Differential Equations
Math 231,Fall 2015, Midterm
Solutions
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No books, notes, or calculators. Read the questions carefully.
Show all your work. Answer the questions in the spaces
provided. If you run out of room for an answer, continue on the
back of the page. There are 6 pages and 5 questions in total,
make sure that your exam paper is complete.
Question Points Score
1
20
2
20
3
20
4
20
5
20
Total
100
1
1. (20 pts) Solve the initial value problem:
tx0 + 2x =
cos t
, x(2) = 0
t
Solution.
The first order linear ODE in standard form is:
2
cos t
x0 + x = 2
t
t
R
Calculating integrating factor µ = e 2/t = e2 ln |t| = t2 , and multiply it to both sides of
the equation, we get:
t2 x0 + 2tx = cos t
So (t2 x)0 = cos t. Integrating the equation, we get:
Z
2
t x = cos tdt = sin t + c
Substitute the initial condition x(2) = 0, get c = − sin 2. So the solution to the initial
value problem is
sin t sin 2
x= 2 − 2 .
t
t
2
2. (20pts) Consider:
(cos t − x) sin t + (x + cos t)x0 = 0, x(0) = −2
(a) (10pts) Solve the initial value problem.
(b) (10pts) Find the interval of existence of the solution.
Solution.
(a) This ODE is not linear or separable, let us check if it is exact. Here,
M = (cos t − x) sin t, N = x + cos t
Calculating Mx , Nt , we have
Mx = Nt = − sin t
So this is an exact equation, that is, we can find some function F (t, x) so that
Ft = M = (cos t − x) sin t
Fx = N = x + cos t
Integrating the first equation, we have:
Z
F = cos t sin t − x sin tdt
=
1 2
sin t + x cos t + h(x)
2
calculating Fx and set it to be equal to x + cos t, we have
h0 (x) = x.
So h(x) = x2 /2, and we get the solution of the exact equation in implicit form:
F (t, x) =
1 2
x2
sin t + x cos t +
=c
2
2
Substituting the initial condition x(0) = −2, we get c = 0. So the solution of the
initial value problem in implicit form is
1 2
x2
sin t + x cos t +
= 0.
2
2
3
(b) To find the interval of existence of the solution, observe that we can find x(t)
explicitly. In particular, we have
p
x = − cos t ± cos2 t − sin2 t
The one that satisfies the initial condition x(0) = −2 is
p
x(t) = − cos t − cos2 t − sin2 t.
The solution is well-defined and differentiable if and only if:
cos2 t − sin2 t > 0
That is,
cos 2t > 0,
Since the initial time is 0, the interval of existence is −π/4 < t < π/4.
4
3. (20pts) Consider the autonomous equation x0 = x2 (x + 1)(2 − x).
(a) Find the equilibrium solutions.
(b) Classify each equilibrium solution as asymptotically stable, unstable or semistable.
(c) Draw the phase line.
Solution.
(a) The equilibrium points are values of x such that the rate function
f (x) = x2 (x + 1)(2 − x) = 0
so the equilibrium points are −1, 0, 2.
(b)(c) To find the stability properties, we use the phase line. Find the sign of f (x) on
each of the intervals divided by the equilibrium points:
x < −1 :
−1 < x < 0 :
0<x<2:
x>2:
x2 (x + 1)(2 − x) < 0
x2 (x + 1)(2 − x) > 0
x2 (x + 1)(2 − x) > 0
x2 (x + 1)(2 − x) < 0
So the phase line is:
−1
0
2
x
Based on the phase line, we know −1 is unstable, 2 is asymptotically stable and
0 is semi-stable.
5
4. (20pts) Suppose x1 (t), x2 (t), x3 (t) are solutions of the second order ODE
x00 + tet x0 + t2 x = 0
and satisfy the following initial conditions:
x1 (0) = 1, x01 (0) = −2;
x2 (0) = −2, x02 (0) = 4;
x3 (0) = 1, x03 (0) = 2;
(a) (5pts)What is the interval of existence for xi (t), i = 1, 2, 3?
(b) (15pts)Which of the following sets form a fundamental set of solutions for the
ODE:
(1) {x1 , x2 }; (2) {x2 , x3 }; (3) {x1 , x3 }? Explain your answers.
Solution.
(a) This is a second order linear equation in standard form, and the coefficient functions tet , t2 are continuous for all t in (−∞, +∞), so the solution is guaranteed
to be valid on the interval (−∞, +∞).
(b) To check if {xi , xj } is a fundamental set of solutions, we use its Wronskians. In
particular, if W (xi , xj )(0) 6= 0, then {xi , xj } is a fundamental set of solutions.
Direct calculation gives:
W (x1 , x2 )(0) = 1 × 4 − (−2) × (−2) = 0;
W (x2 , x3 )(0) = −2 × 2 − 1 × 4 = −8;
W (x1 , x3 )(0) = 1 × 2 − 1 × (−2) = 4.
Thus {x2 , x3 }, {x1 , x3 } are fundamental sets of solutions.
6
5. (20pts) Suppose a periodically forced spring-mass system is described by the ODE
4x00 + x = 3 cos t
(a) (15pts) Find the general solution.
(b) (5pts) Find the solution that satisfies the initial condition: x(0) = 0, x0 (0) = 0.
Solution.
(a) The corresponding homogeneous equation is 4x00 + x = 0, so the characteristic
equation is
4r2 + 1 = 0
which has complex conjugate roots r = ±i/2, so the complementary solution is
given by:
xc = c1 cos(t/2) + c2 sin(t/2).
Use the method of undetermined coefficient, a particular solution for the nonhomogeneous equation has the form
xp = a cos t + b sin t.
Substitute in the equation to determine a, b, we get xp = − cos t. So the general
solution is
x = xc + xp = c1 cos(t/2) + c2 sin(t/2) − cos t.
Note: We can also use the exponential input theorem to find a particular solution.
Observe that cos t = Re(eit ), so a particular solution is given by the real part of
3eit
3eit
=
= −eit
4i2 + 1
−3
Its real part is − cos t, so xp = − cos t.
(b) To find the solution with the initial condition x(0) = 0, x0 (0) = 0, we substitute
to the general solution and solve for c1 , c2 :
x(0) = c1 − 1 = 0
1
x0 (0) = c2 = 0
2
We get c1 = 1, c2 = 0, so the solution to the I.V.P. is
x(t) = cos(t/2) − cos t.
7