Math2065: Intro To Pdes Tutorial Solutions (Week 2)

MATH2065: INTRO TO PDEs
Semester 2, 2009
Tutorial Solutions (Week 2)
1. (a) Try yp = Aex , where A is a constant to be determined. Differentiating this form gives
yp0 = Aex and yp00 = Aex .
Substituting into the ODE we find A = − 13 . A particular solution is therefore
1
yp = − ex .
3
(b) Try yp = A x + B. Then, y 00 = 0, and by substitution, we get
−9(A x + B) = x + 18 ⇒ − 9A = 1 and − 9B = 18.
Thus, A = −1/9 and B = −2, yielding the particular solution

1
yp (x) = − x − 2.
9
(c) Set yp (x) = A cos x+B sin x. Then, yp0 = −A sin x+B cos x and yp00 = −A cos x−B sin x.
Upon substitution into the ODE we get
(−A cos x − B sin x) − (−A sin x + B cos x) + 2 (A cos x + B sin x) = −2 sin x.
Equating coefficients of cos x and sin x, we get the simultaneous equations A − B = 0

and A + B = −2. Solving, A = B = −1, and hence a particular solution is
yp (x) = − cos x − sin x .
(d) Try yp = Ax2 + Bx + C, where A, B and C are constants (note that we must guess
the most general form of a second-order polynomial). Differentiating this form gives
yp0 = 2Ax + B and yp00 = 2A .
Substituting into the ODE and equating coefficients of powers of x gives A = 1, B = 0

and C = −2. A particular solution is therefore
yp (x) = x2 − 2.
(e) Note that e3x is a solution of the homogeneous ODE. So substituting Ae3x as a par-
ticular solution to the inhomogeneous equation will not work, since the left-hand side
will produce a zero. Therefore, we try yp = Axe3x , where A is a constant, using the
modification rule of multiplying the initial guess by a factor of x. Differentiating this
form gives
yp0 = Ae3x (1 + 3x) and yp00 = Ae3x (6 + 9x) .
Substituting into the ODE gives A = 31 . A particular solution is therefore
1
yp = xe3x .
3
2. The ODEs are all second order, linear, inhomogeneous, with constant coefficients.
(a) First find yh . The characteristic equation is λ2 + λ − 6 = 0 ⇒ λ = 2, −3. The general
solution of the homogeneous equation is therefore
yh = C1 e2x + C2 e−3x .
For yp , the form of R(x) suggests we try yp = Aex , where A is a constant. Differenti-
ating this form gives
yp0 = Aex and yp00 = Aex .
Substituting into the ODE gives A = − 45 . The general solution of the inhomogeneous
ODE is therefore
5
y = yh + yp = C1 e2x + C2 e−3x − ex .
4
Now, using the initial conditions, y(0) = 0 ⇒ C1 + C2 − 45 = 0 and y 0 (0) = 2 ⇒
2C1 − 3C2 − 54 = 2. Hence C1 = 57 , C2 = − 20 3
. Thus the particular solution which
satisfies the initial conditions is
7 3 5
y(x) = e2x − e−3x − ex .
5 20 4
(b) First find yh . The characteristic equation is λ2 + 3λ + 2 = 0 ⇒ λ = −2, −1. The
general solution of the homogeneous equation is therefore
yh = C1 e−2x + C2 e−x .
For yp , the form of R(x) suggests we try yp = (Ax + B)e4x , where A and B are
constants. Differentiating this form gives
yp0 = Ae4x + 4(Ax + B)e4x and yp00 = 8Ae4x + 16(Ax + B)e4x .
1 11
Substituting into the ODE gives A = 30
, B = − 900 . The general solution of the
inhomogeneous ODE is therefore
1 11 4x
y = yh + yp = C1 e−2x + C2 e−x + ( x− )e .
30 900
Now, using the initial conditions, y(0) = 0, y 0 (0) = 1 ⇒ C1 = − 36 37
and C2 = 26
25
.
Thus the particular solution which satisfies the initial conditions is
37 −2x 26 −x 1 11 4x
y(x) = − e + e +( x− )e .
36 25 30 900
(c) First find yh . The characteristic equation is λ2 − 4λ + 3 = 0 ⇒ λ = 1, 3. The general
solution of the homogeneous equation is therefore
yh = C1 e3x + C2 ex .
For yp the form of R(x) suggests we try yp = A sin x + B cos x, where A and B are
constants. Differentiating and the substituting into the ODE gives A = 15 , B = 25 . The
general solution of the inhomogeneous ODE is therefore
1 2
y = yh + yp = C1 e3x + C2 ex + sin x + cos x .
5 5
Now, using the initial conditions, y(0) = 1, y 0 (0) = 0 ⇒ C1 = − 52 and C2 = 1. Thus
the particular solution which satisfies the initial conditions is
2 1 2
y(x) = − e3x + ex + sin x + cos x .
5 5 5
3. We know y100 + ay10 + by1 = f1 , and also y200 + ay20 + by2 = f2 . Simply adding the two gives
(y100 + y200 ) + a(y10 + y20 ) + b(y1 + y2 ) = f1 + f2 ,
thus demonstrating that y1 + y2 satisfies the given ODE.
(a) Let f1 = x, and look for a corresponding particular solution y1 of the form y1 = Ax+B,
where A and B are constants. Differentiating and substituting, we find A = 31 , B =
− 10
9
.
Similarly for f2 = cos x, try y2 = C sin x + D cos x, where C and D are constants.
1
Differentiating and substituting, we find C = 10 , D = 0. Hence
1 10 1
yp = x − + sin x .
3 9 10
(b) Let f1 = 2ex , and try y1 = Aex , where A is a constant. Differentiating and substituting,
we find A = 16 .
For f2 = sin x, try y2 = B sin x + C cos x, where B and C are constants. Differentiating
1 1
and substituting, we find B = 10 , C = − 10 . Hence
1 1 1
yp = ex + sin x − cos x .
6 10 10
(c) Let f1 = ex . Since ex is a solution of the homogeneous equation, try y1 = Axex , where
A is a constant. Differentiating and substituting gives A = − 21 .
Similarly for f2 = e2x , try y2 = Ce2x , where C is a constant. Differentiating and
substituting gives C = −1. Hence
1
yp = − xex − e2x .
2
4. The governing differential equation is
ẍ + 4 x = cos ωt .
(a) The homogeneous solution xh obeys ẍ + 4x = 0. This has characteristic equation

λ2 + 4 = 0, and hence λ = ±2i. The corresponding solutions are therefore cos 2t and
sin 2t, leading to
xh (t) = C1 cos (2t) + C2 sin (2t) ,
where C1 and C2 are arbitrary constants. Note that this is an oscillatory function – it
keeps on periodically wiggling.
(b) To find xp (t), we use undetermined coefficients. Based on the form of the inhomogene-
ity, the correct guess is
xp (t) = A cos (ωt) + B sin (ωt) . (1)
However, we note that there is a possible problem: if ω = 2, this is actually part

of the homogeneous solution, and therefore will not work (upon substitution into the
ODE, the left-hand side will yield zero, giving us no information on the undetermined
coefficients A and B). Thus, the above guess is valid only if ω 6= 2. Under this
condition, substitution gives
−Aω 2 cos ωt − Bω 2 sin ωt + 4 (A cos ωt + B sin ωt) = cos ωt .

Equating coefficients of sin ωt, we get B = 0. Equating coefficients of cos ωt, we find
that A = 1/(4 − ω 2 ), and so
1
xp (t) = cos (ωt) (ω 6= 2) .
4 − ω2
If ω = 2, the above does not work. We might have guessed (1) with ω = 2, but since
this is a part of the homogeneous solution, we need to guess t times that; i.e.,
xp (t) = A t cos (2t) + B t sin (2t) .
Substituting into ẍ + 4x = cos 2t (since ω = 2 now), we get, after some algebra (terms
with multiplicative factors of t thankfully cancel, as they must),
−4 A sin (2t) + 4 B cos (2t) = cos (2t) .
Equating coefficients as before, we get A = 0 and B = 1/4. Thus, if ω = 2, then

t
xp (t) = sin (2t) (ω = 2) .
4
(c) Since x(t) = xh (t) + xp (t), we have


1
 C1 cos (2t) + C2 sin (2t) +
 cos (ωt) (if ω 6= 2)
x(t) = 4 − ω2
 C cos (2t) + C sin (2t) + t sin (2t)

(if ω = 2)
1 2
4
If ω 6= 2, the solution is oscillatory, and so the block just keeps on periodically repeat-
ing its motion (it goes back and forth indefinitely – this is reasonable in that we have
set γ = 0 here, and hence there is no friction causing it to slow down). If ω = 2, on
the other hand, the term t sin 2t appears in the solution. This grows (since t does),
while oscillating as well. In fact, it grows without bound, oscillating between the lines
x = ± t. The block will therefore start oscillating wildly.
This phenomenon is called resonance. If the block is excited (forced) at a frequency

which is equal to its natural frequency (the frequency at which it would oscillate if left
to its own devices), the oscillations grow wilder. In this case, the resonance frequency
is 2.

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MATH2065: INTRO TO PDEs

yp0 = Aex and yp00 = Aex .

Substituting into the ODE we find A = − 13 . A particular solution is therefore

−9(A x + B) = x + 18 ⇒ − 9A = 1 and − 9B = 18.

Thus, A = −1/9 and B = −2, yielding the particular solution

(−A cos x − B sin x) − (−A sin x + B cos x) + 2 (A cos x + B sin x) = −2 sin x.

Equating coefficients of cos x and sin x, we get the simultaneous equations A − B = 0

yp (x) = − cos x − sin x .

yp0 = 2Ax + B and yp00 = 2A .

Substituting into the ODE and equating coefficients of powers of x gives A = 1, B = 0

(y100 + y200 ) + a(y10 + y20 ) + b(y1 + y2 ) = f1 + f2 ,

thus demonstrating that y1 + y2 satisfies the given ODE.

4. The governing differential equation is

(a) The homogeneous solution xh obeys ẍ + 4x = 0. This has characteristic equation

xp (t) = A cos (ωt) + B sin (ωt) . (1)

However, we note that there is a possible problem: if ω = 2, this is actually part

−Aω 2 cos ωt − Bω 2 sin ωt + 4 (A cos ωt + B sin ωt) = cos ωt .

xp (t) = A t cos (2t) + B t sin (2t) .

−4 A sin (2t) + 4 B cos (2t) = cos (2t) .

Equating coefficients as before, we get A = 0 and B = 1/4. Thus, if ω = 2, then

(c) Since x(t) = xh (t) + xp (t), we have

This phenomenon is called resonance. If the block is excited (forced) at a frequency

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