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3.4 Second-Order Differential Equations
The second order differential equation with constant coefficient
Consider a second order differential equation of the form
d2y
= m2y,
dx 2
where m2 is a positive constant
(3.4-8)
Assume that the solution is y = Aekx, where A and k are two unknown constants then
dy
d2y
kx
= Ake ,
= Ak2ekx
dx
dx 2
The constant k can be obtained by substituting the solution and its second derivative into Eq.
(3.4-8)
Ak2ekx = Am2ekx
Therefore k = m or k =  m, and the most general solution of Eq. (3.4-8) is
y = A1emx + A2e-mx
(3.4-9)
where A1 and A2 are the two constants of integration. Eq. (3.4-9) can also be expressed in terms
of the hyperbolic functions sinh(mx) and cosh(mx)
sinh(mx) =
e mx  e  mx
e mx  e  mx
, cosh(mx) =
2
2
or
emx = sinh(mx) + cosh(mx), e-mx =  sinh(mx) + cosh(mx)
y = (A1 - A2)sinh(mx) + (A1 + A2)cosh(mx)
y = B1sinh(mx) + B2cosh(mx)
(3.4-10)
Eq. (3.4-9) is usually used when the range of x is infinite and Eq. (3.4-10) is usually used when
the range of x is finite.
When the coefficient of y in Eq. (3.4-8) is negative, that is
d2y
=  m2y = i2m2y, where i2 = 1
dx 2
Then the solution to Eq. (3.4-11) is
68
(3.4-11)
y = A1eimx + A2e-imx
(3.4-12)
Eq. (3.4-11) can be expressed in terms of the real functions by using the Euler’s identity
eimx = cos(mx) + isin(mx) , and e-imx = cos(mx) - isin(mx)
y = A1(cos(mx) + isin(mx)) + A2(cos(mx) - isin(mx))
y = (A1 + A2)cos(mx) + i(A1 - A2)sin(mx)
y = C1cos(mx) + C2sin(mx)
(3.4-13)
The second order nonhomogeneous differential equation with constant coefficient
Eq. (3.4-8) can be written as
d2y
 m2y = 0
dx 2
(3.4-14)
where all the terms that contains the dependent variable y are moved to the left hand side of the
equation. The right hand side is equal to zero. Eq. (3.4-14) is a homogeneous equation. If the
RHS of the differential equation is not equal to zero, it is a nonhomogeneous equation.
d2y
 m2y = f(x)
dx 2
(3.4-15)
Eq. (3.4-15) is a nonhomogeneous with forcing function f(x). Let consider the simplest case
when f(x) is a constant K.
d2y
 m2y = K
dx 2
(3.4-16)
The general solution, y, of Eq. (3.4-16) is simply the sum of the general solution, yc, of Eq. (3.414) and a particular solution, yp, of Eq. (3.4-16).
y = yc + yp
where
yc = A1emx + A2e-mx
yc is called the complimentary solution of Eq. (3.4-15). A particular solution of Eq. (3.4-16) can
be obtained by assuming yp = C = constant since the RHS of Eq. (3.4-16) is a constant.
Substituting yp into Eq. (3.4-16)
 m2C = K or C =  K/m2
Hence the general solution to Eq. (3.4-16) is
y = yc + yp = A1emx + A2e-mx  K/m2
69
(3.4-17)
Eq. (3.4-17) contains two arbitrary constants A1 and A2 to be determined from the boundary
conditions.
Example: Find the general solution to the differential equation
d2y
 4y = 3; y(0) = 0, y (1) = 2
dx 2
Solution:
The complimentary solution of
equation
d2y
 4y = 3 is the general solution to the homogeneous
dx 2
d2y
 4y = 0
dx 2
yc = A1e2x + A2e-2x
The particular solution of
d2y
 4y = 3 is yp = C
dx 2
 4C = 3  C =  0.75
The general solution is then
y = A1e2x + A2e-2x  0.75
at x = 0, y = 0 = A1 + A2  0.75
 A1 + A2 = 0.75
at x = 1, y = 2 = 7.3891 A1 + 0.1353A2  0.75
 7.3891A1 + 0.1353A2 = 2.75
Solve for A1 and A2 to obtain A1 = 0.3651, and A2 = 0.3849. The final solution is
y = 0.3651e2x + 0.3849e-2x  0.75
>> y=dsolve('D2y-4*y=3','y(0)=0','y(1)=2','x')
y=
-3/4-1/4*(-11+3*cosh(2))/sinh(2)*sinh(2*x)+3/4*cosh(2*x)
>> y=simple(y)
y=
1/4*(-3*sinh(2)+11*sinh(2*x)-3*sinh(2*x-2))/sinh(2)
>> x=0:.1:1;
>> [x' (eval(y))']
ans =
0
0
0.1000 0.0111
0.2000 0.0527
0.3000 0.1265
0.4000 0.2355
0.5000
0.6000
0.7000
0.8000
0.9000
1.0000
70
0.3841
0.5782
0.8256
1.1362
1.5225
2.0000
71