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Differential Equations Math 225
Final Exam (Sample)
December 19, 2005
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Open Notes
Calculators Permitted
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1. (10 pts) Solve y’ + y/x = 3 cos(2x)
There are at least two methods we can choose from to solve this equation. First
we notice that this is a linear first order equation, and thus can be solved by
computing an integrating factor. Alternatively, we can observe that this is a
Cauchy-Euler equation, solve the homogeneous equation that way and then use
variation of parameters to find the general solution.
First method: Note that this is a first order linear equation.
Compute the integrating factor µ(x) = e∫(1/x)dx = eln(x) = x
So y(x) = (1/µ(x))∫(µ(x))(3cos(2x))dx = (1/x)∫(x)(3cos(2x))dx
= 3cos(2x)/4x + 3sin(2x)/2 + C/x (integrating by parts)
Second method: Note that this is a Cauchy-Euler equation.
The homogeneous equation is equivalent to xy’ + y = 0
If we assume a solution of form y = xr, we plug in to get x(rxr-1) + xr = 0
Thus the indicial equation is r +1 = 0, with a solution of r = -1.
yH(x) = Ax-1 is the homogeneous solution
We can use variation of parameters to solve for the full solution:
Assume y(x) = u(x) yH(x) = u(x) x-1
y’ = u’ x-1 - u x-2
Plugging in: (u’ x-1 - u x-2) + (1/x)u x-1 = 3 cos(2x)
u’ x-1 = 3 cos(2x)
u’ = 3 x cos(2x)
u(x) = ∫ 3x cos(2x) dx = 3(cos(2x)/4 + x sin(2x)/2) + C
Thus y(x) = u(x) x-1 = 3cos(2x)/4x + 3sin(2x)/2 + C/x
Answer: y(t) = 3cos(2x)/4x + 3sin(2x)/2 + C/x
2. (25 pts) Given y” + 2y’ + y = g(t) with initial conditions y’(0)=1 and y(0)=2 :
a) (5 pts) Find the solution to the homogeneous equation with the specified initial
conditions.
No method was specified, so we can solve this using either the method of
Laplace transforms or the more conventional characteristic equation method.
(s2 Y(s) – sy(0) – y’(0)) + 2(sY(s) – y(0)) + Y(s) = 0
So Y(s) (s2 + 2s + 1) – (s y(0) + (2 y(0) + y’(0)) = 0
Y(s) (s2 + 2s + 1) – (s + 4) = 0
Y(s) = (s + 4)/ (s2 + 2s + 1) = (s + 4)/(s + 1)2
y(t) = L-1{(s + 4)/(s + 1)2 } = L-1{(s + 4)/(s + 1)2 }
= L-1{(s + 1)/(s + 1)2 } + L -1{3/(s + 1)2 }
= e-t + 3te-t
Sample Final, Math 225
1
Fall 2005
A second method would be to guess the solution form y(t) = ert. This gives us
the characteristic equation: r2 + 2r + 1 = 0 or (r + 1)2 = 0, which has (repeated)
roots r = {-1,-1}.
Thus we have solutions y1 (x) = e-t and y2 (t) = te-t
yH(t) = Ae-t + Bte-t and y’H(t) = -Ae-t + B(e-t - te-t)
yH(0) = 1 implies A = 1
y’H(0) = 2 implies –A + B = 2, so B = 3
So yH(t) = -e-t + 3te-t
Answer: yH(t) = e-t + 3te-t
b) (10 pts) Compute the impulse response function of this equation.
In part (a) above, we saw that Y(s) (s2 + 2s + 1) – (s + 4) = 0 in the
homogeneous case, so more generally we have
Y(s) (s2 + 2s + 1) – (s + 4) = G(s), where G(s) is the transform of the driving
force g(t).
Thus Y(s) = (s + 4)/(s + 1)2 + G(s)/(s + 1)2
H(s), the Transfer function, is defined by H(s) = 1/(s + 1)2 , so
Y(s) = (s + 4)/(s + 1)2 + G(s)H(s)
The impulse response function is defined as h(t) = L-1{H(s)} = L-1{1/(s + 1)2 }
Thus, in this case h(t) = L-1{1/(s + 1)2 } = t e-t
Answer: h(t) = t e-t
c) (10 pts) Compute a particular solution for the original equation if the driving
force g(t) is a square pulse with g(t) = 1 for 0<t<1 and g(t) = 0 otherwise.
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We can express this driving function as g(t) = u(t) – u(t-1)
A particular solution can be computed as the convolution of the equation’s
impulse response function h(t) = t e-t with the given g(t).
t
yp (t) = h(t)*g(t) = ∫ 0 g(v)h(t − v)dv
This is most easily computed by considering the two separate cases, t>1 and t<1.
Case t > 1:
t
y p ( t ) = ∫ 0 g(v)h(t − v)dv
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1
1
= ∫ 0 h(t − v)dv = ∫ 0 (t − v)e−( t−v ) dv
1
1
1
1


= e−t t ∫ 0 e v dv − ∫ 0 ve v dv = e−t  t e v
− ve v − e v 
v= 0
v= 0 

= (e-1) t e-t – e-t
(
Case t < 1:
y p (t ) =
=
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∫
t
)
∫
t
0
t
∫ (t − v)e ( ) dv
(t ∫ e dv − ∫ ve dv) = e  t(e
t
0
-t
(
g(v)h(t − v)dv
h(t − v)dv =
0
= e−t
(
v
− t−v
0
t
v
−t
0
= 1 - t e – e-t
v
t
v= 0
(
− ve v − e v
t
v= 0



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Sample Final, Math 225
2
Fall 2005
Answer:
(e −1) te−t − e−t for t > 1
y p (t ) = 
−t
−t
 1− te − e for t < 1
3. (30 pts) Given the equation 9 sin(x) y” + 9y’ - y/x = 0 :
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a) (5 pts) Find all the singular points for the equation and categorize them as
regular or irregular.
Rewrite in normal form as y” + csc(x)y’ - csc(x)y/9x = 0
Thus P(x) = csc(x) and Q(x) = -csc(x)/9x
The function csc(x) = 1/sin(x) has poles of order 1 at each point x = kπ, for all
integers k (as sin(x) has zeros at these points and a Taylor expansion shows that
sin(x) is approximately (x-kπ) near each of these points.)
Thus P(x) has a pole of order 1 at each of these points.
Q(x) has a pole of order 2 at x=0 and order 1 at every other point x = kπ
As the poles of P(x) are all of order no more than 1 and the poles of Q(x) are of
order no more than 2, these singular points are all regular.
Answer: The singular points are x = kπ, for all integers k.
These singular points are all regular
b) (20 pts) Expanding about x = 0, find the terms of degree less than or equal to 5
for some (non-zero) solution.
As the equation has a singularity at x=0 we cannot use a conventional power
series, but as the singularity is regular we may use the method of Frobenius to
compute a series solution.
We can write the equation in normal form as y” + csc(x) y’ – csc(x) y/9x = 0
The indicial equation is r(r-1) + rp0 + q0 = 0
where p0 := lim xP(x) = lim (x/sin(x)) = 1
and q0 := lim x2 Q(x) = lim (x2 /(9 x sin(x))) = 1/9
Thus r2 – 1/9 = 0, with roots r = ±1/3
So we expect to find solutions of the form:
∞
y1/ 3 (x) = x
1/ 3
∞
∑a x = ∑a x
i
i
i= 0
∞
i+1/ 3
i
i= 0
∞
y−1/ 3 (x) = x −1/ 3 ∑ bi x i = ∑ bi x i−1/ 3
i= 0
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i= 0
As we are only asked for a single solution we choose to compute the terms of
y1/3(x). (Computing the terms of y-1/3(x) would be an equivalent computation,
and as the difference of the two roots of the indicial equation is not an integer,
the two form a complete set of solutions).
Recall that the expansion for sin(x) about x=0 is:
∞
sin(x) = ∑ (−1)
2k +1
3
x 2k +1 (2k + 1)! = x − x3! +
x5
5!
7
− x7! + ...
k= 0
So, plugging into the equation we get:
9 sin(x) y1/3” + 9y1/3’ - y1/3/x
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Sample Final, Math 225
3
Fall 2005
∞
(
= 9sin(x)∑ ai i +
i= 0
3

=  x − x3! +

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x5
5!
( ()
1
3
)(i − ) x
2
3
7

− x7! + ... a0

∞
i= 0
( ( )(− ) x
()
1
3
(
+ 9∑ ai i +
i− 5 / 3
2
3
−5 / 3
+ a1
1
3
)x
∞
i− 2 / 3
( )( ) x
4
3
1
3
− ∑ ai x i+1/ 3 x
i= 0
−2 / 3
+ a2
( )( ) x
7
3
4
3
1/ 3
)
+ ...
)
()
+9 a0 13 x −2 / 3 + a1 43 x1/ 3 + a2 73 x 4 / 3 + ... − ( a0 x −2 / 3 + a1 x1/ 3 + a2 x 4 / 3 + ...) = 0
We now collect terms of like degree in x:
x-2/3: 9a0 (1/3)(-2/3) x-2/3 + 9 a0 (1/3) x-2/3 – a0 x-2/3 = 0
-2 a0 + 3 a0 – 1 a0 = 0 (no information – forced by indicial equation)
x1/3: 9a1 (4/3)(1/3) x1/3 + 9 a1 (4/3) x1/3 – a1 x1/3 = 0
4 a1 + 12 a1 – 1 a1 = 0 (so a1 = 0)
x4/3: 9(xa2 (7/3)(4/3)x1/3 + (x3 /3!)a0 (1/3)(-2/3)x-5/3) + 9 a2 (7/3) x4/3 – a2 x4/3 = 0
28 a2 + (1/3) a0 + 21 a2 – 1 a2 = 0
(so a2 = -(1/144) a0 )
x7/3: 9(xa3 (10/3)(7/3)x4/3 - (x3 /3!)a1 (4/3)(1/3)x-2/3) + 9 a3 (10/3) x7/3 – a3 x7/3 = 0
70 a3 - (2/3) a1 + 30 a3 – 1 a3 = 0
(so a3 = (2/297) a1 = 0)
x10/3: 9(xa4 (13/3)(10/3)x7/3 - (x3 /3!)a2 (7/3)(4/3)x1/3 - (x5 /5!)a0 (1/3)(-2/3)x-5/3) +
9 a4 (13/3) x10/3 – a4 x10/3 = 0
130 a4 - (14/3) a2 - (1/60) a0 + 39 a4 – 1 a4 = 0
(so a4 = (1/10080)( a0 + 280 a2 ) = -(17/181440) a0 )
So y1 (x) = a0 x1/3 + a1 x4/3 + a2 x7/3 + a3 x10/3 + a4 x13/3 + …
= a0 x1/3 + (0)x4/3 + (-1/144)a0 x7/3 + (0)x10/3 + (-17/181440)a0 x13/3 + …
= a0 [x1/3 + (-1/144)x7/3 + (-17/181440) x13/3 + …]
Answer: y(t) = = a0 [x1/3 + (-1/144)x7/3 + (-17/181440) x13/3 + …]
(for any value of a0 . Another solution exists for r=-1/3.)
c) (5 pts) What can be said about the radius of convergence of the solution series?
As the points of singularity are x = kπ, the closest singularities to x=0 are
located at x = kπ and x = -kπ. Thus, the radius of convergence (while it could
be larger) is at least π.
Answer: The radius of convergence is at least π.
4. (15 pts) Given the equation y” + 2 y’ + 2 y = g(t)
a) (5 pts) Find the homogeneous solution.
We guess the solution form y(t) = ert. This gives us the characteristic equation:
−2 ± 4 − 8
r2 + 2r + 2 = 0. The roots of this polynomial are r =
={-1 ± i}
2
Thus we have solutions y1 (x) = e-t cos(t) and y1 (x) = e-t sin(t)
yH(t) = c1 e-t cos(t) + c2 e-t sin(t)
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Answer: yH(t) = c1 e-t cos(t) + c2 e-t sin(t)
Sample Final, Math 225
4
Fall 2005
b) (10 pts) Find the solution if g(t) = sin(2t)
As the equation is linear with constant coefficients and the driving function is a
sin function we can use the method of undetermined coefficients.
Assume that we have a solution of form A cos(2t) + B sin(2t), plug it into the
original equation and solve for A and B.
yp = A cos(2t) + B sin(2t)
yp = -2A sin(2t) + 2B cos(2t)
yp = -4A cos(2t) - 4B sin(2t)
Plugging in, we get:
(-4A cos(2t) - 4B sin(2t)) + 2 (-2A sin(2t) + 2B cos(2t)) + 2 (A cos(2t) + B
sin(2t)) = sin(2t)
cos(2x) (-2A + 4B) + sin(2t) (-4B – 2A - 1) = 0
As cos(2t) and sin(2t) are independent functions, we know that the coefficient of
sin(2t) and the coefficient of cos(2t) are both equal to zero:
-2A + 4B = 0
-4A – 2B = 1
So A = -1/5 and B = -1/10
yp = -cos(2t)/5 - sin(2t)/10
Answer: yp = -cos(2t)/5 - sin(2t)/10
c) (5 pts) How does the frequency response change for high frequency driving
functions (g(t) = sin(ωt), for large values of ω)?
The frequency response is given by the expression M(ω) = 1/√((2-ω2 )2 + 22 ω2 )
[NSS, Sect 4.9] which behaves as 1/ω2 for large values of ω.
Answer: The frequency response tends to zero as the frequency becomes
large.
5. (10 pts)
 dx + 3x + y = 0
dt
 derive a differential equation which
a) (5 pts) From the system  dy
 dt + 2y − x = 0
involves y(t), but not x(t).
Write the equations in operator form:
(D+3)[x] +€(1)[y] = 0
(-1)[x] + (D+2)[y] = 0
Now apply (D+3) to the second equation and add it to the first equation:
((D+3) + (D+3)(-1))[x] + ((1) + (D+3)(D+2))[y] = 0
(D2 + 5D + 6 + 1)[y] = 0
Answer:
d 2y
dt 2
+ 5 dy
+ 7y = 0
dt
b) (5 pts) A solution to the equation y” + (4 + sin(x))y = 0 has a zero at x = 0.
Which interval will the next zero for the solution be in?
• 0<x<1
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Sample Final, Math 225
5
Fall 2005
•
•
•
1≤x<2
2≤x<3
x≥3
Note that 3 < (4 + sin(x)) < 5. This observation allows us to compare this
equation to the equations y” + 3 y = 0 and y” + 5 y = 0.
The first equation has a solution with zeros at 0 and π/√3 = 1.814. The second
equation has a solution with zeros at 0 and π/√5 = 1.405.
Thus, using either the mass-spring analogy or Sturm’s Oscillation theorem, we
see that the original equation has a zero in the interval [π/√3, π/√5] = [1.814,
1.405].
Answer: 1 ≤ x < 2
6. (10 pts) (Always, Sometimes or Never)
a) (3 pts) Given a first order linear ODE y’ + a0 (x) y = g(x) with a regular singular
point at x=0 and a solution satisfying a given initial condition, y(1) = a. This
solution is unique.
Never: In general, a solution is not guaranteed to be unique across a singular
point, regular or not. If x0 is an ordinary point of the equation then the solution
is unique for some interval around x0 , but not necessarily globally. For example,
the equation y’ + (1/x)y = 0, with initial condition y(1) = 1 has the unique
solution y(x) = 1/x for x>0, but the solution y(x) = A/x for any value A, for x<0.
This non-uniqueness is caused by the equation’s singular point at x = 0.
b) (3 pts) The equation y” + cos(x) y’ + 5 y = x2 (cos(x))3 has solutions y1 (x)
and y2 (x). The expression a y1 (x) + b y2 (x) is also a solution for a and b both
non-zero.
Never: While solutions to a homogeneous linear equation can be added in
linear combinations to produce additional solutions (superposition principle),
this is not true for inhomogeneous equations. It is true that a solution to an
inhomogeneous equation can be added to a solution of the associated
homogeneous equation to get another solution to the inhomogeneous equation.
In this case, if we construct the combination y(x) = a y1 (x) + b y2 (x), then this
satisfies the equation y” + cos(x) y’ + 5 y = (a+b) x2 (cos(x))3 .
c) (3 pts) A solution to a Cauchy-Euler equation has a discontinuity at x=0.
Sometimes: If the solutions to the indicial equation are unique, the solutions xr
have a discontinuity at x=0 if (and only if) r < 0. If the solutions are repeated on
solution is of form ln(x) xr and will have a discontinuity at x=0.
Sample Final, Math 225
6
Fall 2005