MATH201: Problem set 1
solutions to the marked set
Published in week 4
Corrected 2014/10/14
x
R
1. (d ) Question exp − tan s ds
0
Solution
Z
tan s ds = − ln (|cos s|) + C
Zx
x
tan s ds =
− ln (|cos s|)
= − ln (|cos x|)
0
0


Zx
tan s ds = exp (ln (|cos x|)) = |cos|x
exp −
0

d
exp −
dx
Zx

tan s ds = − sin(x) sgn (|cos(x)|)
0
dy
− x2 = x2 y
2. (j ) Question x3
dx
Answer
Z
Z
dy
y+1
dy
dx
[i] We can separate the variables here
=
, hence
=
so that
dx
x
y+1
x
ln |y + 1| = ln |x| + C, and eventually y = Ax − 1 (which also covers the special solution of
y = −1 for A = 0).
[ii] This equation is not homogeneous as given but can be transformed into the homogeneous
type, as in the training set example 2(e), by introducing new variables for the numerator of
the right-hand side, Y = y + 1, and its denominator, X = x.
Working out in the same
dY
dy dX
dY
dY /dx
Y
way as in that example, we have
=
,
= 1,
=
=
which is of course
dx
dx dx
dX
dX/dx
X
dY
dV
dV
homogeneous. Putting Y = XV , we have
=V +X
= V , hence
= 0. The solution
dX
dX
dX
is straightforward, V = const = Y /X = (y + 1)/x = C and y = Cx − 1, as before. R
[iii] We can write dy/dx − y/x = 1/x, and the equation is linear, P (x) =Z−1/x, P (x) dx =
1
dx
−1
− ln |x| = ln |1/x|, and the integrating factor is I(x) = 1/x. Hence y =
=
+ A, or
2
x
x
x
y = Ax − 1, as before.
d2 y
dy
3. (m) Question
−5
+ 6y = 4xe4x , y(0) = 5, y 0 (0) = 11.
2
dx
dx
Solution
[i] Complementary Function: m2 − 5m + 6 = 0, m = {2, 3},
y = Ae2x + Be3x
.
[ii] Particular Integral: we try (α + βx)e4x .
This gives: (16α + 16βx + 8β)e4x −
4x
4x
4x
5 [4α + 4βx + β] e +6 [α + βx] e = 4xe . Cancelling through by e4x gives 2α+3β+2βx = 4x.
Equating the coefficients of x on both sides gives β = 2 and equating the constant terms gives
α = −3. So the particular integral is
(−3 + 2x)e4x
.
y = Ae2x + Be3x + (2x − 3)e4x
[iii] The General Solution is then
.
[iv] From the initial condidtions: y(0) = 5 = A + B − 3 and y 0 (0) = 11 = 2A + 3B − 10, we get
A = 3 and B = 5. The complete solution is then
y = 3e2x + 5e3x + (2x − 3)e4x
.
d2 y
dy
+ 25y = 10x cos(5x), y(0) = 0, y 0 (0) = 16/45.
(n) Question
−6
2
dx
dx
Solution
[i] Complementary Function: m2 − 6m + 25 = 0, m = 3 ± 4i,
y = e3x [A cos(4x) + B sin(4x)]
.
dy
= α cos(5x) + γ sin(5x) −
dx
5(αx + β) sin(5x) + 5(γx + δ) cos(5x) = (α + 5γx + 5δ) cos(5x) + (γ − 5αx − 5β) sin(5x) and
d2 y
= [10γ − 25αx − 25β] cos(5x) + [−25γx − 25δ − 10α] sin(5x). Substituting these into
dx2
the differential equation gives: [10γ − 25αx − 25β − 6α − 30γx − 30δ + 25αx + 25β] cos(5x)
+[−25γx − 25δ − 10α − 6γ + 30αx + 30β + 25γx + 25δ] sin(5x) = 10x cos(5x). We now equate
coefficients of x cos(5x) on both sides of the equation. We find −30γ = 10 so that γ = −1/3.
We now equate coefficients of x sin(5x) on both sides of the equation. We find 30α = 0. We now
equate coefficients of cos(5x) on both sides of the equation. We find 10γ−25β−6α−30δ+25β = 0
so that δ = −1/9. Finally, equating coefficients of sin(5x) on both sides, we find −25δ − 10α −
1
1
1
6γ + 30β + 25δ = 0, so that β = −1/15. So,
y = − cos(5x) + − x −
sin(5x) .
15
3
9
We can also use the complex method. Try y = (αx+β)e5ix , where here α and β are now complex.
Then y 0 (x) = (α + 5iαx + 5iβ)e5ix and y 00 = (10iα − 25αx − 25β)e5ix . Substituting these into the
differential equation gives: (10iα − 25αx − 25β)e5ix − 6 [α + 5iαx + 5iβ] e5ix + 25[αx + β]e5ix =
10xe5ix . Dividing through by e5ix and collecting terms together, we find −30iαx + (10i − 6)α −
30iβ = 10x. Therefore α = i/3 and β = (10i − 6)(i/3)/(30i) = (10i − 6)/90 = i/9 − 1/15.
The particular integral is then the real part of (ix/3 + i/9 − 1/15)(cos(5x) + i sin(5x)) which is
−(x/3 + 1/9) sin(5x) − (1/15) cos(5x) as before.
[iii] The general solution is therefore:
1
1
1
y = e3x (A cos(4x) + B sin(4x)) −
cos(5x) − x sin(5x) − sin(5x) .
15
3
9
[iv] From the initial conditions: y(0) = 0 = A − 1/15, we see that A = 1/15, and
y 0 (0) = 16/45 = 3A + 4B − 5/9. So that B = 8/45. The solution is therefore
1
8
1
1
1
3x
y=e
cos(4x) +
sin(4x) −
cos(5x) − x sin(5x) − sin(5x) .
15
45
15
3
9
[ii] Particular integral: y = (αx + β) cos(5x) + (γx + δ) sin(5x), so
2