Download 1. Solve the following differential equations. If an explicit solution

1. Solve the following differential equations. If an explicit solution cannot be found, leave the solution in an
implicit form
(a) xv 0 = 21 v + v1 ,
v(1) = 0.
Solution. Use separation of variables
dv
1 v2 + 1 ,
=
dx
2
v
2v
dx
dv =
,
v2 + 1
x
ln(v 2 + 1) = ln x + C,
x
and since
v(1) = 0
Hence
ln
v2 + 1 x
=0
⇒
⇒
C = 0.
√
v = ± x − 1.
v2 + 1
= 1,
x
(b) y 0 + 2t y = t3 , y(2) = 3.
Solution. Use integrating factor
R
e
2
t
2
= e2 ln(t) = eln(t
)
= t2
and obtain
(t2 y)0 = t5 ,
t6
+ C,
6
4
t
4
y(t) =
+ 2.
6
3t
t2 y =
1
4
y(2) = 3 ⇒ C =
,
3
2. Solve the following initial value problem
cos(x) − y dx − (1 + x)dy = 0,
y
π
2
Solution. Use separation of variables to write
dy
+ y − cos x = 0,
dx
dy
(1 + x)
+ y = cos x,
dx
(1 + x)
and use integrating factor
e
R
1
1+x
=1+x
to get
[(1 + x)y]0 = cos(x),
(1 + x)y = sin(x) + C,
y=
and since
y
π
2
sin(x) + C
,
1+x
=0⇒0=
1+C
⇒ C = −1,
1 + π2
then finally
y(x) =
2
sin(x) − 1
.
1+x
= 0.
3. Use the Laplace transform method to solve
y 00 − 4y 0 + 3y = δ(t),
y(0) = y 0 (0) = 0.
where δ(t) is the delta function. Completely determine the solution for all values of t.
Solution.
(a) Using partial fractions
(s2 − 4s + 3)Y (s) = 1,
1
1
1
1
1
Y (s) =
=
−
= L(e3t ) − L(et )
(s − 3)(s − 1)
2(s − 3) 2(s − 1)
2
2
hence
(
y=
0,
e3t
2
−
et
2,
t<0
t ≥ 0.
(b) Using convolutions
(s2 − 4s + 3)Y (s) = 1,
1
= L(e3t )L(e3t ) = L(e3t ? et )
Y (s) =
(s − 3)(s − 1)
and
y(t) = e3t ? et =
Z
t
e3u et−u du = et
0
Z
t
e2u du = et
0
therefore
(
y=
0,
e3t
2
3
−
et
2,
t<0
t ≥ 0.
1 2t
e −1
2
∀t ≥ 0,
4. Use the variation of parameters method to find the particular solution of
y 00 − 4y 0 + 3y = e3t .
Solution. Find the solution corresponding to the homogeneous equations, using the characteristic equation:
λ2 − 4λ + 3 = 0 ⇒ λ1 = 3, λ2 = 1
hence
y1 (t) = e3t ,
y2 (t) = et .
We seek the particular solution of the form
yp (t) = e3t v1 + et v2 ,
where v1 , v2 satisfy
(
e3t v10 + et v20 = 0,
3e3t v10 + et v20 = e3t .
Then
2e3t v10 = e3t ⇒ v10 =
t
1
⇒ v1 = ,
2
2
and
et v20 = −e3t v10 = −
e3t
e2t
e2t
⇒ v20 = −
⇒ v2 = − .
2
2
4
Finally
yp (t) = e3t
e2t t
t
1 3t
+ et −
=
−
e .
2
4
2 4
4
5. Suppose that an electric circuit has a resistor R = 2Ω, a capacitor C = 1F , and voltage E = 100e−t V . If
the initial current is 0A, find the resulting current.
Solution.
RQ0 =
Q
= E(t),
C
dQ
+ Q = 100e−t ,
dt
dQ 1
+ Q = 50e−t ,
dt
2
t 0
t
e 2 Q = 50e− 2 ,
2
t
t
e 2 Q(t) = Q(0) − 100(e− 2 − 1),
t
t
t
1 t
1
Q(t) = Q(0)e− 2 + 100(e− 2 − e−t ) ⇒ Q0 = − Q(0)e− 2 + 100(− e− 2 + e−t )
2
2
and since
I=
dQ
dt
⇒
t
1
1 t
I(t) = − Q(0)e− 2 + 100(− e− 2 + e−t ).
2
2
Now use
I(0) = 0
⇒
Q(0) = 100,
and finally the current is
t
1 t
1
I(t) = − 100e− 2 + 100(− e− 2 + e−t )
2
2
t
= −100e− 2 + 100e−t .
5
6. Use the Laplace transform method to solve the following initial value problem
y 00 − 4y 0 + 4y = e2t ,
y(0) = y 0 (0) = 1.
Solution. Applying the Laplace transform we get
s2 Y (s) − s − 4 sY (s) − 1 + 4Y (s) =
(s2 − 4s + 4)Y (s) − s + 3 =
1
,
s−2
1
,
s−2
hence
1
s−2−1
1
s−3
+
=
+
2
3
2
(s − 2)
(s − 2)
(s − 2)
(s − 2)3
1
1
1
=
+
−
s − 2 (s − 2)2
(s − 2)3
t2 = L e2t − te2t + e2t ,
2
Y (s) =
where we used
L(eat tN ) =
6
N!
.
(s − a)N +1
Rt
7. The convolution of two functions f and g is f ? g(t) = 0 f (u)g(t − u)du. The convolution theorem states
that
L(f ? g) = Lf Lg,
where L denotes the Laplace transform.
Use the convolution theorem to find y(t) such that
L(y) =
s3
1
.
− s2
Evaluate all integrals.
Solution.
L(y) =
1 1
= L(t)L(et ) = L(t ? et ),
s2 s − 1
and then, using integration by parts
Z t
Z t
u=t Z t
y(t) = t ? et =
uet−u du = et
ue−u du = et u(−e−u )
−
−ue−u du
u=0
0
0
0
Z t
= −t + et
e−u du = −t + et (−e−t + 1) = et − t − 1.
0
7
8. Consider the system y0 = Ay where
A=
0
−1
1
.
0
1
Find the solution of the system with the initial value y(0) =
. Express your solution as a real vector
1
2
in R (i.e., no complex numbers).
Solution. The characteristic equation is:
0 = λ2 − T r(A)λ + det(A) = λ2 + 1
λ1,2 = ±i.
The eigenvector corresponding to i:
−i
1
A − iI =
,
−1 −i
−i
1
x1
0
=
−1 −i
x2
0
1
v=
i
The general solution, in complex form:
1
1
0
x(t) = eit
= (cos t + i sin t)
+i
i
0
1
1
0
0
1
=
cos t
− sin t
+ i cos t
+ sin t
0
1
1
0
cos t
sin t
=
+i
.
− sin t
cos t
Therefore
x(t) = C1
cos t
sin t
C1
1
+ C2
⇒ x(0) ≡
=
,
− sin t
cos t
C2
1
and finally
x(t) = C1
8
cos t + sin t
.
− sin t + cos t
9. Consider the function
(
2 + 4x
f (x) =
2 − 4x
if − 1 ≤ x ≤ 0
if −10 ≤ x ≤ 1
Expand f (x) in a Fourier series on [−1, 1].
Solution. L = 1 hence the Fourier series associated with f is
f (x) =
∞
a0 X +
an cos(nπx) + bn sin(nπx) .
2
n=1
Since the function f is even, the coefficients
bn :=
and
1
an :=
L
Z
L
1
L
Z
L
L
−L
L
−L
nπx f (x) cos
nπx f (x) sin
Z
dx ≡ 0,
1
dx ≡ 2
(2 − 4x) cos(nπx)dx,
0
For n = 0 we have
Z
a0 = 2
0
1
1
(2 − 4x)dx = 2(2x − 2x2 ) = 0,
0
and for all n ≥ 1
Z
1
(2 − 4x)
an = 2
0
sin(nπx) 0
nπ
dx
Z 1
1
2
2
(2 − 4x) sin(nπx) −
(−4) sin(nπx) dx
nπ
nπ 0
0
1
8
8
= − 2 2 cos(nπx) = − 2 2 (−1)n − 1 .
n π
n π
0
=
9
n ≥ 0.
10. Consider the following system
x0 = y
y 0 = x3 + y 3 + x
(a) Find all equilibrium points.
Solution. The x-nullcline:
y = 0,
the y-nullcline:
x3 + y 3 + x = 0.
There is only one equilibrium point
(0, 0).
(b) Classify the stability of each equilibrium
points.
0
1
Solution. The Jacobian is: J(x, y) =
. At (0, 0), the Jacobian is:
3x2 + 1 3y 2
0 1
J(0, 0) =
,
1 0
hence the linearization around (0, 0) is
0 u
0
=
v
1
1
0
u
.
v
The characteristic polynomial is
λ2 − T r(J)λ + det(J) ≡ λ2 − 1 = 0,
hence the eigenvalues are
λ1 = 1 > 0,
λ2 = −1 < 0,
1
1
−1
solution decays along v2 =
1
solution grows along v1 =
and the origin (0, 0) is a saddle point (unstable equilibrium point) for the linearized system. Since
saddle points are generic equilibrium points, the equilibrium point for the given nonlinear is also a
saddle point.
(c) Draw the phase portrait of the system in the vicinity of each equilibrium point.
Solution.
10
Table 1: Laplace Transforms
f (t)
F (s)
1
1
s
tn
n!
sn+1
eat
1
s−a
tn eat
n!
(s − a)n+1
sin bt
s2
b
+ b2
cos bt
s
s2 + b2
eat sin bt
b
(s − a)2 + b2
eat cos bt
(s − a)
(s − a)2 + b2
y 0 (t)
sY (s) − y(0)
y 00 (t)
s2 Y (s) − sy(0) − sy 0 (0)
tn f (t)
(−1)n F (n) (s)
Hc (t) = H(t − c)
e−cs
s
Hab (t) = Ha (t) − Hb (t)
e−as − e−bs
s
f (t − c) H(t − c)
e−cs F (s)
δc (t) = δ(t − c)
e−cs
11