Exam
MAT 244 – ODE – Winter 2015
Solutions
#1
Solve the following IVP(5 points):
xy 0
Solution
3y = 3x5
y(1) = 7.5
Rewrite the equation
3
y = 3x4
x
In this form we may use the standard integrating factor:
✓Z
◆
✓ Z
◆
3
1
µ(x) = exp
p(x)dx = exp
dx = exp( 3 ln x) = 3
x
x
y0
We know the general solution is given by
✓
◆
Z
Z
Z
1
3 2
3
4 1
3
3
y(x) =
g(x)µ(x)dx = x
3x 3 dx = x
3xdx = x
x +C
µ(x)
x
2
Plugging in the initial data gives
7.5 =
3
3
+ C =) C = 6 =) y(x) = x5 + 6x3
2
2
# 2 Find the general solution of the equation xy 2 (xy 0 + y) = 1 (hint: find an integrating factor)(5 pt). Then
solve the IVP y(1) = 3 for this equation.(1 pt)
Solution
Lets check if the equation is exact, we rewrite into standard form
xy 3
1 + x2 y 2 y 0 = 0 () (xy 3 1) dx + x2 y 2 dy = 0
|{z}
| {z }
M
We try to check the exactness condition but see
My = 3xy 2
Notice that
My
&
Nx
N
1
Nx = 2xy 2
=
1
x
N
Exam – Winter 2015
MAT 244
is just a function of x, so it will work as an integrating factor. We have
✓Z
◆
✓Z
◆
My N x
dx
=x
µ(x) = exp
dx = exp
N
x
makes the equation exact. Thus we may integrate the ODE after multiplying by the above factor to obtain
Z
Z
2 3
3 2
(x y
x) dx + x y dy = 0 =) F (x, y) = M̃ dx
Ñ dy
|{z}
| {z }
Ñ
M̃
=
Z
(x2 y 3
✓
Z
x)dx
x3 y 2 dy
◆
x2
x3 y 3
2
3
x3 y 3
3
3 3
x y
x2
=
3
2
=
Thus the general solution to the ODE is
x3 y 3
C=
3
x2
=) y(x) =
2
r
3
3
C
+ 3
2x x
The solution to the IVP is found by calculating C, we see
C = 25.5 =) y(x) =
# 3
Find the general solution of the equation x3 y 00
r
3
3
51
+
2x 2x3
2xy = 6 ln x ( hint: the homogeneous part of this
equation can be reduced to an Euler equation)(6 points)
Solution
Rewrite the equation
ln x
x
Notice the homogeneous part is now Euler, so try y(x) = x as the homogeneous solution, we obtain
x2 y 00
x ( (
1)
2y = 6
2) = 0 =) (
2)( + 1) = 0 =)
Thus the homogenous solution is
y(x) = A |{z}
x2 +B
y1
1
x
|{z}
=
1, 2
A, B 2 R
y2
We now may find the non homogeneous solution using the variation of parameters formula, so we calculate the
Wronskian:
W [y1 , y2 ](x) = y1 y20
y2 y10 =
x2
1
x2
2
x=
x
3
Thus
1
x
is the solution, where (don’t forget to divided the ODE by x2 !) (and u = 1/x)
Z
Z
Z
Z 2
u3 ln(u)
y2 g
ln x
u
2
A(x) =
=2
dx = 2 u ln udu =
du =
W
x4
3
3
y(x) = A(x)x2 + B(x)
2
2 ln(x)
3x3
2
+B
9x3
Exam – Winter 2015
MAT 244
and (where u = ln x)
B(x) =
Z
y1 g
=
W
2
Z
ln x
dx =
x
2
Z
udu =
u2 + B =
(ln x)2 + B
Putting it all together we obtain
y(x) = Ax2 +
#4
2 ln x
3x
p
2y 00 + 4y 0 = et sin( 3t).
Find the general solution of y 000
Solution
(ln x)2
x
B
x
First find the homogeneous solution, try e t , we see
e t(
3
2
2
+ 4 ) = 0 =)
(
2
2 + 4) = 0 =)
p
= 0, 1 ± i 3
Thus the homogeneous solution is
p
p
y(t) = A + B et sin( 3t) +C et cos( 3t)
| {z }
| {z }
y1
A, B, C 2 R
y2
To find the non homogeneous solution, we’ll use the method of undetermined coefficients and guess
p
p
yp (t) = Dtet sin( 3t) + Etet cos( 3t) = Dty1 + Ety2
The derivatives are
yp0 =Dy1 + Ey2 + Dty10 + Ety20
yp00 =2Dy10 + 2Ey20 + Dty100 + Ety200
yp000 =3Dy100 + 3Ey200 + Dty1000 + Ety200
Thus we see
yp000
2yp00 + 4yp0 =D(4y1
4y10 + 3y100 ) + E(4y2 4y20 + 3y200 )
p
p
p
p
p
p
=et D( 6 sin( 3t) + 2 3 cos( 3t)) + E( 6 cos( 3t) 2 3 sin( 3t))
p
p
p
p
=( 6D 2 3E)et sin( 3t) + (2 3D 6E)et cos( 3t)
comparing this with the RHS of the ODE, we obtain
6D
p
2 3E = 1 &
p
2 3D
1
6E = 0 =) E = p
8 3
&
D=
1
8
Thus the general solution to the ODE is
p
p
p
3tet sin( 3t) + 3tet cos( 3t)
y(t) = A + Be sin( 3t) + Ce cos( 3t) +
24
t
p
t
p
3
Exam – Winter 2015
#5
MAT 244
a)Find the general solution (3 points)
(
x0 = 2x
y
0
y = x + 4y
0.5 Point - Computing the characteristic equation to find the eigenvalues
P ( ) = det(A
2
I )=
1
1
2
=
4
2 Point - Computing the eigenvector and generalized eigenvector
!
1
1
= 3 =) ker(A 3I) = ker
= span
1
1
3I)~ g = ~ =)
(A
!
1 ~
1
1
1
g
3)2 =)
6 +9=(
1
1
!
1
=
1
!
=) ~ =
=) ~ g =
=3
1
1
!
!
1
0
0.5 Point -Writing the general solution
x(t) = A~ e
t
+ Be t (t~ + ~ g ) =) x(t) = A
!
1
1
e3t + B
1
t
t
!
e3t
A, B 2 R
b) Input initial data (2 points)
x(0) = 1
&
2 point Solve for A and B from the previous part
!
!
1
1
x(0) =
=A
+B
3
1
y(0) = 3
1
0
!
1
=) x(t) =
=) B = 4 &
4t
3 + 4t
!
A=3
e3t
c) If W (t) has the property W (1) = 2 for this system, what is W (3) (2 points)
!
1 1 t
(1) (2)
6t
W (t) = C det(x x ) = Ce det
= Ce6t
1
t
W (1) = 2 =) C =
#6
2e
6
=) W (3) = 2e12
Find the general solution to (6 points)
0
1
2
C
2 1C
Ax
1 2
1
3
B
.
x=B
@0
0
2 Point - Computing the characteristic equation to find the eigenvalues
1
P ( ) = det(A I ) =
3
0
0
2
2
1
1
2
=
⇣
(1+ ) (2
4
)(2
⌘
) 1 =
(1+ )(
3)(
1) =)
= ±1, 3
Exam – Winter 2015
MAT 244
3 Points - Computing the eigenvectors (1 point for each)
=
0
0
B
B
1 =) ker @0
0
0
2
0
4
#7
3
1
1
1
0 1
0 1
5
5
C
B C
B C
~
C
B
C
B
1 A = span @4A =) 1 = @4C
A
1
4
4
2
1
1
0
0 1
5
B C t
B C 3t
t
B
C
B
+ B @ 2 A e + C @4C
Ae
2
4
Consider the nonlinear system
1
0 1
1
B C
B
= @0C
A
0
1
0 1
0 1
2
1
1
C
B C
B C
~
C
B
C
B
1 1A = span @ 2 A =) 1 = @ 2 C
A
1 1
2
2
3
B
= 3 =) ker B
@0
0
0
2
3
B
= 1 =) ker B
@0
0
1 Point - Writing the general solution
0 1
1
B C
B
x(t) = A @0C
Ae
0 1
1
C
B C
~
C
B
1A = span @0C
A =)
3
0
3
(
1
1
x0 = (1
A, B, C 2 R
y)(y + x)
0
y = (2 + y)(x
y)
Describe the locations of all critical points (2 pts). Classify their types (i.e. specify whether they are nodes,
saddles, etc) and stability (3 pts). Sketch the phase portraits near the critical points(2 pts). Sketch the phase
portrait of the whole system(1 pt).
Solution
We compute the solutions to x0 = y 0 = 0 for critical points:
(
y=1
x0 = 0 = (1 y)(y + x) =)
& y 0 = 0 = (2 + y)(x
y= x
y) =)
(
y=
y=x
Thus the only critical points are
(x, y) = (0, 0), (1, 1), (2, 2)
.
To linearize the system, we know z = Jz where z = x
J(x, y) =
1
x0 , thus compute the Jacobian.
!
y
2y + x + 1
2+y
2y + x
2
We have 3 points to check, so let’s start with x0 = (0, 0). We see
!
1 1
J(0, 0) =
2
2
The eigenvalues are found by checking the characteristic equation,
P ( ) = det(J(0, 0)
1 )=
1
1
2
=
2
5
2
+
4 = 0 =)
+
=
2
p
1 ± 17
2
Exam – Winter 2015
MAT 244
The eigenvectors are found by checking the kernel,
!
p
p !
3
17
1
3 + 17
2
p
= span
=)
+ =) ker
3
17
2
4
2
=) ker
p
3+ 17
2
2
1
p
3+ 17
2
!
p
3
= span
17
4
!
=)
+
+
p
3+
=
17
4
p
3
=
4
!
!
17
!
7
⇡
4
!
1
⇡
4
Thus we see this is an unstable saddle. Next up we have x0 = (1, 1),
!
0 0
J(1, 1) =
3
3
The eigenvalues are found by checking the characteristic equation,
P ( ) = det(J(1, 1)
1 )=
0
3
2
=
3
The eigenvectors are found by checking the kernel,
!
0 0
= span
1 =) ker
3
3
2
=) ker
3
0
3
0
!
1
+ 3 = 0 =)
!
=)
1
0
= span
1
!
=)
+
+
1
=
1
0
=
= 0,
1
1
2
=
3
!
!
Thus we see this is an asymptotically stable degenerate node (transiting between a node and a saddle). Lastly
we check x0 = (2, 2).
J(2, 2) =
3
7
1
4
!
The eigenvalues are found by checking the characteristic equation,
P ( ) = det(J(2, 2)
1 )=
3
7
1
=) ker
p
1+ 29
2
1
7
p
1+ 29
2
!
= span
=
4
The eigenvectors are found by checking the kernel,
!
p
1
29
7
2
p
= span
+ =) ker
1
29
1
2
2
1+
p
29
2
p
1
2
7 + 5 = 0 =)
29
!
!
=)
=)
+
+
=
=
±
1+
=
p
29
2
p
1
2
7±
29
!
!
p
29
2
⇡
⇡
4
2
!
6
2
!
Thus this is an unstable node. Since we have all the local information, we know the system looks something
like
6
Exam – Winter 2015
MAT 244
3
2
1
0
-1
-2
-3
-3
# 8 Consider the equation y 00
-2
-1
0
1
2
3
5xy = 0, find the recurrence relation for the series solution of this equation at
x0 = 0, then find the first four terms of the solution with y(0) = 2, y 0 (0) = 3. What is the radius of convergence
of the corresponding series solution? ( 6 pts)
Solution
Try
y(x) =
1
X
an xn
n=0
as a solution, we obtain the following from substituting the solution
1
X
n=0
(n + 2)(n + 1)an+2 xn
5
1
X
an xn+1 = 2a2 +
n=0
1
X
⇥
(n + 2)(n + 1)an+2
5an
n=1
This of course implies that the recurrence relation is
(n + 2)(n + 1)an+2
5an
1
= 0 =) an+2 =
5an 1
(n + 2)(n + 1)
and we also see that a2 = 0 from the equation. We compute the first four terms.
a3 =
5a0
5
=
6
3
&
a4 =
5a1
5
=
12
4
Thus the solution with the first four terms is
y(x) = 2 + 3x +
5x3
5x4
+
+ ...
3
4
The radius of convergence is infinity since the coefficients of the ODE are well behaved.
7
1
⇤
xn = 0