MATH201: Problem set 1
solutions to the training set
Published in week 1
Last corrected 2014/10/15
1. (a) Question sin x2
Answer 2x cos x2
3x2 + 2x3
(b) Question
x+7
(x + 7)(6x + 6x2 ) − (3x2 + 2x3 )(1)
4x3 + 45x2 + 42x
Answer
=
(x + 7)2
(x + 7)2
s
Rx
(c) Question ex + e−s ds
0
−1/2
dp x
1 x
Solution
e + e−x ex + 1 − e−x
e + 1 − e−x =
dx
2
2. (a) Question
1 dy
1
− = 0.
y dx x
Solution
dy
y
=
which is separable, so dy/y = dx/x, ln(|y|) =
dx
x
C
1
ln(|x|) + C1 , y = ±e x = C2 x, where x ∈ (−∞, 0) or x ∈ (0, ∞) (zero x must be excluded as it
is in the denominator in the equation).
[ii] This equation is homogeneous as the right-hand depends only on the combination y/x (in
fact it just is that combination). So we do the standard substitution y = xv, which gives
dy/dx = xdv/dx + v, and then leads to equation xdv/dx = 0 hence v = C and y = Cx as before.
[iii] We can write dy/dx − y/x R= 0, and the equation is linear, with P (x) = −1/x. Hence the
integrating factor is I(x) = exp( P (x) dx) = exp(− ln(x)) = x−1 . Multiplying the equation by
1
y
that factor we get y 0 − 2 = 0, where the left-hand side is a full derivative, (y/x)0 = 0, hence
x
x
y/x = C and y = Cx as before.
dy
(b) Question
= 2 − 3y
dx
Solution
dy
= dx, so that
[i] This equation is separable.
To solve, write:
2
−
3y
Z
Z
dy
1
=
dx = − ln |2 − 3y| = x + A, and |2 − 3y| = eA e−3x . Thus 2 − 3y = Be−3x ,
2 − 3y
3
and y = 2 − Be−3x /3, x ∈ R, where B = ±eA . Note that the special solution happens to be
obtained by the same formula for B = 0.
[ii] This equation is not homogeneous. This is obvious; however if a proof was required, one can
argue that f (1, 1) = −1 and f (2, 2) = −4, hence f (tx, ty) 6= f (x, y) for x = 1, y = 1 and t = 2,
hence function f is not homogeneous of degree 0.
R
[iii] This equation is linear, dy/dx + 3y = 2. Here P (x) = 3 and P (x)dx = 3x so
d 3x dy
the integrating factor is I(x) = e3x . We have:
e y = e3x
+ 3e3x y = 2e3x so that
dx
dx
Z
2
2
ye3x = 2e3x = e3x + A, y = + Ae−3x , in agreement with (a).
3
3
[i] The standard (generic) form is
1
dy
(c) Question
= x3 − y 3
dx
Solution
[i] This equation is not separable: x3 − y 3 can not be factorized into a product of a function of
x and a function of y. This is quite obvious, but also can be proved. Let x1 = 0, x2 = 1, y1 = 0,
y2 , then f (x1 , y1 )f (x2 , y2 ) − f (x1 , y2 )f (x2 , y1 ) = 0 × 0 − (−1) × 1 6= 0 whereas for a separable
equation any such combination must be zero.
[ii] This equation is not homogeneous. For instance, f (2, 1) = 7, f (4, 2) = 60 so the equality
f (x, y) = f (tx, ty) which must be valid for the right-hand side of a homogeneous equation, is not
valid at least for x = 1, y = 1, t = 2.
[iii] This equation is not linear since it contains y 3 .
So this equation can not be solved analytically by any of the three methods.
dy
(d ) Question xy
= x2 + y 2
dx
Solution
[i] Write dy/dx = x/y + y/x. This equation is not separable. To prove that, we can choose say
x1 = 1, x2 = 2, y1 = 1, y2 = 3, and f (x1 , y1 )f (x2 , y2 ) − f (x1 , y2 )f (x2 , y1 ) = (1 + 1)(2/3 + 3/2) −
10 5
(3 + 1/3)(2 + 1/2) = 2 · 13
6 − 3 · 2 6= 0 whereas for a separable equation any such determinant
must be zero.
[ii] This equation is homogeneous. If we write y = xv, we find
dy
dv
1
=v+x
= +v
dx
dx
v
dv
1
= .
dx
v
p
This can be separated to give v 2 /2 = ln(|x|) + C/2 so that y = ±x 2 ln(|x|) + C. These are
so that
x
defined in the intervals x ∈ (e−C/2 , ∞) and x ∈ (−∞, −e−C/2 ).
[iii] This equation is not linear, since it contains the unknown in the denominator.
dy
3x + 4y + 5
(e) Question
=
dx
4x + 3y + 7
Solution
[i] This differential equation is not separable (proof as before).
[ii] This differential equation is not homogeneous as given (proof as before) but can be transformed into one, as follows. Write Y = 3x + 4y + 5 and X = 4x + 3y + 7 (these are the numerator
and the denominator of the right-hand side, of course), then
dY
dy
=3+4
dx
dx
so that
and
dX
dy
=4+3
dx
dx
dy
Y
=
dx
X
and
dY /dx
3 + 4Y /X
dY
=
=
.
dX
dX/dx
4 + 3Y /X
This is now a homogeneous differential equation which we can solve by putting Y /X = V . Then
dY
dV
3 + 4V
=V +X
=
dX
dX
4 + 3V
Thus
Z
(4 + 3V ) dV
1
=
2
1−V
2
Z so thatX
7
1
+
1−V
1+V
dV
3 − 3V 2
=
dX
4 + 3V
Z
dV = 3
dX
X
If we multiply through by 2, we get 6 ln(X) = ln(1 + V ) − 7 ln(1 − V ) + const = ln
so that X 6 = A
(1 + V )
, where ln A = const is the arbitrary constant.
(1 − V )7
2
A(1 + V )
(1 − V )7
This simplifys to
X 6 (1 − V )7 = A(1 + V ) or (X − Y )7 = A(X + Y ). Substituting for X and Y , we get
(x − y + 2)7 = A(7x + 7y + 12).
It is not possible to express this simply in the form y = a function of x. For the same reason,
the intervals of x values in which the solutions are defined is not possible to describe analytically
(but we can anticipate that these intervals will depend on the arbitrary constant A). As in the
classroom example, the special solutions can be obtained from the typical solution as limit cases
A → 0 and A → ∞.
[iii] This differential equation is not linear (the unknown in the denominator).
dy
(f ) Question y 3
+ xy 2 = x2 y 2
dx
Solution
[i] Write dy/dx = (x2 − x)/y. This equation is separable,
Z
Z
p
1 2 x3 x2
y dy = (x2 − x)dx, so that
y =
−
+ A, or y = ± 2x3 /3 − x2 + B.
2
3
2
The intervals where these are defined depend on B in a complicated way. Note that while solving
we divided by y, and the original equation has a special solution y = 0, x ∈ R.
[ii] This equation is not homogeneous (proof as before), and the right-hand side in the standard
form is not a ratio of two linear functions of x and y as in the previous example (the top is a
quadratic functions), so it cannot be made homogeneous by a similar transformation.
[iii] This equation is not linear, as in the standard form we have y in the denominator.
dy
(g ) Question x3
+ y 3 = xy
dx
Answer
[i] dy/dx = y(x − y 2 )/x3 . This can not be factorized into a function of x multiplied by a
function of y, so the equation is not separable (proof as before).
[ii] This equation is not homogeneous (proof as before).
[iii] This equation is not linear since we have y 3 even after we bring it to the standard form.
dy
(h) Question (x + 3y 2 )
+ y = −3x2
dx
Answer
[i] It is not possible to separate the variables (proof as before).
[ii] This equation is not homogeneous (proof as before).
dy
3x2 + y
[iii] The equation is not linear, since in the standard form
=−
we have y 2 in the
dx
x + 3y 2
denominator.
dy
(i ) Question x
+ y − 2x = 0.
dx
Solution
dy
y
2x − y
[i] The standard (generic) form is
=2− =
and the variables do not separate (proof
dx
x
x
as before).
dy
y
[ii] This equation is homogeneous, as is evident from the form
= 2 − . As before, put
dx
x
y = xv, dy/dx = xdv/dx + v which leads to a separable equation xdv/dx = 2(1 − v). Hence
dv/(1 − v) = 2dx/x, − ln(|v − 1| = 2 ln(|x|) + C1 , |v − 1|−1 = C2 |x|2 , (v − 1)−1 = C3 x2 ,
v = 1 + C4 x−2 . Substituting this into y = xv, we get ultimately y = x + C4 x−1 , where x ∈ (0, ∞)
or x ∈ (−∞, 0). The special solution corresponds to C4 = 0.
3
[iii] We can write dy/dx + y/x =
R 2, and the equation is linear, with P (x) = 1/x. Hence the
integrating factor is I(x) = exp P (x) dx = x, hence xy 0 + y = 2x, (xy)0 = 2x, xy = x2 + C
and y = x + C/x as before.
d2 y
dy
3. (a) Question
+6
+ 8y = 105e3x .
2
dx
dx
Solution
[i] To find the Complementary Solution, we substitute y = emx . This gives the characteristic
(a.k.a. auxiliary) equation m2 + 6m + 8 = 0 and the solutions m = −2 and m = −4. Therefore
the linearly independent solutions are e−2x and e−4x , and the Complementary Function is then
C1 e−2x + C2 e−4x
.
[ii] The index in the exponential in the right-hand side γ = 3 is not one of the roots of the
auxiliary equation, so we get the Particular Integral by trying y = Ae3x . Substitution into the
differential equation gives:
LHS = 9Ae3x + 18Ae3x + 8Ae3x = RHS = 105e3x .
Cancel through by e3x . Then 35A = 105 so A = 3. The particular integral is
[iii] The General Solution is therefore
y = 3e3x + C1 e−2x + C2 e−4x
3e3x
.
.
d2 y
dy
(b) Question
+8
+ 15y = −3e−4x , y(0) = 6, y 0 (0) = −25.
dx2
dx
Solution
[i] To find the Complementary Solution, we substitute y = emx . This gives the characteristic
equation m2 + 8m + 15 = 0. This quadratic equation has roots m = −3 and m = −5. Therefore the linearly independent solutions are e−3x and e−5x , and the Complementary Function is
y = Ae−3x + Be−5x
.
[ii] We get the Particular Integral by trying y = αe−4x . On substituting this into the differential
equation and cancelling by e−4x , we get 16α+8(−4α)+15α = −3 so that α = 3 and the Particular
Integral is
y = 3e−4x
.
[iii] The General Solution is therefore
y = Ae−3x + Be−5x + 3e−4x
.
[iv] From the initial conditions: y(0) = 6 = A + B + 3 and y 0 (0) = −25 = −3A − 5B − 12, so
that A + B = 3 and 3A + 5B = 13 so that A = 1 and B = 2. The solution of the initial value
y = e−3x + 2e−5x + 3e−4x
problem is therefore
.
d2 y
dy
(c) Question
+6
+ 8y = 16e−2x .
dx2
dx
Solution
[i] The left-hand side is the same as in Problem 3a, so the complementary function is the same,
C1 e−2x + C2 e−4x
.
[ii] The right-hand side is a quasipolynomial with index γ = −2 which is one of the roots of
the auxiliary equation. So the RHS R(x) is part of the complementary function. Consider: if we
4
try y = Ae−2x , we get zero on the left hand side! As discussed in the lecture, the Modification
Rule applies: multiply the trial solution by the independent variable (γ = −2 is a simple root
of the axiliary equation). So we try Axe−2x .
dy
= Ae−2x − 2xAe−2x ,
dx
d2 y
= 4Axe−2x − 4Ae−2x .
dx2
and
Substituting into the equation gives:
LHS = −4Ae−2x + 4Axe−2x + 6 Ae−2x − 2Axe−2x + 8Axe−2x = RHS = 16e−2x .
We see that the terms in xe−2x cancel out. Collecting the other terms and cancelling through
by e−2x gives 2A = 16 so that A = 8. The Particular Integral is therefore
[iii] The General Solution is
8xe−2x
.
(8x + C1 )e−2x + C2 e−4x
d2 y
dy
(d ) Question
+8
+ 12y = 8e−2x , y(0) = 5, y 0 (0) = −16.
2
dx
dx
Solution
[i] Substitute y = emx . Axiliary equation: m2 + 8m + 12 = 0. Two real roots: m = −2
and m = −6. The linearly independent solutions: e−2x and e−6x . Complementary Function:
y = Ae−2x + Be−6x
.
[ii] We notice that e−2x is in the Complementary Function. This means that the Modification
Rule applies and for the particular integral we have to try y = αxe−2x . On substituting this into
the differential equation, we get α(4x − 4)e−2x + 8 α(1 − 2x)e−2x + 12αxe−2x = 8e−2x . We see
that the xe−2x terms cancel leaving 4α = 8 so that α = 2. Hence we have the particular integral
as
2xe−2x
.
[iii] The General Solution is therefore
y = Ae−2x + Be−6x + 2xe−2x .
[iv] From the initial conditions: y(0) = 5 = A + B and y 0 (0) = −16 = −2A − 6B + 2, so that
y = 3e−2x + 2e−6x + 2xe−2x .
A = 3 and B = 2. The solution is therefore
d2 y
dy
(e) Question
−6
+ 9y = 10e3x .
dx2
dx
Solution
[i] The characteristic equation m2 −6m+9 = 0 has a double root m = 3. So the Complementary
Function is
(C1 + C2 x)e3x
.
[ii] The index in the exponential in the right-hand side γ = 3 is the double root of the characteristic equation, so the Modification Rule suggests suggests a particular integral in the form
y = Ax2 e3x . Substituting into the equation gives:
LHS = 9Ax2 e3x + 12Axe3x + 2Ae3x − 6 3Ax2 e3x + 2Axe3x + 9Ax2 e3x = RHS = 10e3x .
We can see that the x2 e3x terms and the xe3x terms cancel out. We are left with 2A = 10, so
A = 5, and the Particular Integral is
5x2 e3x
5
.
[iii] The General Solution is
(C1 + C2 x + 5x2 )e3x
.
d2 y
dy
(f ) Question
+6
+ 8y = 80 cos(2x)
2
dx
dx
Solution
[i] The left-hand side and therefore the Complementary Function are the same as in Problem 3a,
C1 e−2x + C2 e−4x
.
[ii] We try y = A cos(2x) + B sin(2x). This gives LHS = −4A cos(2x) − 4B sin(2x) +
6 (−2A sin(2x) + 2B cos(2x)) + 8A cos(2x) + 8B sin(2x) = RHS = 80 cos(2x). The terms multiplying cos(2x) on the left should equal the cos(2x) term on the right and the sin(2x) terms on
the left should add up to zero since there are no sin(2x) terms on the right. Thus
4A + 12B = 80
and
Thus A = 2 and B = 6, and the Particular Integral is
[iii] The General Solution is
4B − 12A = 0.
2 cos(2x) + 6 sin(2x)
2 cos(2x) + 6 sin(2x) + C1 e−2x + C2 e−4x
.
.
d2 y
dy
(g ) Question
+ 10
+ 25y = 50 cos(5x), y(0) = 5, y 0 (0) = 0.
2
dx
dx
Solution
[i] Auxiliary equation: m2 + 10m + 25 = 0. This quadratic equation has the double root m =
−5. Therefore the linearly independent solutions are e−5x and xe−5x , and the Complementary
function is
y = (A + Bx)e−5x
.
[ii] We get the particular integral by trying y = α cos(5x) + β sin(5x). On substituting this
into the differential equation, we get −25α cos(5x) − 25β sin(5x) + 10[−5α sin(5x) + 5β cos(5x)] +
25[α cos(5x) + β sin(5x)] = 50 cos(5x). Equating the coefficients of cos and sin on both sides of
the equation gives: −25α + 50β + 25α = 50 and −25β − 50α + 25β = 0. These equations have
the simple solution α = 0 and β = 1, and the Particular Integral
y = sin(5x)
.
Alternatively, we can use complexification. Noting that cos(5x) is the real part of e5ix , we can
try αe5ix and take the real part afterwards. Substituting this into the differential equation gives:
−25αe5ix + 50iαe5ix + 25αe5ix = 50e5ix .
Cancelling through by e5ix leads to 50iα = 50, so that α = 1/i = −i. The solution is therefore
−ie5ix = −i(cos(5x) + i sin(5x)). The real part of this is sin(5x), in agreement with the result
obtained by the other method.
[iii] The General Solution is therefore
y = (A + Bx)e−5x + sin(5x)
.
[iv] From the initial conditions: y(0) = 5 = A and y 0 (0) = 0 = −5A + B + 5, so that A = 5, and
−25 + B + 5 = 0 so that B = 20. The solution is therefore
y = (5 + 20x)e−5x + sin(5x)
d2 y
dy
(h) Question
+4
+ 13y = 136 cos(5x), y(0) = 0, y 0 (0) = 25.
2
dx
dx
Solution
6
.
[i] The auxiliary equation is m2 + 4m + 13 = 0. This has the complex roots m = −2 ± 3i. The
real form of the solution is then y = yh = e−2x (C1 cos(3x) + C2 sin(3x)).
[ii] We now look for the Particular Integral. This could be done by trying a solution x =
a cos(5t) + b sin(5t) — please do it yourself and compare the answer. For now let us exercise the
complexification method. That is we set y = Rez where
d2 z
dz
+4
+ 13z = 136e5ix .
dx2
dx
We try a solution z = Ae5ix . Substituting this into the differential equation we get:
−52 Ae5ix = 4 · 5ie5ix + 13Ae5ix = 136e5ix
The complex exponential cancels and we are left with
(−12 + 20i)A = 136
so
A=
136
34
34(−3 − 5i)
= −3 − 5i
=
=
−12 + 20i
−3 + 5i
32 + 52
and z = (−3 − 5i)e5ix . Hence
y = Rez = Re(−3 − 5i)(cos(5x) + i sin(5x)) =
− 3 cos(5x) + 5 sin(5x)
.
[iii] The complete General Solution is then
e−2x (C1 cos(3x) + C2 sin(3x)) − 3 cos(5x) + 5 sin(5x).
[iv] We now use the initial conditions to determine C1 and C2 . First, y(0) = C1 − 3 = 0.
Therefore C1 = 3. Second, y 0 (0) = 3C2 − 2C1 + 25 = 25 and so C2 = 2. So the solution of the
initial value problem is
e−2x (3 cos(3x) + 2 sin(3x)) − 3 cos(5x) + 5 sin(5x).
d2 y
dy
(i ) Question
+6
+ 8y = 16x2 + 48x − 34.
2
dx
dx
Solution
[i] The left-hand side and therefore the Complementary Function are the same as in Problem 3a,
C1 e−2x + C2 e−4x
.
[ii] The right-hand side is a polynomial, which corresponds to the index of the exponential
γ = 0, and it is not a root of the characteristic equation so there is no resonance. We try a
solution y = Ax2 + Bx + C. Substituting this we get
2A + 6 [2Ax + B] + 8 Ax2 + Bx + C = 16x2 + 48x − 34.
Equating powers of x2 on both sides of the equation gives 8A = 16 so that A = 2. Equating the
coefficients of x on both sides gives 8B + 12A = 48 so that B = 3. Equating the constant terms
2A + 6B + 8C = −34. Thus C = −7. The particular integral is
7
2x2 + 3x − 7
.
[iii] The General Solution is
2x2 + 3x − 7 + C1 e−2x + C2 e−4x
.
d2 y
dy
(j ) Question
+8
+ 25y = 125x2 + 5x − 14, y(0) = 5, y 0 (0) = −11.
2
dx
dx
Solution
[i] Complementary Function: m2 + 8m + 25 = 0, m = −4 ± 3i, and
y = e−4x [A cos(3x) + B sin(3x)]
.
[ii] Again, the right-hand side is a polynomial and γ = 0 is not a root of the characteristic
equation. For the particular integral, we try y = αx2 + βx + γ. Substitution gives 2α + 8[2αx +
β] + 25[αx2 + βx + γ] = 125x2 + 5x − 14. Equating the coefficients of x2 on both sides gives
25α = 125 so that α = 5. Equating the coefficients of x on both sides gives 16α + 25β = 5 so
that β = −3. Equating the constant terms on both sides gives 2α + 8β + 25γ = −14, so that
γ = 0. Hence the particular integral is
[iii] The general solution is then
y = 5x2 − 3x
.
y = e−4x [A cos(3x) + B sin(3x)] + 5x2 − 3x
.
[iv] From the initial conditions: y(0) = 5 = A, so y 0 (x) = −4e−4x [A cos(3x) + B sin(3x)] +
e−4x [−3A sin(3x) + 3B cos(3x)] + 10x − 3, and y 0 (0) = −11 = −4A + 3B − 3 so that B = 4. The
solution is therefore
y = e−4x [5 cos(3x) + 4 sin(3x)] + 5x2 − 3x
.
dy
d2 y
+6
+ 8y = (129 + 70x)e3x .
(k ) Question
2
dx
dx
Answer
y = (2x + 3)e3x + C1 e−2x + C2 e−4x
d2 y
dy
(l ) Question
+6
+ 8y = e3x sin(2x).
2
dx
dx
Answer
1 3x
y=
e (31 sin(2x) − 24 cos(2x)) + C1 e−2x + C2 e−4x
1537
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