MAT2007 ODEs (2008/9)
ES1
Solutions to revision exercises, week 1
The M-file exercise revision.m gives some Matlab commands.
1.
All ODEs in this question are separable. Note that dfield gives a nice overview of the
shape of the solutions.
R
R
(a) Note that dx
dx = t dt . Hence the solutions are
dt = t can be rewritten as
2
x(t) = t2 + C , see Figure 1(a).
R dx
R
(b) The ODE dx
dt. This gives ln x = t + C , thus
dt = x can be rewritten as
x =
t
the solutions are x(t) = A e , see Figure 1(b).
R
R
3t2 +1
(c) The ODE dx
can be rewritten as 2x
dx = (3t2 + 1) dt . This gives
dt =
2x
√
x2 = t3 + t + C , thus the solutions are x(t) = ± t3 + t + C , see Figure 1(c).
R
R
(d) The ODE sin t cos x = sin x cos t dx
tan x dx = tan t dt.
dt can be rewritten as
This gives ln sec x = ln sec t + C , hence solutions are of the form sec x = A sec t or
cos x = B cos t, see Figure 1(d).
(a) Solutions of 1(a), note the similarity of the
solutions (shift in x -direction).
(b) Solutions of 1(b), note the similarity of the
solutions (shift in t direction and reflection).
(c) Solutions of 1(c), note the vertical derivative
near x = 0 and the fact that the solutions can’t
be continued here.
(d) Solutions of 1(d), note the periodicity in both
x and t . The solutions have vertical derivatives
near x = kπ , k ∈ Z and can’t be continued at
those points.
Figure 1: Exercise 1, solutions with dfield.
MAT2007 ODEs (2008/9)
2.
ES2
Again, both ODEs are separable. Checks with Matlab are in the M -file.
R dx
R dt
2 can be written as
(a) The ODE (2t + 1) dx
=
This gives
dt = x
2t+1 .
x2
− x1 = 12 ln(2t + 1) + C . Using the initial condition x(0) = 1, we get that C
satisfies −1 = 21 ln(1) + C , hence C = −1. Thus the solution of the IVP is
¡
¢−1
2
x(t) = 1 − 21 ln(2t + 1)
.
= 2−ln(2t+1)
R −2x
2x can be rewritten as
e
dx = dt
(b) The ODE t2 dx
. This gives − 21 e−2x =
dt = e
t2
1
− t + C . Using the initial condition x(1) = 0, we get that C satisfies − 12 =
1
−2x(t) = 2 − 1 or x(t) =
−1 + C
t
¡ 2, hence
¢ C = 2 . Thus the solution of the IVP is e
1
− 2 ln t − 1 .
3.
These ODEs are of first order and linear and can be solved with the integrating factor
method. Again dfield gives nice pictures of the solutions.
(a) The ODE can be written as dx
dt + p(t) x = q(t) with p(t) =
the integrating factor is
µZ
¶
¡
¢
2
IF = exp
dt = exp ln t2 = t2 .
t
2
t
and q(t) =
3
t2
. Thus
Multiplying the ODE with the integrating factor gives
Z
d ¡2 ¢
3 C
t x = 3, hence t2 x = 3 dt = 3t + C, thus x(t) = + 2 ,
dx
t
t
see Figure 2(a).
−t
(b) The ODE can be written as dx
dt + x = 2t e . Thus the integrating factor is
µZ ¶
IF = exp
dt = et .
Multiplying the ODE with the integrating factor gives
Z
d ¡ t ¢
t
e x = 2t, hence e x = 2t dt = t2 + C,
dx
thus x(t) = (t2 + C) e−t ,
see Figure 2(b)
(c) The integrating factor for the ODE
µZ
¶
IF = exp
2 dt = e2t .
dx
dt
+ 2x = et is
Multiplying the ODE with the integrating factor gives
d ¡ 2t ¢
e x = e3t ,
dx
see Figure 2(c)
hence e2t x =
1 3t
e + C,
3
thus x(t) =
1 t
e + Ce−2t ,
3
MAT2007 ODEs (2008/9)
ES3
(d) The integrating factor for the ODE
µ Z
¶
2
IF = exp −2 t dt = e−t .
dx
dt
− 2xt = t is
Multiplying the ODE with the integrating factor gives
d ³ −t2 ´
2
e x = te−t ,
dx
1 2
2
hence e−t x = − e−t + C,
2
1
2
thus x(t) = − + Cet ,
2
see Figure 2(d)
(a) Solutions of 3(a).
(b) Solutions of 3(b).
(c) Solutions of 3(c).
(d) Solutions of 3(d).
Figure 2: Exercise 3, solutions with dfield.
4.
All ODEs are linear, second order ones. To solve them we use the methods of Linear
Algebra, chapter 5. The plots are in figure 4, the Matlab code is in exercise revision.m .
2
2
(a) The auxiliary quadratic for the homogeneous ODE ddt2x + dx
dt − 30x = 0 is m +
m − 30 = 0, or (m − 5)(m + 6) = 0. Thus the general solution of this ODE is
x(t) = A e5t + B e−6t , see Figure 3(a).
MAT2007 ODEs (2008/9)
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2
2t is
(b) The auxiliary quadratic for the inhomogeneous ODE 2 ddt2x + 3 dx
dt + x = e
2m2 + 3m + 1 = 0 or (2m + 1)(m + 1) = 0 . Thus the complementary function (CF)
is xC (t) = A e−t + B e−t/2 . For the particular integral (PI), we try xP (t) = C e2t .
1
Substitution into the ODE gives (8C + 6C + C)e2t = e2t . Hence C = 15
and the
1 2t
−t
−t/2
general solution for the ODE is x(t) = A e + B e
+ 15 e , see Figure 3(a).
2
(c) The auxiliary quadratic for the inhomogeneous ODE ddt2x + 2 dx
dt + 5x = 5t − 3 is
m2 + 2m + 5 = 0 or (m + 1)2 + 4 = 0 . Thus m = −1 ± 2i and the complementary
function is xC (t) = e−t (A cos 2t + B sin 2t). For the particular integral, we try
xP (t) = C t + D . Substitution into the ODE gives 0 + 2C + 5(Ct + D) = 5t − 3.
Hence C = 1 and D = −1, thus the general solution for the ODE is x(t) =
e−t (A cos 2t + B sin 2t) + t − 1, see Figure 3(a).
(a) Solutions of 4(a).
(b) Solutions of 4(b).
(c) Solutions of 4(c).
Figure 3: Exercise 4, solutions with dfield.
5.
The ODEs are again second order linear ones and can solved in the same way as in the
previous question.
MAT2007 ODEs (2008/9)
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2
(a) The auxiliary quadratic for the inhomogeneous ODE ddt2x − 2 dx
dt + 5x = 10 cos t is
2
2
m − 2m + 5 = 0 or (m − 1) + 4 = 0. Thus m = 1 ± 2i and the complementary
function is xC (t) = et (A cos 2t + B sin 2t). For the particular integral, we try
xP (t) = C cos t + D sin t. Substitution into the ODE gives (−C − 2D + 5C) cos t +
(−D + 2C + 5D) sin t = 10 cos t. Hence C = 2 and D = −1, thus the general
solution for the ODE is x(t) = et (A cos 2t+B sin 2t)+2 cos t−sin t. Using the initial
condition x(0) = 2 , we get A + 2 = 2, thus A = 0 and x0 (0) = 1 gives 2B − 1 = 1,
hence B = 1. Thus the solution of the IVP is x(t) = et sin 2t + 2 cos t − sin t.
2
2t
2
(b) The auxiliary quadratic for the inhomogeneous ODE 2 ddt2x +2 dx
dt −x = 11e is 2m +
√
¡
¢
2
2m − 1 = 0 or 2 m + 21 − 32 = 0. Thus m = 12 (−1 ± 3) and the complementary
√
√
function is xC (t) = A e(−1+ 3)t/2 +B e(−1− 3)t/2 . For the particular integral, we try
xP (t) = C e2t . Substitution into the ODE gives (8C +4C√−C)e2t = 11e2t√, thus C =
1 and the general solution for the ODE is x(t) = A e(−1+ 3)t/2 + B e(−1− 3)t/2 + e2t .
With the initial conditions x(0) = 1 and x0 (0) = 2, we get
1=A+B+1
2=
√
−1+ 3
2
A+
√
−1− 3
2
B+2
Thus A = 0 = B and the solutions of the IVP is x(t) = e2t .
6.
We start pplane by typing pplane7 into Matlab and introduce the ODE. In Figure 4(a),
the input window from pplane is shown. By using the keyboard and/or clicking on points
near (0, 0), (1, 0), (1, 1), (−1/2, 0) and (1/2, 0), the plot is Figure 4(b) is obtained.
The solutions divide the plane in 3 regions, one with periodic solutions and two with
unbounded solutions.
(a) Input of pplane
(b) Some typical solutions of the system
Figure 4: Exercise 6, using pplane.