2: SECOND ORDER LINEAR DIFFERENTIAL EQUATION (LDE)
2.1: Introduction to second order linear differential Equations
General Form:
dy
dny
d n 1 y
an(x)
+
a
(x)
+ …+ a1(x)
+ a0(x)y = f(x) …(1)
n-1
n
n 1
dx
dx
dx
where the coefficients a0(x), a1(x),…, an(x), f(x) is the function of x and an(x) ≠ 0.
If one of the coefficients is not constant, hence (1) is called a Linear Differential Equation
with variable coefficient.
If all of the coefficients are constants, hence (1) could be written as:
an
dy
dny
d n 1 y
+
a
+ … + a1
+ a0y = f(x) … (2)
n-1
n
n 1
dx
dx
dx
(2) is called a Linear Differential Equation with constant coefficient.
If f(x) in (1) and (2) equal to zero, is called a Homogeneous Differential Equation
(HDE).
If f(x) ≠ 0, is called Non Homogeneous Diff. Equation.
Examples:
d2y
a)
+ 20y = 0
HDE with constant coefficient.
dx 2
b) y``- 5y` + 3y = ex
Non HDE with constant coefficient.
2
2
c) x y``+xy`+(x -2)y = 0 HDE with variable coefficient.
d 2 y 2 dy
d)
+
= ln x
Non HDE with variable coefficient.
x dx
dx 2
2.2: Solution to homogeneous equations with constant coefficients.
Consider a second order linear differential equation:
a
dy
d2y
+b
+ cy = 0 where a, b, c constant. …… (3)
2
dx
dx
If y = emx is the solution, hence
dy
d2y
= memx and
= m2emx
dx
dx 2
Substitute into (3), hence
dy
d2y
a 2 +b
+ cy = 0 can be written as:
dx
dx
am2emx + bmemx + cemx = 0.
(am2 + bm + c) emx = 0.
But emx ≠ 0, hence
am2 + bm + c = 0 ……. (4).
(4) is the quadratic equation and called
characteristic equation. The roots of (4) are called the characteristic roots.
Equation (4) has three forms of roots. .
(i) Real and different roots, if b2 – 4ac > 0.
(ii) Real and equal roots, if b2 - 4ac = 0.
(iii) Two complex roots, if b2 - 4ac < 0.
Let m1 and m2 are the characteristic roots of equation (4).
a) If b2 – 4ac > 0 hence m1 ≠ m2. Then y1 = em1x and y2 = em2x are the solutions of
the homogeneous equation. Then the general solution written as:
y = A em1x + B em2x { A, B constants}.
b) If b2 – 4ac = 0 hence m1 = m2. The characteristic equation has only one root, m = b
.
2a
Then the general solution written as:
y = (A + Bx) emx.
{A, B constants}
c) If b2 – 4ac < 0 the characteristic equation has two complex roots, m1 = α + βi and
m2 = α – βi .
Then the general solution written as :
y = C.e(α + βi)x + D.e(α – βi)x {C, D constants}.
By using Euler formula:
eiθ = cosθ + i sinθ and e-iθ = cosθ – i sinθ,
then
y = C.e(α + βi)x + D.e(α – βi)x
= eαx { C.eiβx + D.e-iβx }
= eαx { C(cos βx + i sin βx) + D(cos βx – i sin βx)}
= eαx {(C + D) cos βx + i(C – D) sin βx}
= eαx { A cos βx + B sin βx } where A = C + D and
B = (C – D)i.
Conclusion:
If characteristic equation has two complex roots, m1 = α + βi and m2 = α – βi , then the
general solution could be written as:
y = eαx ( A cos βx + B sin βx )
Hence y1 = eαx cos βx and y2 = eαx sin βx
Exercises:
Determine the general solution from the following equations:
1). y`` - y` - 6y = 0
2). y`` - 4y = 0
3). y`` - 2y` - 3y =0 with conditions y(0) = 2
and y`(0) = 1
4). y`` - 4y`+ 13y = 0, y(0) = -1, y`(0) = 2
5). y`` + 4y`+ 4y = 0, y(0) = 2, y`(0) = 1
6). 9y`` - 6y`+ y = 0, y(0) = 6, y`(0) = 9
2.2: Solution to homogeneous equations with constant coefficients.
dy
d2y
a
+b
+ cy = f(x)
2
dx
dx
dy
d2y
-4
+ 3y = 10 e-2x.
2
dx
dx
has ex expression.
Example: Solve the equation:
Solution: f(x) = 10 e-2x
Let Ce-2x is the solution.
Thus: y = Ce-2x ;
dy
d2y
= -2Ce-2x ;
= 4Ce-2x.
2
dx
dx
dy
d2y
-4
+ 3y = 10e-2x. Substitute :
2
dx
dx
4Ce-2x -4(-2Ce-2x) + 3Ce-2x = 10e-2x.
15Ce-2x = 10e-2x.
C = 2/3.
2
Hence y = e-2x satisfied the given equation and is called the particular
3
integral.
The other solution which could be obtain from homogenous equation is
dy
d2y
-4
+ 3y = 0.
2
dx
dx
Characteristic equation: m2 – 4m + 3 = 0.
→ m = {1, 3}.
The solution of HDE: yc = Aex + Be3x.
General Solution: y = Aex +Be3x +
2 -2x
e (A, B constants).
3
Definition:
i) The general solution of equation: a
dy
d2y
+b
+ cy = 0 is yc , called complementary
2
dx
dx
function..
dy
d2y
ii) The solution of : a 2 + b
+ cy = f(x) is yp, called particular integral.
dx
dx
Theorem:
dy
d2y
+b
+ cy = 0 and yp is
2
dx
dx
dy
d2y
the particular integral for non homogenous equation a 2 +b
+ cy = f(x), hence the
dx
dx
general solution of the non homogenous equation is given by: y = yc + yp.
If yc is the complementary function for diff. equation a
2.3.1: Method of Undetermined Coefficients.
Consider : ay``+ by` + c = f(x), a ≠ 0. ………. (i).
The basic idea behind this approach is as follows.
a) f(x) a polynomial of degree n.
b) f(x) an exponential form Ceαx , (α, C constants).
c) f(x) = C cosβx or C sin βx, (C, β constants).
Case a:
f(x) = Anxn + An-1xn-1 + … + A1x + Ao .
(An , An-1 , … , A1 , Ao constants).
Suppose: yp = Bnxn + Bn-1xn-1 + … + B1x + B0. ……(ii).
(Bn , Bn-1 , … , B1 , Bo constants).
Differentiate (ii) for yp`, yp``, … , yp(n) and substituting into (i). Equate the coefficients
of corresponding powers of x, and solve the resulting equations for undetermined
coefficients, then we get: B1 , B2 , … , B1, Bo .
Example: Solve the diff. equation: y`` + 3y` + 2y = 5x2.
Solution:
f(x) = 5x2
Suppose:
yp = ax2 + bx + c.
yp` = 2ax + b
yp`` = 2a
→
y``+ 3y`+ 2y = 5x2.
2a + 3(2ax + b) + 2(ax2+ bx + c) = 5x2.
2ax2 + (6a + 2b)x + (2a + 3b + 2c) = 5x2.
Hence: 2a = 5
→ a=
5
.
2
15
.
2
35
2a + 3b + 2c = 0 → c =
.
4
5
15
35
→ yp = x2 x+
.
2
2
4
6a + 2b = 0
→ b=-
Consider:
y``+ 3y`+ 2 = 0. (HDE).
Characteristic eq. : m2 + 3m + 2 = 0.
(m + 1)(m + 2) = 0.
m = {-1, -2}.
→ yc = Ae-x + Be-2x.
y = yc + yp. or:
y = Ae-x + Be-2x +
5 2 15
35
x x+
.
2
2
4
Example: Solve the equation: y`` - 2y` + y = x2 – 3x.
Solution: f(x) = x2 – 3x.
Suppose: yp = ax2 + bx + c then
yp`= 2ax + b and
yp``= 2a
y``- 2y`+ y = x2 – 3x.
2a – 2(2ax + b) + ax2 + bx + c = x2 – 3x
Hence: a = 1; b = 1; c = 0.
→ yp = x2 + x.
Consider: y`` - 2y`+ y = 0 (HDE)
Characteristic equation: m2 – 2m + 1 = 0
m=1
→ yh = (A + Bx).ex.
General solution: y = (A+Bx)ex + x2 + x.
Case b: f(x) = Ceαx , (C, α constants).
Then: ay``+ by`+ cy = Ceαx. ….. (iii)
Suppose : yp = k.eαx, then,
yp`= αkeαx.
yp``= α2keαx.
=
By substituting yp , yp`, and yp`` into (iii), then [a(α2k) + b(αk) + ck].eαx = Ceαx.
or: aα2k+ bαk + ck = C .
Example: Solve
y`` - y` - 2y = 2e3x.
Solution :
f(x) = 2e3x
Suppose yp = ke3x.
yp`= 3ke3x
yp``= 9ke3x
y`` - y` - 2y = 2e3x
9ke3x – 3ke3x – 2ke3x = 2e3x
4ke3x = 2e3x
1
k=
2
1 3x
Then: yp = e .
2
Consider:
Charac.eq:
y``- y`- 2y = 0.
m2 – m – 2 = 0
m = {2, -1}
Thus : yc = Ae2x + Be-x
The general solution is:
Case c:
y = Ae2x + Be-x +
1 3x
e .
2
f(x) = C cos αx or C sin αx. (C, α constants)
Then ay``+ by` + cy = C cos αx
or ay``+ by` + cy = C sinαx
For the two expressions, suppose
yp = P cos αx + Q sin αx
yp` = -αP sin αx + αQ cos αx
yp``= -α2P cos αx – α2Q sin αx.
Substituting yp , yp` dan yp`` into the given equation, then equate the coefficient of
corresponding sin αx or cos αx.
Example. Find the general solution of the equation y``+ 9y = cos 2x.
Solution. The characteristic equation of the homogeneous equation is m2+ 9 = 0 and its
roots are m = ± 3i.
The complementary function is
yc = A cos 3x + B sin 3x.
We choose the particular integral is
yp = p cos 2x + q sin 2x
yp`= -2p sin 2x + 2q cos 2x.
yp``= -4p cos 2x – 4q sin 2x.
Substituting in the given equation we get
y``+ 9y = -4p cos 2x – 4q sin 2x
+ 9(p cos 2x + q sin 2x)
= 5p cos 2x + 5q sin 2x = cos 2x.
1
5
5q = 0 → q = 0
1
Then yp =
cos 2x.
5
→ 5p = 1 → p =
The general solution: y = A cos 3x + B sin 3x +
1
cos 2x.
5
Exercises: Solve the equation.
a) y``+ y` - 6y = 52 cos2x.
b) y``- y`- 2y = cos x+ 3 sin x.
Case d: f(x) = f1(x) ± f2(x) ± f3(x) ± … ± fn(x).
For this case, suppose:
yp = yp1 + yp2 + yp3 + … + ypn , where
yp1 is the particular integral for ay``+ by`+ cy = f1(x)
yp2 is the particular integral for ay``+ by`+ cy = f2(x)
.
Ypn is the particular integral for ay``+ by` + cy = fn(x)
General Solution: y = yc + yp .
Example: Solve the differential equation
y``+ 2y`+ 2y = x2 + sin x.
Solution:
Characteristic equation: m2 + 2m + 2 = 0
m = -1 ± i.
-x
yc = e (A cos x + B sin x).
(i) Suppose yp1 is particular integral for y``+ 2y`+ 2y = x2.
Then yp1 = ax2 + bx + c
yp1` = 2ax + b and yp1``= 2a .
2a + 2(2ax + b) + 2(ax2 + bx + c) = x2.
→ a = ½ , b = -1 , c = ½ .
yp1 = ½ x2 – x + ½ = ½ (x – 1)2.
(ii) yp2 is particular integral for y``+ 2y`+ 2y = sin x .
Then yp2 = p cos x + q sin x.
yp2`= -p sin x + q cos x.
yk2``= -p cos x – q sin x.
y``+ 2y` + 2y = sin x.
(-p cos x – q sin x) + 2(-p sin x + q cos x)
+ 2(p cos x + q sin x)
= sin x.
(-2p + q)sin x + (p + 2q)cos x = sin x.
→ -2p + q = 1
p + 2q = 0
p = -2/5and q = 1/5
2
1
1
cos x + sin x = (sin x – 2kos x).
5
5
5
1
1
Hence: y = e-x(Acos x + Bsin x)+ (x-1)2 + (sin x – 2cosx)
2
5
yp2 = -
Case e: f(x) = g(x).v(x)
f(x)
Pn(x).eαx
Pn(x).cosβx
Pn(x).sin βx
Ceαx.cos βx
or
αx
Ce sin βx
Pn(x)eαxsin βx
Pn(x)eαxcos βx
Yp
xr(Bnxn + Bn-1xn-1 + … + B1x + Bo).eαx
xr(Bnxn + Bn-1xn-1 + … + B1x + B0).cos βx
xr(Bnxn + Bn-1xn-1 + … + B1x + B0).sin βx
xr.eαx(p cos βx + q sin βx)
xr(Bnxn + Bn-1xn-1 + … + B0).eαxsin βx
xr(Bnxn + Bn-1xn-1 + … + B0).eαxcos βx
r is the smallest non negative integer.
Example: Find the general solution of the equation
y`` - 2y` + 3y = ex sin 2x.
Solution.: Characteristic equation: m2 – 2m + 3 = 0
m=1±i 2.
x
yc = e (A cos √2 x + B sin √2 x)
f(x) = ex sin 2x.
yp = ex(p cos 2x + q sin 2x).
yp` = ex{(p + 2q)cos 2x + (-2p + q)sin 2x}.
yp``= ex{(-3p + 4q)cos 2x – (4p + 3q) sin 2x}.
y``- 2y`+ 3y = ex{-2p cos 2x – 2q sin 2x).
→ ex{-2p cos 2x – 2q sin 2x} = ex sin 2x.
Then: p = 0 dan q = - ½ .
Hence: yk = - ½ ex sin 2x.
General solution: y = yc + yp or:
y = ex(A cos
2 x + B sin
2 x – ½ sin 2x).
2.3.2: The Method of Variation of Parameter.
This method can be used in solving non homogeneous differential equation:
dy
d2y
a 2 +b
+ cy = f(x), (a, b, c constans) and
dx
dx
1
f(x) = tan x, cot x, sec x, cosec x, n , ln x.
x
In this method , the general solution is in the form:
y = uy1 + vy2 where u = u(x) and v = v(x) and y1 , y2 are independent solution
respectively.
The method of solution as follows.
Given: ay``+ by` + cy = f(x).
i)
Determine a and f(x).
ii)
Determine y1 and y2, the independent solution for homogeneous linier
equation.
y1
iii) Find Wronskian: W =
i)
Obtain: u = - ∫
y2
'
1
y2'
y
 aW .
y 2 f ( x)
y f ( x)
dx + A and v = ∫ 1
dx + B.
aW
aW
***or can use Cramer Rule
w 
0
w 
y1
y1
1
2
ii)
w
y f ( x)
 
dx
y
W
aW
w
y f ( x)
0

dx
 y f ( x) then v 
f ( x)
W
aW
y2
'
2
f ( x)
 y2 f ( x) then
u
1
2
1
Hence the general solution is: y = uy1 + vy2.
Example: Solve the following differential equations:
(i) y``+ y = cot x.
(ii) y`` + 6y`+ 8y = e-2x.
Soluton (i): y`` + y` = cot x.
i) a = 1, f(x) = cot x.
ii) The characteristic equation is m2 + 1 = 0.
thus m = ± i and yc = Acos x + Bsin x
hence y1 = cos x and y1` = - sin x.
y2 = sin x and y2` = cos x.
cos x
 sin x
iii)
W=
iv)
u=-∫
v=∫
sin x
cos x
= cos2x + sin2x = 1.
sin x. cot x
y f ( x)
dx = -∫
dx = - sin x + A.
1
aW
2
y f ( x)
dx =
aW
1
2
cos x
∫cos x.cot x dx = ∫
dx
sin x
= ∫(cosec x – sin x)dx
2
1
= ln[cosec x – cot x] + cos x + B.
**can use Cramer rule
0
sin x
 sin x cot x
cot x
cos x
cos x
0
 cos x cot x
w 
 sin x
cot x
w
u
  sin x cot xdx   cos xdx   sin x  A
W
w 
1
2
1
2
w
cos
v
  cos cot xdx  
xdx  ln[csc x  cot c]  cos x  B
W
sin x
2
v)
General solution: y = uy1 + vy2.
y = (-sinx+A)kosx+(ln[cosecx–cotx]+cosx+B)sinx.
2.4: Solution to Cauchy-Euler Equations.
A differential equation in the form of
dny
d n 1 y
dy
n-1
anx dx n + an-1x dx n 1 + … + a1x + a0x = f(x)
dx
where a0, a1, … , an are constants, is known as Euler’s equation of nth order.
n
A second order Euler’s equation can be written as :
d2y
dy
ax dx 2 + bx
+ cy = f(x) [a, b and c constants] … (1)
dx
2
The method of solution.
Substitute x = et, or, equivalent t = ln x and
dt 1

dx x
dy dy dt dy 1
dy
dy
 .  . or x
=
. … (2)
dx dt dx dt x
dx
dt
d
dy
d dy
(x ) =
( )
dx dx
dx dt
dy
d dy dt
d2y
d2y 1
( )
x 2 +
=
=
.
dx
dt dt dx
dx
dt 2 x
d2y
dy
dy
dy
d2y
2
x dx 2 =
-x
[x
=
]
2
dx
dx
dt
dx
d2y
d 2 y dy
2
→ x dx 2 =
… (3)
dt
dx 2
Substitute (2) and (3) into (1) and then
[
dt 1
 ]
dx x
dy
d 2 y dy
a( 2 
)+b
+ cy = f(et) or
dt
dt
dt
2
dy
d y
a 2 + (b – a)
+ cy = f(et) … (4)
dt
dt
(4) is the Euler’s equation with constant coefficients.
d2y
dy
Example: Solve the equation x dx 2 - 2x
- 4y = x2.
dx
Solution: a = 1, b = -2, c = -4.
2
Substitute x = et, then t= ln x and
dy
1
=
dt
x
dy
d2y
+ (b - a)
+ cy = f(et).
2
dt
dt
dy
d2y
→
-3
- 4y = e2t.
2
dt
dt
yc = Ae4t + Be-t
a
yp = ke2t, yp` = 2ke2t , yp``= 4ke2t
4ke2t – 6ke2t – 4ke2t = e2t.
1
k=6
1
yp = - e2t
6
1
→ y = Ae4t + Be-t - e2t and substitute et=x then
6
B 1 2
y = Ax4 +
- x.
x 6