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Differential Equations with Boundary
Value Problems
Authors: Dennis G. Zill, Michael R. Cullen
Exercise 1.1
In Problems 1–8 state the order of the given ordinary differential equation. Determine whether
the equation is linear or nonlinear.
1.
1
4
5
cos
⋯
A differential equation is linear if it is in the form
.
If we compare given differential equation with the standard form a linear differential
equation, we see that it is linear.
0
2.
It is non-linear because of 4th power of
6
3.
It is linear. Note that
0
denotes fourth derivative and not
raised to power 4.
cos
4.
It is non-linear due to the term cos
5.
.
.
1
It is non-linear because of two things. (1) Presence of dependent variable under radical sign.
(2) Square term of derivative
6.
It is non-linear. It was going to be linear if
was
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7.
cos
sin
2
It is linear.
1
8.
0
It is non-linear due to square term of a derivative which is . Note that in this question
denotes second derivative while
denotes square of first derivative. For better
understanding, we can write given differential equation as ′′
1
0
In Problems 9 and 10 determine whether the given first-order differential equation is linear in the
indicated dependent variable.
1
9.
0; in ; in
Solution:
1
1
Clearly it in not linear in .
Write
as
or
0 or or
We see that it is linear in .
0; in ; in
10.
Solution:
1
1
1
0
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1
Clearly it is linear in .
Write
as
or
.
We see that it is clearly not linear in .
In Problems 11 –14 verify that the indicated function is an explicit solution of the given
differential equation. Assume an appropriate interval I of definition for each solution.
11. 2
/
0;
L. H. S
2
/
/
Since L.H.S. = R.H.S., therefore
12.
1
2
2
/
/
/
is a solution of 2
/
/
0
R. H. S
0
dy
6 6
 20 y  24 ; y   e 20t
dt
5 5
d  6 6 20t 
 6 6 20t 
  e   20   e 
dt  5 5

5 5

6
  (20e 20t )  24  24e 20t
5
 24e20t  24  24e 20t
 24  R.H.S.
L.H.S. 
Since L.H.S. = R.H.S., therefore y 
6 6 20t
dy
is a solution of
 e
 20 y  24 .
dt
5 5
13. y //  6 y /  13 y  0 ; y  e3 x cos 2 x
L.H.S.  y //  6 y /  13 y
d 2 3x
d
e cos 2 x   6  e3 x cos 2 x   13e3 x cos 2 x
2 
dx
dx
d
  3e3 x cos 2 x  2e3 x sin 2 x   6  3e3 x cos 2 x  2e3 x sin 2 x   13e3 x cos 2 x
dx
 9e3 x cos 2 x  6e3 x sin 2 x  6e3 x sin 2 x  4e3 x cos 2 x  18e3 x cos 2 x  12e3 x sin 2 x  13e3 x cos 2 x

 0  R.H.S.
Since L.H.S. = R.H.S., therefore y  e3 x cos 2 x is a solution of y //  6 y /  13 y  0 .
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In Problems 33– 36 use the concept that
, ∞
∞ is a constant function if and only
if ′ 0 to determine whether the given differential equation possesses constant solutions.
33. 3
5
10
Substitute
for , we get
0 in given differential equation, we get 3 0
5
10 or 5
10. Solving
2. Therefore, given differential equation has one constant solution
2.
34.
2
3
Substitute
0 in given differential equation, we get 0
2
3. Now, if we solve
given equation for , then we get two real solutions
3 and
1 which are the constant
solutions of given differential equation.
1
1
35.
0, we get 0
If we substitute
has no constant solution.
4
35.
6
0 and
Substitute
6
1, which is not true. Therefore, given differential equation
10
0 since derivatives of constant equals zero. Doing so, we get
10. Solving for , we get the constant solution
.
Exercise 1.2
In problems 1 and 2,
1/ 1
is a one parameter family of solutions of the first-order
. Find a solution of the first-order IVP consisting of this differential equation and
DE
the given initial condition.
0
1.
when
0. Substitute in
3⟹
4 . Therefore, solution of given IVP is
i.e.
1
3
1
3
, we get
1
1
1
1
1
2.
1/ 1
1
2
1/ 1
4
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2 at
1. Substitute in
1/ 1
Therefore, solution of given IVP is
1/ 1
i.e.
2
2
2
2
, we get
1/ 1
1
1
1
1
2
1
2
1
1/2
Exercise 2.1
In Problems 1-22 solve the given differential equation by separation of variables.
1.
dy
 sin 5 x
dx
dy  sin 5 xdx
 dy   sin 5 xdx
y
2.
cos 5 x
C
5
dy
2
  x  1
dx
dy   x  1 dx
2
 dy    x  1 dx
2
y
 x  1
3
3
C
3. dx  e3 x dy  0
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e3 x dy  dx
1
dy   3 x dx
e
dy  e3 x dx
 dy   e dx
3x
e3 x
y
C
3
4. dy  ( y  1) 2 dx  0
dy  ( y  1) 2 dx
1
dy  dx
( y  1) 2
1
 ( y  1)2 dy   dx
1

 xC
y 1
1

 y 1
xC
1
1
y
xC
1
y 1
xC
5. x
dy
 4y
dx
1
4
dy  dx
y
x
1
4
 y dy   x dx
ln y  4 ln x  C1
y  e4ln xC1
y  e4ln x eC1
y  eln x eC1
4
y  Cx 4
 eC1  C
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6.
dy
 2 xy 2  0
dx
dy
 2 xy 2
dx
1
 2 dy  2 xdx
y
1
  2 dy   2 xdx
y
1
 x2  C
y
1
y 2
x C
7.
dy
 e3 x  2 y
dx
dy
 e3 x e 2 y
dx
1
dy  e3 x dx
2y
e
e 2 y dy  e3 x dx
e
2 y
dy   e3 x dx
e 2 y e3 x


 C1
2
3
 e3 x

 C1 
e 2 y  2 
 3

2
e 2 y   e3 x  2C1
3
2
 2C1  C
e 2 y   e3 x  C
3
 2

2 y  ln   e3 x  C 
 3

1  2

y   ln   e3 x  C 
2  3

8. e x y
dy
 e  y  e 2 x y
dx
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dy
 e y  e 2 x e  y
dx
dy
 e y 1  e 2 x 
ex y
dx
1  e 2 x
y

dy
dx
e y
ex
ye y dy   e x  e3 x  dx
ex y
 ye dy    e  e  dx
x
y
3 x
1
ye y  e y  e  x  e 3 x  C
3
dx  y  1 

9. y ln x

dy  x 
y ln x
2
dx ( y  1) 2

dy
x2
x 2 ln xdx 
( y  1) 2
dy
y
x 2 ln xdx 
y2  2 y  1
dy
y

1
x 2 ln xdx   y  2   dy
y


1
 x ln xdx    y  2  y  dy
2
y2
1 3
1
 2 y  ln y  C
x ln x  x3 
3
9
2
10.
dy  2 y  3 


dx  4 x  5 
2
dy (2 y  3) 2

dx (4 x  5) 2
1
1
dy 
dx
2
(2 y  3)
(4 x  5) 2
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1
1
 (2 y  3) dy   (4 x  5) dx
2

2
1
1

 C1
2(2 y  3)
4(4 x  5)


1
1
 2  
 C1 
2y  3
 4(4 x  5)

1
1

 2C1
2 y  3 2(4 x  5)
1
1

C
2 y  3 2(4 x  5)
1
 2y  3
1
2(4 x 5)  C
2y 
1
1
2(4 x 5)  C
1
1
y  1

2  2(4 x5)  C
 2C1  C
3

 3


11. csc y dx  sec2 x dy  0
sec2 x dy   csc y dx
1
1
dy  
dx
2
cos x
sin y
 sin y dy  cos 2 xdx
1 1

 sin y dy    cos 2 x  dx
2 2

1 1

  sin y dy     cos 2 x  dx
2 2

1
1
cos y  x  sin 4 x  C
2
4
1
1

y  cos 1  x  sin 4 x  C 
4
2

12.