EXERCISES FOR CHAPTER 7: Differential Equations
dx
dv
y
dy
=
= F( ) becomes the separable equation
F(v) v x
x
dx
after the substitution y = vx .
Solution
y
dy
dy dv
dv
= F( ) becomes
y = vx =
x + v . Hence
x + v = F(v) , i.e. the separable
x
dx
dx dx
dx
equation
dv
x = F(v) v
dx
dx
dv
=
F(v) v x
1. Prove that the equation
2. Find the solution to (x + y)
y = 1 when x = 1 .
Solution
dy dv
y = vx =
x + v . Hence
dx dx
dy
= y where x and y are restricted to be positive and
dx
dv
x + v) = vx
dx
dv
x(1 + v)( x + v) = v
dx
dv
(1 + v)( x + v) = v
dx
dv
(1 + v) x = v 2
dx
dx
(1 + v)
dv = 2
x
v
(x + vx)(
dx
(1 + v)
dv = 2
x
v
1
+ ln v = ln x + C
v
y
x
+ ln = ln x + C 1 + 0 = 0 + C C = 1
x
y
x
y
x
Hence + ln = ln x 1 ln y = 1 .
y
x
y
3. Find the solution to x
dy
= x + y given that y = ln 2 when x = 1 .
dx
92
K. A. Tsokos: Series and D.E.
Solution
y = vx dy dv
=
x + v . Hence
dx dx
dv
x + v) = x + xv
dx
dv
x + v = 1+ v
dx
dv
=1
x
dx
dx
dv =
x
v = ln x + ln C
x(
y
= ln(C x )
x
y = x ln(C x )
Then ln 2 = ln(C) C = 2 . The solution is thus y = x ln(2 x ) .
dy
dy
4. Find the solution to (a) y = x + 2y and (b) x
= x + 2y .
dx
dx
Solution
dy dv
y = vx =
x + v . Hence,
dx dx
dv
vx( x + v) = x + 2vx
dx
dv
v( x + v) = 1 + 2v
dx
dv
vx v 2 = 1 + 2v
dx
dv
= 1 + 2v + v 2
vx
dx
v
dx
dv =
2
(1 + v)
x
To do the v integral, write
v
(1 + v)
2
1+ v 1
dv =
(1 + v)
=
(1 + v)
2
1+ v
2
dv = ln(1 + v) +
so that
dv
1
1+ v
1
dv
(1 + v)2
93
K. A. Tsokos: Series and D.E.
1
= ln x + ln C
1+ v
y
1
ln (1 + ) = ln C x
y
x
1+
x
y+ x
x
ln (
)
= ln C x
x
x+y
ln (1 + v) y+ x
x
)+
= ln C x .
x
x+y
dy dv
(b) y = vx =
x + v . Hence,
dx dx
or ln (
dv
x + v) = x + 2vx
dx
dv
x + v = 1 + 2v
dx
dv
= 1+ v
x
dx
dx
1
dv =
x
1+ v
ln (1 + v) = ln C x
x(
ln 1 +
y
= ln C x
x
y = x ± Cx 2
5. Find the solution to (a) (x y)
dy
dy
= x + y and (b) x 2
= x 2 + xy + y 2 .
dx
dx
Solution
(a) y = vx dy dv
=
x + v . Hence,
dx dx
94
K. A. Tsokos: Series and D.E.
dv
x + v) = x + vx
dx
dv
(1 v)( x + v) = 1 + v
dx
dv
= 1 + v v(1 v)
(1 v)x
dx
dv
= 1 + v2
(1 v)x
dx
dx
(1 v)
dv =
2
x
1+ v
1
v
dx
(
)dv =
2
2
1+ v 1+ v
x
1
arctan v ln (1 + v 2 ) = ln x + ln C
2
y 1
x 2 + y2
arctan ln (
) = ln C x
x 2
x2
(x vx)(
This can be simplified to
y
arctan x
y
arctan x
y
arctan x
(b) y = vx 1
1
ln(x 2 + y 2 ) + ln(x 2 ) = ln C x
2
2
1
ln(x 2 + y 2 ) + ln x = ln C x
2
1
ln(x 2 + y 2 ) = ln C
2
dy dv
=
x + v . Hence,
dx dx
x2 (
dv
x + v) = x 2 + vx 2 + v 2 x 2
dx
dv
x + v) = 1 + v + v 2
dx
dv
x = 1 + v2
dx
dx
1
dv =
2
x
1+ v
arctan v = ln(C x )
(
v = tan(ln(C x ))
y = x tan(ln(C x ))
6. Find the solution to xy
dy
2
2
=x y .
dx
Solution
y = vx dy dv
=
x + v . Hence,
dx dx
95
K. A. Tsokos: Series and D.E.
dv
x + v) = x 2 (vx)2
dx
dv
v( x + v) = 1 v 2
dx
dv
vx
= 1 2v 2
dx
dx
v
dv =
2
x
1 2v
1
ln(1 2v 2 ) = ln x + ln C
4
y2
1
ln(1 2 2 ) = ln(C x )
x
4
2
y
y2
1
1 Thus, ln(1 2 2 ) = 4 ln(C x ) = ln 4 4 , implying 1 2 2 = 4 4 and finally
C x x
x
C x
vx 2 (
y=
x2 k
where k is a constant.
2 x2
7. Find the solution to x(y x)
dy
= y(x + y) where x and y are positive.
dx
Solution
y = vx dy dv
=
x + v . Hence,
dx dx
y
dv
y
y
( 1)( x + v) = (1 + )
x
dx
x
x
dv
(v 1)( x + v) = v(1 + v)
dx
v(1 + v)
dv
x=
v
v 1
dx
2v
dv
x=
v 1
dx
dx
(v 1)dv
x=
x
2v
v 1
ln v = ln Cx
2 2
v ln v = ln C 2 x 2
y
y
= ln C 2 x 2
x
x
y
= ln(C 2 xy)
x
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K. A. Tsokos: Series and D.E.
8. Solve the equations (a) 2x 2
dy
dy
= x 2 + y 2 and (b) x 2
= 3x 2 + xy .
dx
dx
Solution
(a) y = vx dy dv
=
x + v . Hence,
dx dx
dv
x + v) = x 2 + x 2 v 2
dx
dv
2x 2 ( x + v) = x 2 (1 + v 2 )
dx
dv
2( x + v) = (1 + v 2 )
dx
dv
= 1 + v 2 + 2v
2x
dx
dx
2dv
=
2
x
(v + 1)
2
= ln x ln C
v +1
2
= ln C x
v +1
2x
= ln C x
y+ x
2x 2 (
The solution is thus y =
(b) y = vx 2x
x.
ln C x
dy dv
=
x + v . Hence,
dx dx
x2 (
dv
x + v) = 3x 2 + x 2 v
dx
dv
x=3
dx
dx
dv = 3
x
v = 3ln(C x )
y = 3x ln(C x )
dy 2x + y 3
=
. Change variables to
dx x 3y 5
x = X + a and y = Y + b . (a) Choose a and b so that the equation becomes a
homogeneous equation in the variables X and Y . (b) Hence solve the equation.
9. Consider the non-homogeneous equation
Solution
dY 2X + Y + 2a + b 3
=
dX X 3Y + a 3b 5
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K. A. Tsokos: Series and D.E.
2a + b = 3
a 3b = 5
Hence a = 2, b = 1 .
dv
dY 2X + Y
dY
. Write Y = vX so that
. Then,
=
= v+ X
dX
dX X 3Y
dX
dv 2 + v
dv 2 + v v(1 3v) 2 + 3v 2
=
X
=
=
dX 1 3v
dX 1 3v
1 3v
1 3v
dX
1 3v
dv = 2
X
3v + 2
dX
1 6v 1
2
dv = 2
X
3v + 2 2 3v + 2
v+ X
v 3 1
1
arctan
ln(3v 2 + 2) = ln X + C
2 2
2
Y 3 1 3Y 2
1
arctan
ln( 2 + 2) = ln X + C
X
X 2 2
2
Finally,
(y + 1) 3 1 3(y + 1)2
1
arctan
ln(
+ 2) = ln(x 2) + C
(x 2)2
(x 2) 2 2
2
is the required
solution.
dy
dy
d
+ y = cos x , realize that x + y = (xy) and
dx
dx
dx
hence provide the solution of the equation given that y = 0 when x = .
Solution
dy
dy
d
(xy) = x + y and so x + y = cos x becomes
dx
dx
dx
d
(xy) = cos x
dx
10. Given the differential equation x
xy = cos x dx
xy = sin x + C
sin x C
+
y=
x
x
C
sin x
Setting x = we must get y = 0 i.e. 0 = 0 + C = 0 . Hence y =
.
x
11. Given
the
differential
equation
x dy
+ ln y = sin x ,
y dx
realize
that
d
x dy
+ ln y =
(x ln y) and hence provide the solution of the equation given that
dx
y dx
when x = / 2 , y = 1 .
98
K. A. Tsokos: Series and D.E.
Solution
x dy
d
x dy
(x ln y) =
+ ln y and so
+ ln y = e x becomes
y dx
y dx
dx
d
(x ln y) = sin x
dx
x ln y = sin x dx = cos x + C
cos x C
+
x
x
cos x
cos / 2
C
C
Then, ln1 = +
0=0+
C = 0 . Thus y = e x .
/2
/2
/2
y dy
12. Using the method used above solve the equation xe
+ e y = 2x 2 , given that when
dx
x = 1 , y = ln 2 .
Solution
d
dy
y dy
+ e y = 2x 2 becomes
(xe y ) = xe y
+ e y and so xe
dx
dx
dx
d
(xe y ) = 2x 2
dx
2x 3
y
2
xe = 2x dx =
+C
3
2x 2 C
ey =
+
x
3
2
2x
2
4
4
Then, 2 = + C C = . Thus y = ln + .
3
3
3x 3
ln y = dy
+ xy = x given that when x = 0 , y = 2 . (b) Find the
dx
dy
x
general solution to x y =
.
dx
x 1
13. (a) Solve the equation
Solution
2
xdx
(a) The integrating factor is e = e x /2 . Thus
2
2
d
(ye x /2 ) = xe x /2
dx
ye x
2
/2
ye x
2
/2
= xe x
= ex
2
/2
y = 1 + Ce x
2
2
/2
dx
+C
/2
The initial condition requires that 2 = 1 + C C = 1 . Hence y = 1 + e x
2
/2
.
99
K. A. Tsokos: Series and D.E.
dx
1
dy
x
1
dy y
hence
. The integrating factor is e x = e ln x = .
(b) x y =
=
x
dx
x 1
dx x x 1
1
1
1
d y
1
( )=
=
and so:
Hence
. Using partial fractions
x(x 1)
dx x
x(x 1) x 1 x
1
y
1
= dx
x 1 x
x
y
x 1
C(x 1)
= ln
+ ln C = ln
x
x
x
y = x ln
14. Solve the equation
C(x 1)
x
2
dy
xy = e x /2 given that when x = 0 , y = 2 .
dx
Solution
2
xdx
The integrating factor is e = e x /2 . Thus
2
d
(ye x /2 ) = 1
dx
ye x
2
/2
ye x
2
/2
= dx
= x+C
y = (x + C)e x
2
/2
The initial condition requires that 2 = (0 + C) C = 2 . Hence y = (x + 2)e x
15. (a) Solve the equation
general solution to
2
/2
.
dy
+ ysin x = ecos x given that when x = 0 , y = 0 . (b) Find the
dx
dy
+ 2y = e2 x cos x .
dx
Solution
sin xdx
= e cos x . Thus
(a) The integrating factor is e d
(ye cos x ) = 1
dx
ye cos x = dx
ye cos x = x + C
y = (x + C)ecos x
The initial condition requires that 0 = (0 + C)e C = 0 . Hence y = xecos x .
100
K. A. Tsokos: Series and D.E.
2dx
(b) The integrating factor is e = e2 x . Hence,
d
(ye2 x ) = cos x
dx
ye2 x = sin x + C
y = e2 x (sin x + C)
2
dy y
+ = x given that when x = 1 , y = .
3
dx x
1
dx
The integrating factor is e x = eln x = x . Thus
16. Solve the equation
d
(yx) = x 2
dx
yx = x 2 dx
x3
yx =
+C
3
x2 C
y=
+
3 x
The initial condition requires that
17. Solve the equation
1
1
1
2 1 C
= + C = . Hence y = (x 2 + ) .
3
3
x
3 3 1
xy
dy
+
= 1.
dx 1 x 2
Solution
x
The integrating factor is e
1 x 2 dx
=e
1
ln(1 x 2 )
2
=
1
1 x2
. Thus
y
1
d
(
)=
2
dx 1 x
1 x2
1
y
=
dx
2
1 x2
1 x
= arcsin x + C
y = 1 x 2 (arcsin x + C )
18. Solve the equation
xy
dy
= 1 given that when x = 0 , y = 1 .
dx 1 x 2
Solution
The integrating factor is e
x
1 x 2 dx
1
= e2
ln(1 x 2 )
= 1 x 2 . Thus
101
K. A. Tsokos: Series and D.E.
d
(y 1 x 2 ) = 1 x 2
dx
y 1 x2 =
1 x 2 dx
To do the integral use x = sin u dx = cosu du so that
1 sin u cosu du
= cos u du
1 x 2 dx =
2
2
cos 2u + 1
du
2
sin 2u u
=
+
4
2
sin u cosu u
=
+
2
2
=
x 1 x 2 arcsin x
=
+
+C
2
2
Hence
x 1 x 2 arcsin x
+
+C
2
2
C
x arcsin x
+
y= +
2 2 1 x2
1 x2
y 1 x2 =
The initial condition requires that 1 = C . Hence y =
x arcsin x
1
.
+
+
2 2 1 x2
1 x2
19. Verify the claims made at the end of section 7.4.
Solution
dy
dy
= 1 y = dx ln 1 y = x + c 1 y = e x c . As x increases, e x c 0
1 y
dx
and so y 1 .
dy
20. Construct the slope field diagram for the equation
= y for x, y [1,1] and
dx
confirm that it gives the curves y = ce x as solutions.
Solution
The slope field is shown below. The solutions are:
dy
= dx ln y = x + k y = e x + k = ce x as required. The solutions with c > 0 are in
y the upper half-plane, those with c < 0 in the lower half-plane and that with c = 0 is
along the x axis.
102
K. A. Tsokos: Series and D.E.
1
0.5
-1
-0.5
0.5
1
-0.5
-1
x
dy
= for x, y [1,1] and
y
dx
2
2
confirm that it gives the circles y + x = c as solutions.
21. Construct the slope field diagram for the equation
Solution
x
dy
= is separable and becomes
y
dx
2
2
y
x
2
2
ydy = xdx 2 = 2 + c y + x = const , as required. The slope field diagram
is shown below indicating that the solutions in the range x, y [1,1] are indeed parts of
a circle centered at the origin.
The differential equation
103
K. A. Tsokos: Series and D.E.
1
0.5
-1
-0.5
0.5
1
-0.5
-1
22. Describe qualitatively the solutions to the differential equation whose slope field is
given below.
10
8
6
4
2
2
4
6
8
10
Solution
The solutions appear to grow from the initial value of y to a limiting value of 10. If the
initial value is 0 the value of y stays 0. If it is initially 10 it stays at 10.
104
K. A. Tsokos: Series and D.E.
23. Shown below are 4 slope field diagrams. Match each one with one of the equations
dy
dy
dy
dy
in the list: (A)
= 1 , (B)
= x 2 , (C)
= x y and (D)
= xy .
dx
dx
dx
dx
4
1
3
0.5
2
-1
-0.5
1
1 -2
-1
1
2
2
-1
1
0.5
1
-1
1
-0.5
2
-2
0.5
1
-1
2
-0.5
0.5
1
-0.5
-1
-1
3
-2
4
Solution
Slope field number 4 has a constant negative slope and so A 4 .
Slope field number 2 has always a positive slope and so B 2 .
Equation (C) has zero slope when y = x and this matches slope field 3, C 3 .
Equation D should have a positive slope everywhere in thr first quadrant and negative in
the second so D 1.
24. Use Euler’s method to find an approximate value for f (1) given that the function
df
f (x) satisfies the equation
= sin(x 2 ) and f (0) = 0 .
dx
105
K. A. Tsokos: Series and D.E.
Solution
xn
yn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0
0
0.0009999
0.004997
0.013987
0.029919
0.054659
0.089886
0.136949
0.196669
0.269097
The exact answer is 0.3103.
25. Use Euler’s method to find an approximate value for f (1) given that the function
2
df
f (x) satisfies the equation
= e x and f (0) = 1 .
dx
Solution
xn
yn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.
1
1.1
1.199
1.29508
1.38648
1.47169
1.54957
1.61934
1.6806
1.73333
1.77782
The exact answer is 1.7468.
26. (a) Use Euler’s method to find an approximate value for f (2) given that the
1
df
= 2
function f (x) satisfies the equation
and f (1) = 0 , using a step size of
dx x + f 2
0.2. (b) Sketch the slope field for the equation in (a) for x and y in the interval
[2, 2] . (c) Join segments on the slope field diagram starting at the point (1, 0) to
predict the value of f (2) .
106
K. A. Tsokos: Series and D.E.
Solution
xn
1
1.2
1.4
1.6
1.8
2.0
yn
0
0.2
0.335135
0.431646
0.504470
0.561703
(b)
2
1
-2
-1
1
2
-1
-2
(c) Following the arrows starting at (1, 0) we end up at x = 2 with approximately
f (2) 1.15 , substantially above what the Euler method gave in (a).
df
= f (x, y) that gives rise to the
dx
following slope field diagram. (There is obviously no unique answer – just find any
one simple possibility, justifying your choice.)
27. Make the simplest guess for the form of f (x, y) in
2
1
-2
-1
1
2
-1
-2
The arrows are horizontal (indicating that f (x, y) = 0 ) whenever y = ±x . Thus f (x, y)
has at least y x and y + x as factors i.e. y 2 x 2 .