Download Solutions to Homework 6, Introduction to Differential Equations

Solutions to Homework 6, Introduction to Differential Equations, 3450:335-003, Dr.
Montero, Spring 2009
Problem 1. Find a particular solution to
y ′′ + 6y ′ + 9y = −
25e−3t
.
2(1 + t2 )
Solution: First we find the solutions to the homogeneous equation
y ′′ + 6y ′ + 9y = 0.
The polynomial associated to this equation is
λ2 + 6λ + 9 = (λ + 3)2 .
The root is 3 with multiplicity 2. The solutions to the homogeneous problem are
y1 = e−3t
and
y2 = te−3t .
We will need the Wronskian for these two functions. We compute
y (t) y (t) e−3t
te−3t
1
2
=
W (t) = ′
y1 (t) y2′ (t) −3e−3t e−3t − 3te−3t
this to obtain
= e−6t .
We seek a particular solution of the form
yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t).
For this to work we need
u′1 (t) =
25t
25 2t
−te−3t × (−25)e−3t
=
=
.
2
−6t
2
2(1 + t )e
2(1 + t )
4 1 + t2
We integrate this to obtain
u1 (t) =
We also need
u′2 =
25
ln(1 + t2 ).
4
e−3t × 25e−3t
25 1
=
.
2
−6t
2(1 + t )e
2 1 + t2
We find that
u2 =
25
arctan(t).
2
The particular solution we seek is
yp (t) =
25
25
ln(1 + t2 )e−3t +
arctan(t)te−3t .
4
2
Problem 2. Find the solution of the equation
x2 y ′′ − 3xy ′ − 45y = 0
that also satisfies
y(1) = 10 and y ′ (1) = −3.
Solution: The polynomial associated to this equation is
m(m − 1) − 3m − 45 = m2 − 4m − 45 = (m − 9)(m + 5).
The general solution of this equation is
y(x) = α1 x9 + α2 x−5 .
Next we determine the value of the constants. We have
y(0) = α1 + α2 = 10,
and
y ′ (0) = 9α1 − 5α2 = −3.
We solve this system to find
14α1 = 47, so α1 =
47
,
14
14α2 = 93 so α2 =
93
.
14
and
The solution is
y(x) =
47 9 93 −5
x + x .
14
14
Problem 3. Find y if
x2 y ′′ − 3xy ′ + 4y = 0
and
y(1) = 6 and y ′(1) = 2.
Solution: We could solve this as usual, but we can also use the change of variables x = et .
We recall that this change of variables transforms the equation
α2 x2
d2 y
dy
+ α1 x + α0 y = 0
2
dx
dx
into the equation
dy
d2 y
α2 2 + (α1 − α2 ) + α0 y = 0.
dt
dt
This means that the equation
x2
d2 y
dy
− 3x + 4y = 0
2
dx
dx
transforms into
d2 y
dy
− 4 + 4y = 0.
2
dt
dt
The polynomial associated to this equation is
λ2 − 4λ + 4 = (λ − 2)2 .
The solutions of this equation are y1 = e2t and y2 = te2t .
Next we notice that the change of variables x = et transforms the initial data
y(1) = 6 and y ′(1) = 2,
written in terms of x, into the initial conditions
y(0) = 6 and y ′(0) = 2
which are conditions for t. The general solution of the equation, in the t variable, is
y(t) = α1 e2t + α2 te2t .
We use our initial data to obtain
y(0) = α1 = 6.
Next,
y ′ (0) = 2α1 + α2 = 2,
so
α2 = −10.
The solution is
y(t) = 6e2t − 10te2t .
Now we change back to the x variables. For this we use t = ln(x), which yields
y(x) = 6x2 − 10 ln(x)x2 .
Problem 4. Find the solution of
x2 y ′′ − 4xy ′ − 14y = x4
and
y(1) = 1, y ′ (1) = 1.
Solution: First we seek the roots of the polynomial
m(m − 1) − 4m − 14 = m2 − 5m − 14 = (m + 2)(m − 7).
The roots are m = −2 and m = 7. The solutions of the homogeneous problem are
y1 = x−2 and y2 = x7 .
Next, we seek a particular solution. We can use the method of variation of parameters, but
in this case we can at least try
yp (x) = αx4 .
Plugging this into the equation we obtain
x2 yp′′ − 4xyp′ − 14yp = (12 − 16 − 14)αx4 = x4 .
1
This means that α = − 18
. This says that the general solution is
y(x) = α1 x−2 + α2 x7 −
We then use the initial data:
y(1) = α1 + α2 −
1
=1
18
and
y ′(1) = −2α1 + 7α2 −
We solve this system to find
α1 =
The solution is
y(x) =
1 4
x.
18
2
= 1.
9
10
37
and α2 = .
54
27
1
37 −2 10 7
x + x − x4 .
54
27
18
Problem 5. Find the function y(x) if
x2 y ′′ − 9xy ′ + 25 = x6
and
y(1) = −9, y ′ (1) = 5.
Solution: The polynomial is
m(m − 1) − 9m + 25 = (m − 5)2 .
There is only one root, with multiplicity 2. The solutions of the homogeneous problem are
y1 (x) = x5 and y2 (x) = x5 ln(x).
Next, we seek a particular solution. First we compute the Wronskian:
y (x) y (x) x5
5
x
ln(x)
1
2
=
W (x) = ′
= x9 .
y1 (x) y2′ (x) 5x4 5x4 ln(x) + x4 Next, we seek a particular solution of the form
y(x) = u1 y1 + u2 y2 .
We compute
u′2 =
x5 x4
= 1, so u2 = x.
x9
Also
−x5 ln(x)x4
= − ln(x) so u1 = −(x ln(x) − x).
x9
The particular solution is
u′1 =
yp = −(x ln(x) − x)x5 + x × x5 = x6 .
Finally we impose the initial data. The general solution of our equation is
y(x) = α1 x5 + α2 x5 ln(x) + x6 .
From here we find that
y(1) = −9 = α1 + 1, so α1 = −10,
and
y ′(1) = 5 = 5α1 + α2 + 6, so α2 = 49.
The solution we seek is
y(x) = −10x5 + 49x5 ln(x) + x6 .
Problem 6. Find a particular solution of the equation
x2 y ′′ + 13xy ′ + 61y = x−6 .
Solution: The polynomial is
m(m − 1) + 13m + 61 = (m + 6)2 + 25.
The roots are m = −6 ± 5i, so the solutions of the homogeneous problem are
y1 (x) = x−6 cos(5 ln(x)) and y2 (x) = x−6 sin(5 ln(x)).
We need the Wronskian of these two functions:
x−6 cos(5 ln(x))
x−6 sin(5 ln(x))
W (x) = −6x−7 cos(5 ln(x)) − 5x−7 sin(5 ln(x)) −6x−7 sin(5 ln(x)) + 5x−7 cos(5 ln(x))
Next, we find
u′2 =
x−6 cos(5 ln(x))x−8
1 cos(5 ln(x))
sin(5 ln(x))
=
, so u2 (x) =
,
−13
5x
5
x
25
and
1 sin(5 ln(x))
cos(5 ln(x))
x−8 × x−6 sin(5 ln(x))
=−
, so u1 (x) =
.
−13
5x
5
x
25
The solution we seek is
x−6
.
yp = u1 y1 + u2 y2 =
25
Note that we could have also guessed that
u′1 = −
yp = αx−6 .
Plugging this into the equation leads to the exact same solution.
= 5x−13 .