hw24
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moienMath 21b: Linear Algebra Spring 2016
Homework 24: Differential equations I
This homework is due on Friday, April 7, respectively on Tuesday, April 12, 2016.
1 a) Solve the differential equation dx
dt = 3/x ,with x(0) = 1.
b) Solve the differential equation dx 2
dt = 1 + x ,with x(0) = 0.
c) Solve the differential equation dx
dt = 1/ cos(x) ,with x(0) = 0.
2 Solve the system
dx 3 8
= Ax, A =
dt 7 4
−5
with initial condition x(0) = .
4
dx p −q
3 For which p, q is the system = x(t) stable?
dt q p
4 The interaction of two animal species is modeled by the equations
dx
= 1.5x − 1.2y
dt
dy
= 0.8x − 1.4y
dt
a) Interpret the system. Is it a symbiosis, competition or predator-
prey?
b) Sketch the phase portrait in the first quadrant.
c) What happens in the long term? Does it depend on the initial
population? If so, how?
5 A door opens on one side only. A spring mechanism closes the door
which forms an angle θ(t) with the frame. The angular velocity
dθ
is ω(t) = dt (t). The differential equations are
dθ
= ω
dt
dω
= −2θ − 3ω
dt
The first equation is the definition, the second incorporates the
force −2θ of the spring and the friction −3ω.
Sketch a phase portrait for the system and use this to answer
the question, for which initial conditions, the door slams (reaches
θ = 0 with negative ω).
Differential Equations I
A system dx dt = f (x) is a differential equation. One can often
solve them by separation of variables. For example, if dtd x =
x2/t, x(0) = 0, then we get tdt = x2dx and integrate both sides
to get t2/2 = x3/3 + c so that x(t) = (3(t2/2 − c))1/3. As
x(0) = 0 we have c = 0 and x(t) = (3t2/2)1/3. The linear
differential equation dx kt
dt = kx has the solution x(t) = e x(0).
For k > 0, this gives exponential growth. For k < 0, exponential
decay. A linear system of differential equations is dx dt = Ax.
If x(0) = v is an eigenvector with eigenvalue λ, then x(t) is
always a multiple of v, say x(t) = c(t)v where dc dt = λv. Thus if
x(0) = c1v1 + . . . + cnvn writes an initial condition as a sum of
eigenvectors, then x(t) = c1eλ1tv1 + . . . + cneλntvn is the explicit
solution of the system. A system is asymptotically stable, if
x(t) → 0 for all initial conditions x(0). We have asymptotic
stability if Re(λj ) < 0 for all j. Be sure to compare all this with
the case of discrete dynamical systems.