1. No category

Problem 1
Consider the Euler equation
t2 x00 + αtx0 + βx = 0
for the function x = x(t) with t > 0, and with α and β two real parameters.
• Show that the change of variables t = eu transforms the Euler equation into a second order
linear equation with constant coefficients for the function x = x(u).
• Describe the behavior of the solutions of the equations obtained in this way, depending on
the real parameters α and β.
Solution. Observe that eu is a one-to-one map from R to R+ and thus a viable substitution for
d2 x
t > 0. We want to compute du
2 . First, by the chain rule,
dx
dx dt
dx
=
= eu ,
du
dt du
dt
We differentiate again to find
d2 x
= eu
du2
dx
d2 x
+ eu 2
dt
dt
=
dx
d2 x
+ e2u 2 .
du
dt
2
Now we recall that eu = t and use the ODE to replace t2 ddt2x
d2 x
dx
dx
u dx
− αe
+ βx = (1 − α)
− βx.
=
du2
du
dt
du
This linear second order ODE indeed has constant coefficients. We know how to solve this using
the characteristic polynomial r2 + (α − 1)r + β. Let
2
1−α
∆=
−β
2
1−α √
± ∆.
r± =
2
If ∆ > 0, the two linear independent solutions to the transformed equation are er+ u , er− u . Undoing
the change-of-variables, we obtain the general solution
C1 tr+ + C2 tr−
to the original Euler equation in this case. We see that, as t → ∞, a generic solution will go to
infinity/a constant/zero if r+ is positive/zero/negative. Similarly, if ∆ = 0, so r+ = r− , we obtain
the general solution
C1 tr+ + C2 tr+ log t
and as t → ∞, a generic solution will go to infinity/zero if r+ is non-negative/negative. When
∆ < 0, we obtain
C1 t(1−α)/2 cos(∆ log t) + C2 t(1−α)/2 sin(∆ log t)
whose growth at infinity is determined by the sign of 1 − α. Note also that these last solutions are
oscillatory.
Problem 2
Problem 2.1. Solve the following differential equation:
y 00 − 2y 0 + y = et /(1 + t2 )
Solution. The characteristic equation of the corresponding homogenous equation is r2 − 2r + 1 =
(r−1)2 = 0, which has the single repeated root r = 1, and so y1 (t) = et , y2 (t) = tet are fundamental
solutions. Observe that W (y1 , y2 )(t) = et (et + tet ) − tet et = e2t . The method of variation of
parameters shows that a particular solution of the stated (nonhomogeneous) differential equation
is
1
2
Z
y2 (t) · et /(1 + t2 )
y1 (t) · et /(1 + t2 )
dt + y2 (t)
dt
W (y1 , y2 )(t)
W (y1 , y2 )(t)
Z
dt
t
t
dt
+
te
=
2
1+t
1 + t2
t
2
e log(t + 1)
= tet arctan t −
2
(with constant terms omitted).
We conclude that the general solution of the given equation is
Z
Y (t)
=
−y1 (t)
Z
−et
y(t) = c1 et + c2 tet + tet arctan t −
et log(t2 + 1)
,
2
where c1 , c2 are constants.
Problem 2.2. Solve the following differential equation:
y 00 − y 0 − 2y = 2e−t
Solution. The characteristic equation of the corresponding homogeneous equation is r2 − r − 2 =
(r − 2)(r + 1), which has roots r = −1, 2, and so the general solution of this equation is
y1 (t) = c1 e−t + c2 e2t ,
for constants c1 , c2 .
We now use the method of ”undetermined coefficients”. Since −1 is a simple root of the
characteristic equation, we are looking for a particular solution of the form Y (t) = Ate−t with A
some constant A. Then
Y 0 (t) = Ae−t − Ate−t ,
Y 00 (t) = −Ae−t − Ae−t + Ate−t = Ate−t − 2Ae−t ,
and so
Y 00 (t) − Y 0 (t) − 2Y (t)
= Ate−t − 2Ae−t − Ae−t + Ate−t − 2Ate−t
= −3Ae−t .
It follows that if Y 00 (t) − Y 0 (t) − 2Y (t) = 2e−t , then −3A = 2, whence A = −2/3.
Indeed, it is easily verified that Y (t) = − 23 te−t is a solution to the differential equation, and so
the general solution is
2
y(t) = c1 e−t + c2 e2t − te−t ,
3
for constants c1 , c2 .
Problem 3
Let y1 (t) and y2 (t) be two solutions of the homogeneous second order equation
y 00 + p(t)y 0 + q(t)y = 0
where p(t) and q(t) are continuous on an interval t ∈ I = (α, β).
• If the Wronskian of the two solutions is constant, what can one say about p(t) and q(t)?
• Show that if y1 (t) and y2 (t) vanish at the same point in the interval I, or if they have
a maximum or a minimum at the same point, then they are not a fundamental set of
solutions.
Solution. Assume the Wronskian of the two solutions y1 (t), y2 (t) is a constant c. If c = 0, then
y1 (t), y2 (t) are linearly dependent, and there is nothing we can say about p(t), q(t). If c 6= 0, then
by Abel’s theorem W 0 + p(t)W = 0, thus we have p(t)c = 0, that is p(t) = 0.
Next, suppose that y1 (t) and y2 (t) vanish at the same point t1 , then
W (y1 , y2 )(t1 ) = y1 (t1 )y20 (t1 ) − y10 (t1 )y2 (t1 ) = 0.
3
Suppose y1 (t) and y2 (t) have a maximum or a minimum at the same point t1 , then y10 (t1 ) =
y20 (t1 ) = 0, and then W (y1 , y2 )(t1 ) = 0. In either case the Wronskian is 0 at t1 , therefore y1 , y2 do
not form a fundamental set of solutions.
Problem 4
For the following differential equations describe the equilibrium solutions and the asymptotic
behavior of the other solutions, for different choices of the initial condition y(0) = y0 :
• dy/dt = ey − 1, with initial conditions −∞ < y0 < ∞;
• dy/dt = y(a − y 2 ), for different possible values of the parameter a > 0, a = 0, or a < 0,
and with initial conditions −∞ < yo < ∞.
Problem 4.1.
Solution. First, we find the equilibrium solutions. Solve
dy
= ey − 1 = 0
dt
which has solution y = 0, thus this is the only equilibrium solution. Since dy/dt > 0 and is
increasing for y > 0, if y0 > 0, then y → ∞ as t → ∞. Similarly, dy/dt < 0 and is decreasing for
y < 0, so if y0 < 0, then y → −∞ as t → ∞. So y = 0 is an unstable equilibrium solution.
Problem 4.2.
Solution. If a ≤ 0,
dy
= y(a − y 2 ) = 0 ⇒ y = 0.
dt
So y = 0 is the unique equilibrium solution. Note dy/dt > 0 for y < 0 and dy/dt < 0 for y > 0. So
y → 0 as t → ∞ for all initial conditions. The equilibrium solution y = 0 is asymptotically stable.
If a > 0,
√
dy
= y(a − y 2 ) = 0 ⇒ y = 0, ± a.
dt
√
So y = 0, ± a are equilibrium solutions. We plot dy/dt against y:
From the plot, we see that
√
√
(1) for initial conditions y0√< − a, y → − a as
√ t → ∞,
(2) for initial conditions − a < y0√< 0, y →
−
a as t → ∞,
√
(3) for initial conditions 0 < y√
<
a,
y
→
a
as
t → ∞,
0
√
(4) for initial conditions y0 > a, y → a as t → ∞.
√
So the equilibrium solutions y = ± a are asymptotically stable, while y = 0 is unstable.