check that the complementary function is ycf = Ae^3 x^ + Be−^2 x. Note that the first of these terms has the same form as the inhomogeneous term, e^3 x, on the right-hand side of the differential equation.
You should verify for yourself that trying a particular integral of the form yp(x) = αe^3 x^ will not work in a case like this. Can you see why?
Instead, try a particular integral of the form yp(x) = αxe^3 x. Verify that
dyp dx
= αe^3 x(3x + 1) and d^2 yp dx^2
= αe^3 x(9x + 6).
Substitute these expressions into the differential equation to find α = 15.
Finally, the particular integral is yp(x) = 15 xe^3 x^ and so the general solution to the differential equation is:
y = Ae^3 x^ + Be−^2 x^ + 15 xe^3 x
This shows a generally effective method - where the inhomogeneous term f (x) appears in the com- plementary function use as a trial particular integral x times what would otherwise be used.
Key Point 10
When solving a d^2 y dx^2
- b dy dx
- cy = f (x) if the inhomogeneous term f (x) appears in the complementary function use as a trial particular integral x times what would otherwise be used.
HELM (2008): Section 19.3: Second Order Differential Equations
