De Higher Order Ldes

DIFFERENTIAL
EQUATIONS
TOPIC 4 HIGHER-ORDER LINEAR
DIFFERENTIAL EQUATIONS
LEARNING OUTCOMES
• At the end of the discussion, the students
should be able to:
• distinguish homogeneous from nonhomogeneous linear differential
equations given the standard forms;
• solve homogeneous linear differential equations with constant
coefficients;
• apply the method of undetermined coefficients in solving
nonhomogeneous linear differential equations with constant
coefficients;
• utilize the principle of variation of parameters in solving
nonhomogeneous differential equations with constant coefficients; and
• solve higher order differential equations using computer.
joseemalyn.wordpress.com 2
R E V I E W
• Review ……Factoring
Review ……
Review ……Synthetic Division
Review …
Review …Quadratic Formula
R E V I E W cont.
• Synthetic Division cont.

• 8m4 + 36m3 − 66m2 + 35m − 6 = 0 -6 8 36 -66 35 -6
-48 72 -36 6
8m3 − 12m2 + 6m − 1 = 0 1/2 8 -12 6 -1 0
4 -4 1
8m2 − 8m + 2 = 0 1/2 8 -8 2 0
4 -2
8m − 4 = 0 1/2 8 -4 0
4
8 0
R E V I E W cont.
• Synthetic Division
-6 8 36 -66 35 -6 8 36 -66 35 -6 -6
-48 72 -36 6 -48 72 -36 6
1/2 8 -12 6 -1 0 8 -12 6 -1 0 1/2
4 -4 1 or 4 -4 1
1/2 8 -8 2 0 8 -8 2 0 1/2
4 -2 4 -2
1/2 8 -4 0 8 -4 0 1/2
4 4
8 0 8 0
R E V I E W cont.
• Quadratic Formula
−b ± b2 −4ac
x=
2a
➢ when ax 2 + bx + c = 0
a, b, c = constants; a ≠ 0
x = the unknown
Linear Differential Equations with
Constant Coefficients
• Standard Form
dn y dn−1 dy
an n + an−1 n−1 y + ⋯ + a1 + a0 y = R(x)
dx dx dx
In terms of the differential operator D
an Dn + an−1 Dn−1 + ⋯ + a1 D + a0 y = R x
dy d dy 𝑑2 y
where = Dy, = = D Dy = 𝐷2 y, …
dx dx dx 𝑑𝑥 2
dn y
In general, = Dn y
dxn
Linear Differential Equations with
Constant Coefficients cont.
Standard Solution
y = complementary function + any particular solution
y = yc + y p
Homogeneous Linear Differential
Equations with Constant Coefficients
Standard Form
dn y dn−1 y dy
an n + an−1 n−1 + ⋯ + a1 + a0 y = 0
dx dx dx
In terms of the differential operator D
an Dn + an−1 Dn−1 + ⋯ + a1 D + a0 y = 0
Standard Solution
y = yc
HLDEs with Constant Coefficients cont.
Auxiliary or characteristic equation (A.E.)
an mn + an−1 mn−1 + ⋯ + a1 m + a0 = 0
where
emx is a solution of the standard form if and only if
m is a root of the auxiliary equation
Case 1. Roots of auxiliary equation are real and distinct

For roots m1 , m2 , m3 , … , mp
y = c1 em1x + c2 em2x + c3 em3x + ⋯ + cp empx
where c1 , c2 , c3 , … , cp = arbitrary constants
Examples
Solve the following differential equations
1. D2 + 5D + 4 2D2 + 5D + 2 y = 0
Solution
Auxiliary Equation (A.E.)
m2 + 5m + 4 2m2 + 5m + 2 = 0
m + 4 m + 1 2m + 1 m + 2 = 0
m + 4 = 0; m + 1 = 0; 2m + 1 = 0; m+2=0
1
m = -4 m = -1 m= − m = -2
2
′ 1
∴ roots, m s = −4, −1, − , −2
2
1
−2x
y1 = e−4x y3 = e
y2 = e−x y4 = e−2x
General solution (g.s.)
1
−2x
y = c1 e−4x + c2 e−x + c3 e + c4 e−2x Answer
2. 4D3 − 13D + 6 y = 0 -2 4 0 -13 6

-8 16 -6
Solution
1/2 4 -8 3 0
Auxiliary Equation 2 -3
4m3 − 13m + 6 = 0 3/2 4 -6 0

6
1 3
roots, m′ s = −2, , 4 0
2 2
1 3
x x
y = c1 e−2x + c2 e 2 + c3 e 2 g.s. Answer
d3 x d2 x dx
3. − 2 2 − 3 =0
dt3 dt dt
Solution
(D3 − 2D2 − 3D)x = 0
Auxiliary Equation
m3 − 2m2 − 3m = 0 m(m2 − 2m − 3) = 0
roots, m′ s = 0, −1, 3 m(m +1)(m-3) = 0
x = c1 e0 + c2 e−t + c3 e3t
or x = c1 + c2 e−t + c3 e3t g.s. Answer
4. 4D3 − 13D − 6 y = 0 2 4 0 -13 -6
8 16 6
Solution -3/2 4 8 3 0
Auxiliary Equation -6 -3
-1/2 4 2 0
4m3 − 13m − 6 = 0
-2
3 1
roots, m′ s = 2, − , − 4 0
2 2
3 1
− x − x
y = c1 e2x + c2 e 2 + c3 e 2 g.s. Answer
-2 1 0 -5 -2
5. D3 − 5D − 2 y = 0
-2 4 2
Solution
1 -2 -1 0
Auxiliary Equation
m3 − 5m − 2 = 0 m2 − 2m − 1 = 0
−(−2)± (−2)2 −4(1)(−1)
m=
2(1)
2 ± 4+ 4 2± 8 2 ±2 2
m= = = =1± 2
2 2 2
roots, m′ s = −2, 1 + 2, 1 − 2
y = c1 e−2x + c2 e 1+ 2 x + c3 e 1− 2 x g.s. Answer
6. D3 − 3D2 − 3D + 1 y = 0 -1 1 -3 -3 1
-1 4 -1
Solution
1 -4 1 0
Auxiliary equation
m3 − 3m2 − 3m + 1 = 0 m2 − 4m + 1 = 0
−(−4)± (−4)2 −4(1)(1)
m=
2(1)
4 ± 16− 4 4 ± 12 4 ±2 3
m= = = =2± 3
2 2 2
roots, m′ s = −1, 2 + 3, 2 − 3
y = c1 e−x + c2 e 2+ 3 x + c3 e 2− 3 x g.s. Answer
Case 2. Roots of auxiliary equation are real and repeated
For a root m, appearing p times
y = c1 emx + c2 xemx + ⋯ + Cp x p−1 emx
Examples
1. D2 D2 + 2D + 1 3 D2 + 3D + 2 y = 0
Solution
Auxiliary Equation
m2 m2 + 2m + 1 3 m2 + 3m + 2 = 0
m2 m + 1 6 m + 1 m + 2 = 0
m1 = 0 (double)
m2 = −1 (7 times)
m3 = −2
m’s = 0, 0, -1, -1, -1, -1, -1, -1, -1, -2 Cases 1 and 2
y = c1 + c2 x + c3 e−x + c4 xe−x + c5 x 2 e−x + c6 x 3 e−x + c7 x 4 e−x +
c8 x 5 e−x + c9 x 6 e−x + c10 e−2x
y = c1 + c2 x + (c3 + c4 x + c5 x 2 + c6 x 3 + c7 x 4 + c8 x 5 + c9 x 6 )e−x
+ c10 e−2x g.s. Answer
2. D2 + 6D + 9 y = 0
Solution
Auxiliary Equation
m2 + 6m + 9 = 0
(m + 3)(m + 3) = 0
m+3=0 m = −3
roots, m′ s = −3, −3
y = c1 e−3x + c2 xe−3x
y = c1 + c2 x e−3x g.s. Answer
3. 4D3 − 27D + 27 y = 0 -3 4 0 -27 27
-12 36 -27
Solution
3/2 4 -12 9 0
Auxiliary Equation 6 -9
4m3 − 27m + 27 = 0 3/2 4 -6 0
3 3 6
roots, m′ s = −3, , 4 0
2 2
3 3
x x
y = c1 e−3x + c2 e 2 + c3 xe 2
3x
y = ce−3x + c2 + c3 x e 2 g.s. Answer
4. 4D4 − 4D3 − 23D2 + 12D + 36 y = 0
Solution 4 -4 -23 12 36 2
8 8 -30 -36
Auxiliary Equation 4 4 -15 -18 0 2
4m4 − 4m3 − 23m2 + 12m + 36 = 0 8 24 18
3 3 4 12 9 0 -3/2
roots, m′ s = 2, 2, − , −
2 2 -6 -9
4 6 0 -3/2
-6
4 0
3 3
− x − x
y = c1 e2x + c2 xe2x + c3 e 2 + c4 xe 2
3
− x
y = c1 + c2 x e2x + c3 + c4 x e 2 g.s. Answer
5. D4 − 5D2 − 6D − 2 y = 0 -1 1 0 -5 -6 -2
-1 1 4 2
Solution -1 1 -1 -4 -2 0
Auxiliary Equation -1 2 2
m4 − 5m2 − 6m − 2 = 0 1 -2 -2 0
m2 − 2m − 2 = 0
−(−2)± (−2)2 −4(1)(−2)
m=
2(1)
2 ± 4+8 2 ± 12 2 ±2 3
m= = = =1± 3
2 2 2
roots, m′ s = −1, −1, 1 + 3, 1 − 3
y = c1 e−x + c2 xe−x + c3 e 1+ 3 x + c4 e 1− 3 x
y = c1 + c2 x e−x + c3 e 1+ 3 x
+ c4 e 1− 3 x
g.s. Answer
6. D4 + 3D3 + 2D2 y = 0; when x = 0, y = 0, y ′ = 4, y ′′ = −6,
Solution y ′′′ = 14
Auxiliary Equation
m4 + 3m3 + 2m2 = 0
m2 (m2 + 3m + 2) = 0
m2 m + 1 m + 2 = 0
roots, m′ s = 0, 0, −1, −2
y = c1 e0 + c2 xe0 + c3 e−x + c4 e−2x
y = c1 + c2 x + c3 e−x + c4 e−2x g.s.
y = c1 + c2 x + c3 e−x + c4 e−2x g.s. rewritten
when x = 0, y = 0 : 0 = c1 + c3 + c4
when x = 0, y ′ = 4 : y ′ = c2 − c3 e−x − 2c4 e−2x
4 = c2 − c3 − 2c4
when x = 0, y ′′ = −6 : y ′′ = c3 e−x + 4c4 e−2x
−6 = c3 + 4c4
when x = 0, y ′′′ = 14 : y ′′′ = −c3 e−x − 8c4 e−2x
14 = −c3 − 8c4
Solve for c1 , c2 , c3 , and c4
c1 = 0, c2 = 2, c3 = 2, c4 = −2
y = c1 + c2 x + c3 e−x + c4 e−2x g.s. rewritten
y = 2x + 2e−x − 2e−2x
y = 2 x + e−x − e−2x p.s. Answer
Case 3. Auxiliary equation has complex (imaginary)
and distinct roots
For m = a ± bi
y = eax c1 cos bx + c2 sin bx
Examples
1. D − 3 D2 + 4D + 4 D2 + 2D + 3 y = 0
Solution
Auxiliary Equation
m − 3 m2 + 4m + 4 m2 + 2m + 3 = 0
m − 3 m + 2 2 m2 + 2m + 3 = 0
For m2 + 2m + 3 = 0
−(2) ± (2)2 −4(1)(3)
m=
2(1)
−2 ± 4−12 −2 ± −8 −2 ±2 2 i
m= = = = −1 ± 2 i
2 2 2
roots, m’s = 3, -2, -2, −1 ± 2 i
y = c1 e3x + c2 e−2x + c3 xe−2x + e−x c4 cos 2x + c5 sin 2x
y = c1 e3x + (c2 +c3 x)e−2x + e−x c4 cos 2x + c5 sin 2x
g.s. Answer
2. D2 − 2D + 5 y = 0 Verify your answer.
Solution
Auxiliary Equation
m2 − 2m + 5 = 0
− −2 ± (−2)2 −4(1)(5) 2 ± 4 − 20
m= =
2(1) 2
2± −16 2±4i
m= =
2 2
roots, m′ s = 1 ± 2 i
y = ex (c1 cos 2x + c2 sin 2x) g.s. Answer
Verify the answer
Obtain first y′ and y′′
y ′ = ex −2c1 sin 2x + 2c2 sin 2x + ex c1 cos 2x + c2 sin 2x
y ′′ = ex −3c1 cos 2x − 3c2 sin 2x + ex −4c1 sin 2x + 4c2 cos 2x
Substitute the equations into the given d.e.
−3c1 ex cos 2x − 3c2 ex sin 2x − 4c1 ex sin 2x + 4c2 ex cos 2x + 4c1 ex sin 2x −
4c2 ex sin 2x − 2c1 ex cos 2x − 2c2 ex sin 2x + 5c1 ex cos 2x + 5c2 ex sin 2x = 0
0 = 0 Verified!
3. D2 + 9 y = 0 Verify your answer.
Solution
Auxiliary Equation
m2 + 9 = 0
m2 = −9
m = −9
roots, m′ s = ±3 i
y = e0 (c1 cos 3x + c2 sin 3x)
y = c1 cos 3x + c2 sin 3x g. s. Answer
Verify the answer
y = c1 cos 3x + c2 sin 3x rewritten
y ′ = −3c1 sin 3x + 3c2 cos 3x
y ′′ = −9c1 cos 3x − 9c2 sin 3x
Substitute the equations into the given differential equation
D2 + 9 y = 0 rewritten
−9c1 cos 3x − 9c2 sin 3x + 9c1 cos 3x + 9c2 sin 3x = 0
0=0
Verified!
4. D2− 4D + 7 y = 0 −(−4) ± (−4)2 −4(1)(7)
m=
2(1)
Solution
4 ± 16−28 4 ± −12
Auxiliary Equation = =
2 2
m2 − 4m + 1 = 0 4 ±2 3 i
=
roots, m′ s =2± 3i 2
y = e2x c1 cos 3x + c2 sin 3x g.s. Answer
5. D3 + 7D2 + 19D + 13 y = 0; when x = 0, y = 0, y ′ = 2, y ′′ = −12
Solution -1 1 7 19 13
Auxiliary Equation -1 -6 -13

1 6 13 0
3 2
m + 7m + 19m + 13 = 0
m2 + 6m + 13 = 0
−6± (6)2 −4(1)(13) −6 ± 36−52 −6 ± −16 −6 ±4i

m= = = = = −3 ± 2i
2(1) 2 2 2
m′ s = −1, −3 ± 2i
y = c1 e−x + e−3x c2 cos 2x + c3 sin 2x g.s.
when x = 0, y = 0 : 0 = c1 + c2
when x = 0, y ′ = 2 : y ′ = −c1 e−x + e−3x (−2c2 sin 2x +
2c3 cos 2x) − 3e−3x c2 cos 2x + c3 sin 2x
2 = −c1 + 2c3 − 3c2
when x = 0, y ′′ = −12 : y ′′ = c1 e−x + e−3x (5c2 cos 2x +
5c3 sin 2x) − e−3x −12c2 sin 2x + 12c3 cos 2x
−12 = c1 + 5c2 − 12c3
Solve for c1 , c2 , and c3
c1 = c2 = 0, c3 = 1
y = e−3x sin 2x p.s. Answer
6. 2D3 − D2 + 36D − 18 y = 0 1/2 2 -1 36 -18
Solution 1 0 18
2 0 36 0
Auxiliary Equation
2m3 − m2 + 36m − 18 = 0 2m2 + 36 = 0
1
m′ s = , ±3 2 i m2 + 18 = 0
2
m2 = −18 m = −18 m = ±3 −2
1
x
y = c1 e 2 + e0 c2 cos 3 2 x + c3 sin 3 2 x
1
x
y = c1 e 2 + c2 cos 3 2 x + c3 sin 3 2 x
Case 4. Auxiliary equation has complex (imaginary) and
repeated roots
For m = a ± bi, appearing p times
y = eax c1 cos bx + c2 sin bx + xeax (c3 cos bx +
c4 sin bx) + ⋯ + x p eax c2p−1 cos bx + c2p sin bx
Examples
1. D4 + 18D2 + 81 y = 0
Solution
Auxiliary Equation
m4 + 18m2 + 81 = 0
(m2 + 9) (m2 + 9) = 0
m′ s = ±3i, ±3i
y = e0 (c1 cos 3x + c2 sin 3x) + e0 x c3 cos 3x + c4 sin 3x
y = (c1 + c3 x)cos 3x + (c2 +c4 x)sin 3x g.s. Answer
2. D6 + 9D4 + 24D2 + 16 y = 0
Solution
Auxiliary Equation
m6 + 9m4 + 24m2 + 16 = 0
(m2 + 1)(m2 + 4)(m2 +4) = 0
m′ s = ±i, ±2i, ±2i
y = e0 (c1 cos x + c2 sin x) + e0 c3 cos 2x + c4 sin 2x + e0 x c5 cos 2x + c6 sin 2x
y = c1 cos x + c2 sin x + c3 + c5 x cos 2x + c4 + c6 x sin 2x
g.s. Answer
3. D2 + 2D + 4 3 D2 + 1 2 y = 0
Solution
Auxiliary Equation
m2 + 2m + 4 3 m2 + 1 2
=0
m2 + 2m + 4 = 0 m2 + 1 = 0
m′ s = ± i, ± i
For m2 + 2m + 4 = 0
−2± (−2)2 −4(1) 4 −2± 4−16 −2± −12 −2±2 −3
m= = = =
2(1) 2 2 2
m′ s = −1 ± 3 i, −1 ± 3 i, −1 ± 3 i
For the given d.e.

m′ s = ± i, ± i, −1 ± 3 i, −1 ± 3 i, −1 ± 3 i
• y = c1 cos x + c2 sin x + x c3 cos x + c4 sin x + e−x ൫c5 cos 3x +

c6 sin 3x൯ + xe−x c7 cos 3x + c8 sin 3x + x 2 e−x ൫c9 cos 3x +
c10 sin 3x൯
• y = (c1 + c3 x) cos x + (c2 + c4 x) sin x + e−x (c5 + c7 x + c9 x 2 ) cos 3x
+ e−x (c6 +c8 x + c10 x 2 )sin 3x g.s. Answer
Nonhomogeneous Linear Differential Equations
(NHLDEs) with Constant Coefficients
• Standard Form
dn y dn−1 dy
an n + a𝑛−1 n−1 y + ⋯ + a1 + a0 y = R(x)
dx dx dx
or an Dn + a𝑛−1 Dn−1 + ⋯ + a1 D + a0 y = R x
• Standard Solution
y = complementary function + any particular solution
y = yc + y p
Trial Particular Solutions
𝐑(𝐱) Form of 𝐲𝐩
1. 1 (any constant) A
2. 5x + 7 Ax + B
3. 3x 2 − 2 Ax 2 + Bx + C
4. x 3 − x + 1 Ax 3 + Bx 2 + Cx + E
5. sin 4x A cos 4x + B sin 4x
6. cos 4x A cos 4x + B sin 4x
7. e5x Ae5x
8. 9x − 2 e5x Ax + B e5x
9. x 2 e5x Ax 2 + Bx + C e5x
10. e3x sin 4x Ae3x cos 4x + Be3x sin 4x
11. 5x 2 sin 4x Ax 2 + Bx + C cos 4x + Ex 2 + Fx + G sin 4x
12. xe3x cos 4x Ax + B e3x cos 4x + Cx + E e3x sin 4x
Method of Undetermined Coefficients (MUCs)
➢ applicable if R x is a particular solution of a homogeneous
linear differential equation
➢ i.e., the roots of the auxiliary equation can be obtained from R(x)
Examples
Solve the following d.e.
1. D3 − 3D + 2 y = −7
Solution
y = yc + yp
Auxiliary Equation m3 − 3m + 2 = 0
MUCs cont.
m3 − 3m + 2 = 0 rewritten
m’s = -2, 1, 1 Cases 1 and 2
yc = c1 e−2x + c2 ex + c3 xex -2 1 0 -3 2
-2 4 -2
R(x) = -7; n = 0 1 1 -2 1 0
yp = A 1 -1
Dyp = 0 1 1 -1 0
1
2
D yp = 0 1 0
D3 yp = 0
MUCs cont.
Substitute yp into the given d.e.
D3 − 3D + 2 y = −7
0 + 0 + 2A = -7
−7
A=
2
−7
yp =
2
y = yc + yp
−2x x x 7
∴ y = c1 e + c2 e + c3 xe − g.s. Answer
2
MUCs cont.
2. D2 − D − 2 y = 1 − 2x − 9e−x
Solution
y = yc + yp
Auxiliary Equation
m2 − m −2 = 0
(m −2)(m+1) = 0
m’s = 2, −1
yc = c1 e−x + c2 e2x
MUCs cont.
R(x) = 1 − 2x − 9e−x
n’s = 0, 0, -1
−1, appears two times
yp = A + Bx + Cxe−x
y ′p = B + C −xe−x + e−x
y ′′p = C xe−x − e−x − e−x
y ′′p = C(xe−x − 2e−x )
MUCs cont.
Substitute yp into the given differential equation
D2 − D − 2 y = 1 − 2x − 9e−x rewritten
C xe−x − 2e−x − B + C −xe−x + e−x − 2(A + Bx + Cxe−x ) = 1 − 2x − 9e−x
Cxe−x − 2Ce−x − B + Cxe−x − Ce−x − 2A − 2Bx − 2Cxe−x = 1 − 2x − 9e−x
−3Ce−x − B − 2A − 2Bx = 1 − 2x − 9e−x

coefficients of e−x : −3C = −9 C=3
coefficients of x: −2B = −2 B=1
constants: −B − 2A = 1; B = −2A − 1
1 = −2A − 1; A = −1
MUCs cont.
yp = A + Bx + Cxe−x
yp = −1 + x + 3xe−x
➢ yc = c1 e−x + c2 e2x
y = c1 e−x + c2 e2x − 1 + x + 3xe−x g.s. Answer
METHOD OF VARIATION OF
PARAMETERS (MVPs)
➢ applicable as long as the complementary function is known
Example
Solve D2 + 1 y = sec x
Solution
y = yc + yp
Auxiliary Equation
m2 + 1 = 0 m’s = ±i
MVPs cont.
yc = c1 cos x + c2 sin x
yp is obtained from yc by replacing c1 and c2 by A x and
B(x), respectively
yp = A cos x + B sin x A and B = f(x)
Dyp = y ′p = −A sin x + A′ cos x + B cos x + B′ sin x
Let
A′ cos x + B′ sin x = 0 eq. 1
Dyp = y ′p = −A sin x + B cos x
D2 yp = y ′′p = −A cos x − A′ sin x − B sin x + B′ cos x
MVPs cont.
D2 + 1 y = sec x rewritten
−A cos x − A′ sin x − B sin x + B′ cos x + A cos x + B sin x = sec x
−A′ sin x + B′ cos x = sec x eq. 2
Solve for A′ and B′
B′ sin x
from eq. 1, A′ = −
cos x
MVPs cont.
Substitute the above equation into eq. 2
−A′ sin x + B′ cos x = sec x eq. 2 rewritten
B′ sin x
sin x + B′ cos x = sec x
cos x
B′ sin2 x+B′ cos2 x
= sec x
cos x
=1
B’ (sin2 x+cos2 x
= sec x
cos x
B’ = sec x (cos x) = 1
MVPs cont.
B = න dx = x
′ B′ sin x
from eq. 1, A = −
cos x
(1) sin x sin x
A′ = − = −
cos x cos x
sin x dx
A = −න = ln cos x
cos x
MVPs cont.
yp = A cos x + B sin x rewritten
yp = cos x ln cos x + x sin x
y = yc + yp
➢ yc = c1 cos x + c2 sin x
∴ y = c1 cos x + c2 sin x + cos x ln cos x + x sin x g.s. Answer
NHLDEs with Constant Coefficients
SUPPLEMENTARY
EXERCISES …
MUCs cont.
1. D4 + D2 y = −12
D2 D2 + 1 y = −12
Solution
y = yc + yp
Auxiliary Equation
m2 m2 + 1 = 0 m’s = 0, 0, ±i Cases 2 and 3
MUCs cont.
Complementary function, yc
yc = c1 e0 + c2 xe0 + e0 (c3 cos 𝑥 + c4 sin 𝑥)
yc = c1 + c2 x + c3 cos x + c4 sin x
R(x) = -12; n=0
yp = Ax 2
Dyp = 2Ax
D2 yp = 2A
D3 yp = 0
D4 yp = 0
MUCs cont.
D4 + D2 y = −12 rewritten
0 + 2A = −12
−12
A=
2
−12 2
yp = x = −6x 2
2
y = y c + yp
∴ y = c1 + c2 x + c3 cos x + c4 sin x − 6x 2 g.s. Answer
MUCs cont.
2. D2 + D y = sin x
Solution
y = yc + yp
Auxiliary Equation
m2 + m = 0
m(m+1) = 0
m′ s = 0, −1
yc = c1 + c2 e−x
MUCs cont.
R(x) = sin x n′ s = ±i
yp = A cos x + B sin x
D2 yp = y ′′p = −A cos x − B sin x
D2 + D y = sin x rewritten
−A cos x − B sin x − A sin x + B cos x = sin x
MUCs cont.
−A cos x − B sin x − A sin x + B cos x = sin x
Solve for A and B
coefficients of cos x: −A + B = 0; A = B
coefficients of sin x: −B − A = 1; −A − A = 1
−2A = 1
1
A= −
2
1
B=A= −
2
MUCs cont.
1 1 1
yp = − cos x − sin x = − (cos x + sin x)
2 2 2
y = yc + yp
➢ yc = c1 + c2 e−x
−x 1
y = c1 + c2 e − cos x + sin x g.s. Answer
2
MUCs cont.
3. D2 − 3D − 4 y = 2e−x + cos x
Solution
y = yc + yp
Auxiliary Equation m2 − 3m − 4 = 0
(m – 4)(m + 1) = 0
Roots (left) = m′ s = 4, −1
yc = c1 e4x + c2 e−x
MUCs cont.
R x = 2e−x + cos x ; n′ s = −1, ±i
Note: m′ s = 4, −1 n′ s = −1, ±i
−1, appears two times
yp = Axe−x + B cos x + C sin x
Dyp = y ′p = A −xe−x + e−x − B sin x + C cos x
D2 yp = y ′′p = A xe−x − e−x − e−x − B cos x − C sin x
D2 yp = y ′′p = A xe−x − 2e−x − B cos x − C sin x
MUCs cont.
Substitute yp into the given differential equation to determine the
values of A, B and C
D2 − 3D − 4 y = 2e−x + cos x rewritten
A xe−x − 2e−x − B cos x − C sin x − 3[A −xe−x + e−x − B sin x
+ C cos x] − 4 Axe−x + B cos x + C sin x = 2e−x + cos x
Axe−x − 2Ae−x − B cos x − C sin x + 3Axe−x − 3Ae−x + 3B sin x
− 3C cos x − 4Axe−x − 4B cos x − 4C sin x = 2e−x + cos x
−5Ae−x − 5B cos x − 3C cos x − 5C sin x + 3B sin x = 2e−x + cos x
MUCs cont.
−5Ae−x − 5B cos x − 3C cos x − 5C sin x + 3B sin x = 2e−x + cos x rewritten
−x 2
coefficients of e ∶ −5A = 2 A= −
5
5C
coefficients of cos x: −5B − 3C = 1; −5 − 3C = 1
3
coefficients of sin x: −5C + 3B = 0
3B = 5C
5C
B=
3
MUCs cont.
5C
• −5 − 3C = 1
3
25C
• − − 3C = 1
3
−25C−9C
• =1
3
3
• −34C = 3 C=−
34
5C 5 3 5
• B= = − B= −
3 3 34 34
MUCs cont.
yp = Axe−x + B cos x + C sin x rewritten
2 5 3
yp = − xe−x − cos x − sin x
5 34 24
y = yc + yp
➢yc = ce4x + c2 e−x
2 5 3
∴ y = c1 e4x + c2 e−x − xe−x − cos x − sin x g.s. Answer
5 34 24
MUCs cont.
4. D2 + 4 y = 5ex − 4x
Solution
y = yc + yp
Auxiliary Equation
m2 + 4 = 0
m′ s = ±2i
yc = c1 cos 2x + c2 sin 2x
MUCs cont.
R(x) = 5ex − 4x; n’s = 1, 0, 0
yp = A + Bx + Cex
Dyp = y ′p = B + Cex
D2 yp = y ′′p = Cex
D2 + 4 y = 5ex − 4x rewritten
Cex + 4(A + Bx + Cex ) = 5ex − 4x
MUCs cont.
Cex + 4A + 4Bx + 4Cex = 5ex − 4x
5Cex + 4A + 4Bx = 5ex − 4x
Comparing coefficients of the left-hand side part with right-hand
side part of the above equation
coefficients of ex : 5C = 5 C =1
coefficients of x: 4B = −4 B = −1
constant: 4A = 0 A=0
yp = A + Bx + Cex
yp = 0 – x + ex
MUCs cont.
yp = – x + ex
y = yc + yp
➢ yc = c1 cos 2x + c2 sin 2x
∴ y = c1 cos 2x + c2 sin 2x − x + ex g.s. Answer
MVPs cont.
5. D2 + 1 y = csc x
Solution
y = yc + yp
Auxiliary Equation
m2 + 1 = 0
m’s = ±i
yc = c1 sin x + c2 cos x
yp = A sin x + B cos x A and B = f(x)
Dyp = y′p = A cos x + A′ sin x − B sin x + B′ cos x
MVPs cont.
Let
A′ sin x + B′ cos x = 0 eq. 1
Dyp = y′p = A cos x − B sin x = 0
D2 yp = y′′p = −A sin x + A′ cos x − B cos x − B′ sin x
D2 + 1 y = csc x rewritten
−A sin x + A′ cos x − B cos x − B′ sin x + A sin x + B cos x = csc x
′ ′
1
A cos x − B sin x = csc x =
sin x
MVPs cont.
B′ cos x
from eq. 1, A′ = −
sin x
1
A′ cos x − B′ sin x = csc x = rewritten
sin x
B′ cos x ′ 1
− cos x − B sin x =
sin x sin x
−B′ cos2 x − B′ sin2 x 1
=
sin x sin x
=1
B′ (cos2 x + sin2 x) 1
− =
sin x sin x
MVPs cont.
B′ 1
− =
sin x sin x
′
B = −1
B = −x
′ B′ cos x −1 cos x
A = − = −
sin x sin x
′
cos x
A =
sin x
A = ln sin x
yp = A sin x + B cos x
yp = ln sin x sin x − x cos x = sin x ln sin x − x cos x
MVPs cont.
➢ yc = c1 sin x + c2 cos x
y = c1 sin x + c2 cos x − x cos x + sin x ln sin x g.s. Answer
MVPs cont.
6. D2 + 1 y = tan x
Solution
y = yc + yp
Auxiliary Equation
m2 + 1 = 0
m’s = ±i
MVPs cont.
Let
MVPs cont.
D2 + 1 y = tan x rewritten
−A cos x − A′ sin x − B sin x + B′ cos x + A cos x + B sin x = tan x
−A′ sin x + B′ cos x = tan x eq. 2
B′ sin x
from eq. 1, A′ = −
cos x
MVPs cont.
−A′ sin x + B′ cos x = tan x eq. 2 rewritten
B′ sin x
sin x + B′ cos x = tan x
cos x
= tan x
cos x
=1
B’ (sin2 x+cos2 x
= tan x
cos x
MVPs cont.
=1
B’(sin2 x+cos2 x sin x
=
cos x cos x
sin x
B’ = cos x
cos x
B’ = sin x
B = ‫ ׬‬sin x dx
B = − cos x
MVPs cont.
B ′ sin x sin x sin x
′
A =− =−
cos x cos x
2
′
sin x
A =−
cos x
(1−cos2 x) 1
A′ = − =− − cos x = −sec x + cos x
cos x cos x
A = න − sec x + cos x dx
A = −ln sec x + tan x + sin x

MVPs cont.
➢ yp = A cos x + B sin x
= −ln sec x + tan x + sin x cos x + (−cos x) sin x
= −cos x ln sec x + tan x
∴ y = c1 cos x + c2 sin x −cos x ln sec x + tan x g.s. Answer
MVPs cont.
7. D2 + 1 y = sec x csc x
Solution
y = yc + yp
Auxiliary Equation
m2 + 1 = 0
m’s = ±i
MVPs cont.
Let
MVPs cont.
D2 + 1 y = sec x csc x rewritten
−A cos x − A′ sin x − B sin x + B′ cos x + A cos x + B sin x = sec x csc x
−A′ sin x + B′ cos x = sec x csc x eq. 2
B′ sin x
from eq. 1, A′ = −
cos x
MVPs cont.
−A′ sin x + B ′ cos x = sec x csc x eq. 2 rewritten
B′ sin x
sin x + B′ cos x = sec x csc x
cos x
= sec x csc x
cos x
=1
B’(sin2 x+cos2 x
= sec x csc x
cos x
MVPs cont.
B′ = cos x (sec x csc x)
B′ = csc x
B = ‫ ׬‬csc x dx
B = −ln csc x + cot x
B′ sin x
A′ = − B′ = csc x
cos x
csc x sin x 1
A′ = − = − = − sec 𝑥
cos x cos x
A = −ln sec x + tan x
MVPs cont.
➢ yp = A cos x + B sin x
= (−ln sec x + tan x ) cos x + (−ln csc x + cot x )sin x
y = yc + yp
y = c1 cos x + c2 sin x −cos x ln sec x + tan x − sin x ln csc x + cot x g.s. Answer
EXERCISES
• Solve the following differential equations
1. D3 + 6D2 + 12D + 8 y = 0; when x = 0, y = 1, y ′ = −2, y ′′ = 2
2. D4 − 2D3 + 5D2 − 8D + 4 y = 0
3. D5 + D4 − 9D3 − 13D2 + 8D + 12 y = 0
4. 4D3 + 28D2 + 61D + 37 y = 0
5. D5 − 2D3 − 2D2 − 3D − 2 y = 0
EXERCISES cont.
d2 x dx dx
6. + 4 + 5x = 10; when t = 0, x = 0, =0
dt2 dt dt
7. xሷ + 4xሶ + 5x = 8 sin t; when t = 0, x = 0, xሶ = 0
8. y ′′ + 2y ′ + y = x; at x = 0, y = −2, y ′ = 2
9. D2 + 2D + 2 y = e−x csc x
10. D2 + 1 y = tan2 x
REFERENCES
Rainville, E. D., Bedient, P. E., & Bedient, R. E.
(2002). Elementary differential equations (8th ed.).
Singapore: Pearson Education Asia Pte. Ltd.
Zill, D. G. & Wright, W. S. (2013). Differential equations
with boundary-value problems (8th ed.). Boston:
Brooks/Cole, Cengage Learning.

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DIFFERENTIAL

• Synthetic Division cont.

Case 1. Roots of auxiliary equation are real and distinct

2. 4D3 − 13D + 6 y = 0 -2 4 0 -13 6

4m3 − 13m + 6 = 0 3/2 4 -6 0

y = e2x c1 cos 3x + c2 sin 3x g.s. Answer

Auxiliary Equation -1 -6 -13

−6± (6)2 −4(1)(13) −6 ± 36−52 −6 ± −16 −6 ±4i

Solve for c1 , c2 , and c3

y = e−3x sin 2x p.s. Answer

For the given d.e.

• y = c1 cos x + c2 sin x + x c3 cos x + c4 sin x + e−x ൫c5 cos 3x +

−3Ce−x − B − 2A − 2Bx = 1 − 2x − 9e−x

A = −ln sec x + tan x + sin x

∴ y = c1 cos x + c2 sin x −cos x ln sec x + tan x g.s. Answer

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