linear differential equations1
TRANSCRIPT
LINEAR DIFFERENTIAL EQUATIONS
SEBASTIAN VATTAMATTAM
1. Homogeneous Equations
Example 1.1. Solve
d2y
dx2+ a
dy
dx+ by = 0, a, b ∈ R....(1)
Suppose y = emx....(2) is a solution of (1).Then,
m2emx + amemx + bemx = 0⇔ m2 + am+ b = 0....(3)
So, (2) is a solution of (1) iff m is a root of (3).Let m1,m2 be the roots of (3).
Case 1: m1,m2 are real and distinct.
y = c1em1x + c2e
m2x
Case 2: m1 = p+ iq,m2 = p− iq
y = epx(c1 cos qx+ c2 sin qx)
Case 3: m1 = m2 = m
(c1 + c2x)emx
Example 1.2. Solve
y′′ + y′ − 2y = 0
The auxiliary equation is
m2 +m− 2 = 0⇒ m1 = −1,m2 = 2
Date: 15 June 2010.Key words and phrases. Complementary Function, Particular Integral, Complete Solution.
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2 SEBASTIAN VATTAMATTAM
General Solution is
y = c1e−x + c2e
2x
Example 1.3. Solve
y′′ − 2y′ + 10y = 0
The auxiliary equation is
m2 − 2m+ 10 = 0⇒ m1 = 1 + 3i,m2 = 1− 3i
General Solution is
y = ex(c1 cos 3x+ c2 sin 3x
Example 1.4. Solve
y′′ − 2y′ + 10y = 0 given that y(0) = 4, y′(0) = 1
The auxiliary equation is
m2 − 2m+ 10 = 0⇒ m1 = 1 + 3i,m2 = 1− 3i
General Solution is
y(x) = ex(c1 cos 3x+ c2 sin 3x)
y(0) = c1 ⇒ c1 = 4....(1)
y′(x) = ex(c1 cos 3x+ c2 sin 3x− 3c1 sin 3x+ 3c2 cos 3x)
Therefore
y′(0) = c1 + 3c2 ⇒ c1 + 3c2 = 1....(2)
From (1) and (2), c1 = 4, c2 = −1 Therefore, the solutionis
y(x) = ex(4 cos 3x− sin 3x)
Rememberd
dxexf(x) = ex[f(x) + f ′(x)]
Example 1.5. Solve 1
y′′ + n2y = 0
1Similar to Ex 4, p.190 in [1]
ORDINARY DIFFERENTIAL EQUATIONS 3
The auxiliary equation is
m2 + n2 = 0⇒ m1 = ni,m2 = −niGeneral Solution is
y = c1 cosnx+ c2 sinnx
Example 1.6. Solve
y′′ + 8y′ + 16y = 0
The auxiliary equation is
m2 + 8m+ 16 = 0⇒ m1 = −4,−4
General Solution is
y(x) = (c1 + c2x)e−4x
Example 1.7. Solve
y′′ − 4y′ + 4y = 0, y(0) = 3, y′(0) = 1
General Solution is
y(x) = (c1 + c2x)e2x
y(0) = c1 ⇒ c1 = 3....(1)
y′(x) = c2e2x + 2(c1 + c2x)e2x
Therefore
y′(0) = c2 + 2c1 ⇒ c2 + 2c1 = 1....(2)
From (1) and (2), c1 = 3, c2 = −5 Therefore, the solutionis
y(x) = (3− 5x)e2x
Example 1.8. Solve 2
2y′′′ − 5y′′ − 4y′ + 3y = 0
The auxiliary equation is
2m3 − 5m2 − 4m+ 3 = 0
2Ex 1, p.88 of [1]
4 SEBASTIAN VATTAMATTAM
Solution by synthetic division:
Trym = 11 2 -5 -4 3
2 -3 -72 -3 -7 [-4]
Trym = −1-1 2 -5 -4 3
-2 7 -32 -7 3 [0]
2m3−5m2−4m+3 = (m+1)(2m2−7m+3) = (m+1)(2m−1)(m−3) = 0
⇒ m = 1/2,−1, 3
General Solution is
y = Aex/2 +Be−x + Ce3x
Example 1.9. Solve 3
y′′′ + 3y′′ + 3y′ + 2y = 0
The auxiliary equation is
m3 + 3m2 + 3m+ 2 = 0
Solution by synthetic division:
Trym = −1-1 1 3 3 2
-1 -2 -11 2 1 [1]
Trym = −2-2 1 3 3 2
-2 -2 -21 1 1 [0]
m3 + 3m2 + 3m+ 2 = 0 = (m+ 2)(m2 +m+ 1) = 0
⇒ m = −2,−1/2± i1/2General Solution is
y = Ae−2x + e−x/2(C cos
√3x
2+D sin
√3x
2)
2. Non-homogeneous Equations
Inverse Operators
(1)1
Dφ(x) =
∫φ(x)dx
3Ex 3, p.190 in [1]
ORDINARY DIFFERENTIAL EQUATIONS 5
(2)1
D − aφ(x) = eax
∫φ(x)e−axdx
(3)1
f(D)eax =
1
f(a)eax, if f(a) 6= 0
(4) If f(a) = 0, then
1
f(D)eax = x
1
f ′(a)eax, if f ′(a) 6= 0
(5) If f ′(a) = 0, then
1
f(D)eax = x2
1
f ′′(a)eax, if f ′′(a) 6= 0
(6)
1
f(D2)sin(ax+ b) =
1
f(−a2)sin(ax+ b), if f(−a2) 6= 0
(7) If f(−a2) = 0, then
1
f(D2)sin(ax+ b) = x
1
f ′(−a2)sin(ax+ b), if f ′(−a2) 6= 0
(8)1
f(D)eaxφ(x) = eax
1
f(D + a)φ(x)
Example 2.1. Solve
(D − 2)2y = 8(e2x + sin 2x+ x2)
(1) Find the Complementary Function (CF)
(m− 2)2 = 0⇒ m = 2, 2
CF = (c1 + c2x)e2x
(2) Find Particular Integrals(PI)(a)
PI1 =1
(D − 2)2e2x = x
1
2(D − 2)e2x = x2
1
2e2x =
x2
2e2x
Note that f(2) = f ′(2) = 0
6 SEBASTIAN VATTAMATTAM
(b)
PI2 =1
(D − 2)2sin 2x =
1
D2 − 4D + 4sin 2x =
1
−4− 4D + 4sin 2x = −1
4
1
Dsin 2x = −1
4
∫sin 2x =
cos 2x
8
(c)
PI3 =1
(D − 2)2x2 =
1
(2−D)2x2 =
1
4(1−D/2)2x2 =
1
4(1−D/2)−2x2 =
1
4[1+
2D
2+−2×−3
2!
D2
4...]x2 =
1
4[x2+2x+
3x2
4]
(3) Complete solution is
y = CF + PI1 + PI2 + PI3
References[1] P. Kandasamy etc,Engineering Mathematics- For First Year ,S. Chand & Company Ltd
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