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Discrete real roots: If roots m1 , m2 , · · · , mk are distinct
real roots with multiplicity 1, then the general solution has
form:
y = C1 em1 x + C2 em2 x + · · · Ck emk x +
·|· {z
· · ·}·
comes from other roots
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Repeated real roots: If the root m1 has multiplicity j,
then its contribution to the general solution:
y = (C1 em1 x + C2 xem1 x + C3 x2 em1 x + · · · + Cj xj em1 x )
+
·|· {z
· · ·}·
comes from other roots
In this slide I did a correction in the last line!
Section 3.3
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Complex conjugate roots of multiplicity 1: Recall
complex roots come in pairs. Let m1 = α + iβ and
m2 = α − iβ are two roots of multiplicity 1. Then its
contribution to the general solution:
y = eαx {C1 cos (βx) + C2 sin (βx)} +
·|· {z
· · ·}·
comes from other roots
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Complex conjugate roots of multiplicity j > 1: Let
m1 = α + iβ and m2 = α − iβ are two roots of multiplicity
j > 1. Then its contribution to the general solution:
y = eαx {C1 cos (βx) + C2 sin (βx)
+ C3 x cos (βx) + C4 x sin (βx)
+ · · · · · ·} +
·|· {z
· · ·}·
comes from other roots
Section 3.3
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Exercise: Find the general solution of 12y 00 − 5y 0 − 2y = 0.
Section 3.3
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Exercise: Find the general solution of y 000 − 4y 00 − 5y 0 = 0.
Section 3.3
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Exercise: Find the general solution of y 000 − y = 0.
Section 3.3
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Exercise: Find the general solution of
y (5) + y (4) − 2y (3) − 2y (2) + y 0 + y = 0
.
Section 3.3
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Section 3.3
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Exercise: Find the general solution of
y (7) − 3y (6) + 5y (5) − 7y (4) + 7y 000 − 5y 00 + 3y 0 − y = 0
.
Section 3.3
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Section 3.3
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Section 3.4 - Non-homogeneous Linear DE with Constant
Coefficients
Section 3.4
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Consider the non-homogeneous linear DE with constant
coefficients:
an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = g(x)
where g(x) 6= 0.
The general solution of this DE looks like y = yc + yp where yp
is a particular solution of the given DE and yc is the general
solution of the homogeneous DE
an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = 0.
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So you already know how to get yc from Section#3.3.
I
yp comes from trial and error or experience!
Section 3.4
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A table - Trial Particular Solutions:
g(x)
any constant
2x + 5
2
2x + 5x + 3
2x3 − 5X 2 + 3x − 8
sin (3x)
cos (3x)
e7x
(2x + 5)e−3x
x3 e−3x
Section 3.4
Form of yp
A
Ax + B
Ax2 + Bx + C
3
Ax + Bx2 + Cx + D
A cos (3x) + B sin (3x)
A cos (3x) + B sin (3x)
Ae7x
(Ax + B)e−3x
(Ax3 + Bx2 + Cx + D)e−3x
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g(x)
e sin (3x)
e2x cos (3x)
3x2 sin (3x)
2x
3x2 cos (3x)
xe2x sin (3x)
Form of yp
Ae cos (3x) + Be2x sin (3x)
Ae2x cos (3x) + Be2x sin (3x)
(Ax2 + Bx + C) cos (3x)
+(Dx2 + Ex + F ) sin (3x)
(Ax2 + Bx + C) cos (3x)
+(Dx2 + Ex + F ) sin (3x)
(Ax + B)e2x cos (3x)
+(Cx + D)e2x sin (3x)
2x
Remark: To avoid common type of terms in-between yc and
yp , you must multiple the usual form of yp by the smallest
positive power of x.
Section 3.4
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