Second Order Differential Equations - salfordphysics.com
[Pages:42]Differential Equations
SECOND ORDER (inhomogeneous)
Graham S McDonald A Tutorial Module for learning to solve 2nd order (inhomogeneous) differential equations
q Table of contents q Begin Tutorial
c 2004 g.s.mcdonald@salford.ac.uk
Table of contents
1. Theory 2. Exercises 3. Answers 4. Standard derivatives 5. Finding yCF 6. Tips on using solutions
Full worked solutions
Section 1: Theory
3
1 Theory
This Tutorial deals with the solution of second order linear o.d.e.'s with constant coefficients (a, b and c), i.e. of the form:
d2y dy
a dx2
+
b dx
+
cy
=
f (x)
()
The first step is to find the general solution of the homogeneous equation [i.e. as (), except that f (x) = 0]. This gives us the "complementary function" yCF .
The second step is to find a particular solution yP S of the full equation (). Assume that yP S is a more general form of f (x), having undetermined coefficients, as shown in the following table:
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Section 1: Theory
4
f (x) k (a constant)
linear in x quadratic in x k sin px or k cos px
kepx sum of the above product of the above
Form of yP S C
Cx + D Cx2 + Dx + E C cos px + D sin px
C epx sum of the above product of the above
(where p is a constant)
Note: If the suggested form of yP S already appears in the complementary function then multiply this suggested form by x.
Substitution of yP S into () yields values for the undetermined coefficients (C, D, etc). Then,
General solution of () = yCF + yP S .
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Section 2: Exercises
5
2 Exercises
Find the general solution of the following equations. Where boundary conditions are also given, derive the appropriate particular solution.
Click on Exercise links for full worked solutions (there are 13 exercises in total)
d2y
dy
Notation:
y = dx2 ,
y= dx
Exercise 1. y - 2y - 3y = 6
Exercise 2. y + 5y + 6y = 2x
Exercise 3. (a) (b) (c) (d)
y + 5y - 9y = x2
y + 5y - 9y = cos 2x y + 5y - 9y = e4x y + 5y - 9y = e-2x + 2 - x
Exercise 4. y - 2y = sin 2x q Theory q Answers q Derivatives q Finding yCF q Tips
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Section 2: Exercises
6
Exercise 5. y - y = ex
Exercise 6. y + y - 2y = e-2x Exercise 7. y - 2y + y = ex Exercise 8. y + 8y + 17y = 2e-3x ; Exercise 9. y + y - 12y = 4e2x ;
y(0) = 2 and y = 0
2 y(0) = 7 and y (0) = 0
Exercise 10. y + 3y + 2y = 10 cos 2x ; y(0) = 1 and y (0) = 0
Exercise 11. y + 4y + 5y = 2e-2x ; y(0) = 1 and y (0) = -2
Exercise
12.
d2x dx d 2 +4 d +3x
=
e-3
;
1 dx x = and = -2 at = 0
2 d
d2y dy
Exercise
13.
d 2
+4 d
+ 5y
=
6 sin
q Theory q Answers q Derivatives q Finding yCF q Tips
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Section 3: Answers
7
3 Answers
1. y = Ae-x + Be3x - 2 ,
2.
y
=
Ae-2x
+
Be-3x
+
x 3
-
5 18
,
3. General solutions are y = yCF + yP S
where yCF = Aem1x + Bem2x
m1,2
=
-
5 2
?
1 2
61
and
(a)
yP S
=
-
1 9
x2
-
10 81
x
-
68 729
(b)
yP S
=
-
13 269
cos 2x
+
10 269
sin 2x
(c)
yP S
=
1 27
e4x
(d)
yP S
=
-
1 15
e-2x
+
1 9
x
-
13 81
,
4.
y = Ae+x + Be-x -
sin 2x 4+2
,
5.
y = Aex + Be-x +
1 2
xex
,
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Section 3: Answers
8
6.
y = Aex + Be-2x -
1 3
xe-2x
,
7.
y = (A + Bx)ex +
1 2
x2ex
,
8. y = e-4x(A cos x + B sin x) + e-3x ;
y
=
e-4x(cos x
-
e 2
sin x)
+
e-3x
,
9.
y
=
Ae3x
+
Be-4x
-
2 3
e2x
;
y
=
32 7
e3x
+
65 21
e-4x
-
2 3
e2x
,
10.
y = Ae-2x + Be-x -
1 2
cos
2x
+
3 2
sin
2x
;
y
=
3 2
e-2x
-
1 2
cos 2x
+
3 2
sin 2x
,
11. y = e-2x(A cos x + B sin x) + 2e-2x ; y = e-2x(2 - cos x) ,
12.
x = Ae-3
+ Be-
-
1 2
e-3
;
x
=
1 2
(1
-
)e-3
,
13.
y = e-2 (A cos
+
B
sin
)
-
3 4
(cos
-
sin
).
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