Download DIRECT INTEGRATION

Differential Equations
DIRECT INTEGRATION
Graham S McDonald
A Tutorial Module introducing ordinary
differential equations and the method of
direct integration
● Table of contents
● Begin Tutorial
c 2004 [email protected]
Table of contents
1.
2.
3.
4.
5.
6.
Introduction
Theory
Exercises
Answers
Standard integrals
Tips on using solutions
Full worked solutions
Section 1: Introduction
3
1. Introduction
d2 y
+
dx2
dy
dx
3
= x7
is an example of an ordinary differential equa-
tion (o.d.e.) since it contains only ordinary derivatives such as
∂y
and not partial derivatives such as ∂x
.
dy
dx
The dependent variable is y while the independent variable is x (an
o.d.e. has only one independent variable while a partial differential
equation has more than one independent variable).
The above example is a second order equation since the highest ord2 y
der of derivative involved is two (note the presence of the dx
2 term).
Toc
JJ
II
J
I
Back
Section 1: Introduction
4
An o.d.e. is linear when each term has y and its derivatives only
appearing to the power one. The appearance of a term involving the
dy
would also make an o.d.e. nonlinear.
product of y and dx
In the above example, the term
ear.
dy
dx
3
makes the equation nonlin-
The general solution of an nth order o.d.e. has n arbitrary constants that can take any values.
In an initial value problem, one solves an nth order o.d.e. to find
the general solution and then applies n boundary conditions (“initial values/conditions”) to find a particular solution that does not
have any arbitrary constants.
Toc
JJ
II
J
I
Back
Section 2: Theory
5
2. Theory
An ordinary differential equation of the following form:
dy
= f (x)
dx
can be solved by integrating both sides with respect to x:
Z
y=
f (x) dx .
This technique, called DIRECT INTEGRATION, can also be applied when the left hand side is a higher order derivative.
In this case, one integrates the equation a sufficient number of times
until y is found.
Toc
JJ
II
J
I
Back
Section 3: Exercises
3. Exercises
Click on Exercise links for full worked solutions (there are 8
exercises in total)
Exercise 1.
Show that y = 2e2x is a particular solution of the ordinary
d2 y
dy
differential equation:
− 2y = 0
−
dx2
dx
Exercise 2.
Show that y = 7 cos 3x − 2 sin 2x is a particular solution of
d2 y
+ 2y = −49 cos 3x + 4 sin 2x
dx2
● Theory ● Answers ● Integrals ● Tips
Toc
JJ
II
J
I
Back
6
Section 3: Exercises
7
Exercise 3.
Show that y = A sin x + B cos x, where A and B are arbitrary
d2 y
constants, is the general solution of
+y =0
dx2
Exercise 4.
Derive the general solution of
Exercise 5.
Derive the general solution of
dy
= 2x + 3
dx
d2 y
= − sin x
dx2
Exercise 6.
Derive the general solution of
d2 y
= a, where a = constant
dt2
● Theory ● Answers ● Integrals ● Tips
Toc
JJ
II
J
I
Back
Section 3: Exercises
8
Exercise 7.
Derive the general solution of
d3 y
= 3x2
dx3
Exercise 8.
Derive the general solution of e−x
d2 y
=3
dx2
● Theory ● Answers ● Integrals ● Tips
Toc
JJ
II
J
I
Back
Section 4: Answers
9
4. Answers
2
d y
dy
1. HINT: Work out dx
2 and dx and then substitute your results,
along with the given form of y, into the differential equation ,
2
d y
2. HINT: Show that dx
2 = −63 cos 3x + 8 sin 2x and substitute
this, along with the given form of y, into the differential
equation ,
3. HINT: Show that
d2 y
dx2
= −A sin x − B cos x ,
4. y = x2 + 3x + C ,
5. y = sin x + Ax + B ,
6. y = 12 at2 + Ct + D ,
7. y =
1 5
20 x
+ C 0 x2 + Dx + E ,
8. y = 3ex + Cx + D .
Toc
JJ
II
J
I
Back
Section 5: Standard integrals
10
5. Standard integrals
f (x)
n
x
1
x
x
e
sin x
cos x
tan x
cosec x
sec x
sec2 x
cot x
sin2 x
cos2 x
R
f (x)dx
xn+1
n+1
(n 6= −1)
ln |x|
ex
− cos x
sin x
− ln
|cos x|
ln tan x2 ln |sec x + tan x|
tan x
ln |sin x|
x
sin 2x
2 −
4
x
sin 2x
2 +
4
Toc
JJ
R
f (x)
n
0
[g (x)] g (x)
g 0 (x)
g(x)
x
a
sinh x
cosh x
tanh x
cosech x
sech x
sech2 x
coth x
sinh2 x
cosh2 x
II
J
f (x)dx
[g(x)]n+1
n+1
(n 6= −1)
ln |g (x)|
ax
(a > 0)
ln a
cosh x
sinh x
ln cosh x ln tanh x2 2 tan−1 ex
tanh x
ln |sinh x|
sinh 2x
− x2
4
sinh 2x
+ x2
4
I
Back
Section 5: Standard integrals
f (x)
1
a2 +x2
√ 1
a2 −x2
√
a2 − x2
11
R
f (x) dx
f (x)
1
a
tan−1
1
a2 −x2
R
(a > 0)
1
x2 −a2
f (x) dx
a+x 1
2a ln a−x (0 < |x| < a)
x−a 1
ln
2a
x+a (|x| > a > 0)
sin−1
x
a
√ 1
a2 +x2
√
2
2
ln x+ aa +x (a > 0)
(−a < x < a)
√ 1
x2 −a2
√
2
2
ln x+ xa −a (x > a > 0)
a2
2
−1
sin
√
+x
Toc
x
a
x
a
a2 −x2
a2
JJ
√
i √
a2 +x2
a2
2
x2 −a2
II
a2
2
J
h
h
sinh−1
− cosh−1
I
x
a
i
√
x a2 +x2
2
a
i
√
2
2
+ x xa2−a
+
x
a
Back
Section 6: Tips on using solutions
12
6. Tips on using solutions
● When looking at the THEORY, ANSWERS, INTEGRALS or
TIPS pages, use the Back button (at the bottom of the page) to
return to the exercises.
● Use the solutions intelligently. For example, they can help you get
started on an exercise, or they can allow you to check whether your
intermediate results are correct.
● Try to make less use of the full solutions as you work your way
through the Tutorial.
Toc
JJ
II
J
I
Back
Solutions to exercises
13
Full worked solutions
Exercise 1.
dy
dx
=
2 · 2e2x = 4e2x
d2 y
dx2
=
2 · 4e2x = 8e2x .
− 2y
=
8e2x − 4e2x − 2 · e2x
=
(8 − 8)e2x
=
0
=
RHS
We need:
∴
d2 y
dx2
−
dy
dx
Return to Exercise 1
Toc
JJ
II
J
I
Back
Solutions to exercises
14
Exercise 2.
To show that y = 7 cos 3x − 2 sin 2x is a particular solution of
d2 y
dx2 + 2y = −49 cos 3x + 4 sin 2x, work out the following:
dy
dx
d2 y
dx2
∴
d2 y
dx2
+ 2y
= −21 sin 3x − 4 cos 2x
= −63 cos 3x + 8 sin 2x
= −63 cos 3x + 8 sin 2x + 2(7 cos 3x − 2 sin 2x)
= (−63 + 14) cos 3x + (8 − 4) sin 2x
= −49 cos 3x + 4 sin 2x
= RHS
Notes • The equation is second order, so the general solution
would have two arbitrary (undetermined) constants.
• Notice how similar the particular solution is to the
Right-Hand-Side of the equation. It involves the
same functions but they have different coefficients i.e.
Toc
JJ
II
J
I
Back
Solutions to exercises
15
y is of the form
“a cos 3x + b sin 2x”
a=7
b = −2
Return to Exercise 2
Toc
JJ
II
J
I
Back
Solutions to exercises
16
Exercise 3.
dy
dx
We need:
=
A cos x + B · (− sin x)
d y
dx2
=
−A sin x − B cos x
+y
=
(−A sin x − B cos x) + (A sin x + B cos x)
=
0
=
RHS
2
∴
2
d y
dx2
Since the differential equation is second order and the solution has
two arbitrary constants, this solution is the general solution.
Return to Exercise 3
Toc
JJ
II
J
I
Back
Solutions to exercises
17
Exercise 4.
dy
This is an equation of the form
= f (x), and it can be solved by
dx
direct integration.
Integrate both sides with respect to x:
Z
Z
dy
dx =
(2x + 3)dx
dx
Z
Z
i.e.
dy =
(2x + 3)dx
i.e.
y
i.e.
y
1
2 · x2 + 3x + C
2
2
= x + 3x + C,
=
where C is the (combined) arbitrary constant that results from
integrating both sides of the equation. The general solution must
have one arbitrary constant since the differential equation is first
order.
Return to Exercise 4
Toc
JJ
II
J
I
Back
Solutions to exercises
18
Exercise 5.
This is of the form
integration.
d2 y
dx2
= f (x) , so we can solve for y by direct
Integrate both sides with respect to x:
dy
dx
Z
= −
sin xdx
= −(− cos x) + A
Integrate again:
y
=
sin x + Ax + B
where A, B are the two arbitrary constants of the general solution
(the equation is second order).
Return to Exercise 5
Toc
JJ
II
J
I
Back
Solutions to exercises
19
Exercise 6.
Integrate both sides with respect to t:
Z
dy
dt
dy
i.e.
dt
=
a dt
= at + C
Integrate again:
Z
y
=
(at + C)dt
i.e. y
=
1 2
at + Ct + D ,
2
where C and D are the two arbitrary constants required for the
general solution of the second order differential equation.
Return to Exercise 6
Toc
JJ
II
J
I
Back
Solutions to exercises
20
Exercise 7.
Integrate both sides with respect to x:
d2 y
dx2
d2 y
i.e.
dx2
d2 y
i.e.
dx2
Z
=
3x2 dx
=
1
3 · x3 + C
3
= x3 + C
Integrate again:
i.e.
Toc
JJ
dy
dx
dy
dx
II
Z
=
(x3 + C)dx
=
x4
+ Cx + D
4
J
I
Back
Solutions to exercises
21
x4
+ Cx + D dx
4
1
1 1 5
· x + C · x2 + Dx + E
4 5
2
1 5
0 2
x + C x + Dx + E
20
Z Integrate again:
y
=
i.e.
y
=
i.e.
y
=
where C 0 = C2 , D and E are the required three arbitrary constants
of the general solution of the third order differential equation.
Return to Exercise 7
Toc
JJ
II
J
I
Back
Solutions to exercises
22
Exercise 8.
Multiplying both sides of the equation by ex gives:
d2 y
= ex · 3
dx2
d2 y
= 3ex
i.e.
dx2
d2 y
x
This is now of the form dx
2 = f (x), where f (x) = 3e , and the
solution y can be found by direct integration.
ex · e−x
Integrating both sides with respect to x:
Z
dy
= 3ex dx
dx
dy
i.e.
= 3ex + C .
dx
Z
y = (3ex + C)dx
Integrate again:
Toc
JJ
II
J
I
Back
Solutions to exercises
23
i.e. y = 3ex + Cx + D ,
where C and D are the two arbitrary constants of the general
solution of the original second order differential equation.
Return to Exercise 8
Toc
JJ
II
J
I
Back