Math 231.04, Spring 2010: Practice Exam 3 Solutions
Math 231.04
1
Practice Exam 3 Solutions
NAME:
#1.)
/20
#2.)
/20
#3.)
/10
#4.)
#5.)
/20
#6.)
/10
#7.)
/10
Total:
/10
/100
Instructions: There are 5 pages, each containing one problem worth 20 points, for a total of 100
points on the exam. You must show all necessary work to get credit. You may not use your book,
notes, or calculator. Partial credit will only be given for progress toward a correct solution.
1.) (20 points) Solve
𝑦 ′′ − 4𝑦 ′ + 5𝑦 = 0,
𝑦(0) = 3,
𝑦 ′ (0) = 4.
We guess a solution of the form 𝑦 = 𝑒𝑟𝑡 , which leads to the characteristic equation
𝑟2 − 4𝑟 + 5 = 0.
We cannot factor this quadratic, so using the quadratic formula we get
√ √
√
√
4 ± 16 − 4 ⋅ 5
4 ± −4
4 ± 4 −1
4 ± 2𝑖
=
=
=
= 2 ± 𝑖.
𝑟=
2
2
2
2
This implies that the general solution is
𝑦 = 𝐶1 𝑒2𝑡 cos 𝑡 + 𝐶2 𝑒2𝑡 sin 𝑡.
Then the condition 𝑦(0) = 3 yields
3 = 𝐶1 𝑒0 cos 0 + 𝐶2 𝑒0 sin 0 = 𝐶1 + 0 = 𝐶1 .
Hence
𝑦 = 3𝑒2𝑡 cos 𝑡 + 𝐶2 𝑒2𝑡 sin 𝑡.
Next, we calculate 𝑦 ′ using the product rule:
𝑦 ′ = 6𝑒2𝑡 cos 𝑡 − 3𝑒2𝑡 sin 𝑡 + 𝐶2 2𝑒2𝑡 sin 𝑡 + 𝐶2 𝑒2𝑡 cos 𝑡.
Hence the condition 𝑦 ′ (0) = 4 yields
4 = 6 − 0 + 0 + 𝐶2 .
Hence 𝐶2 = 4 − 6 = −2. So finally
𝑦 = 3𝑒2𝑡 cos 𝑡 − 2𝑒2𝑡 sin 𝑡.
Math 231.04, Spring 2010: Practice Exam 3 Solutions
2
2.) (20 points) Assuming the fact that the function 𝑦1 = 𝑒𝑡 is a solution of the equation
(
)
(
)
2
2
′′
′
(★)
𝑦 − 2+
𝑦 + 1+
𝑦 = 0,
𝑡
𝑡
find the general solution of (★).
To get the general solution, we need two linearly independent solutions. Since we have one
solution, we use the method of reduction of order. We guess a second solution of the form 𝑦2 = 𝑣𝑦1 ,
where 𝑣 is a function of 𝑡. Here 𝑦2 = 𝑣𝑒𝑡 = 𝑒𝑡 𝑣; hence by the product rule, 𝑦2 ′ = 𝑒𝑡 𝑣 + 𝑒𝑡 𝑣 ′ and
𝑦2 ′′ = 𝑒𝑡 𝑣 + 𝑒𝑡 𝑣 ′ + 𝑒𝑡 𝑣 ′ + 𝑒𝑡 𝑣 ′′ = 𝑒𝑡 𝑣 + 2𝑒𝑡 𝑣 ′ + 𝑒𝑡 𝑣 ′′ . Substituting this in the equation, we get
)
(
)
(
2
2
′
′′
𝑦2 + 1 +
𝑦2
𝑦2 − 2 +
𝑡
𝑡
(
)
(
)
2
2 𝑡
𝑡
𝑡 ′
𝑡 ′′
𝑡
𝑡 ′
= 𝑒 𝑣 + 2𝑒 𝑣 + 𝑒 𝑣 − 2 +
(𝑒 𝑣 + 𝑒 𝑣 ) + 1 +
𝑒𝑣
𝑡
𝑡
2
2
2
= 𝑒𝑡 𝑣 + 2𝑒𝑡 𝑣 ′ + 𝑒𝑡 𝑣 ′′ − 2𝑒𝑡 𝑣 − 2𝑒𝑡 𝑣 ′ − 𝑒𝑡 𝑣 − 𝑒𝑡 𝑣 ′ + 𝑒𝑡 𝑣 + 𝑒𝑡 𝑣
𝑡
𝑡
𝑡
(
)
(
)
2
2
2
= 𝑒𝑡 𝑣 ′′ + 2𝑒𝑡 − 2𝑒𝑡 − 𝑒𝑡 𝑣 ′ + 𝑒𝑡 − 2𝑒𝑡 − 𝑒𝑡 + 𝑒𝑡 + 𝑒𝑡 𝑣
𝑡
𝑡
𝑡
2
= 𝑒𝑡 𝑣 ′′ − 𝑒𝑡 𝑣 ′ .
𝑡
𝑡 ′′
Setting this equal to 0 and we get the equation 𝑒 𝑣 − 2𝑡 𝑒𝑡 𝑣 ′ = 0. Dividing through by 𝑒𝑡 we get
2
𝑣 ′′ − 𝑣 ′ = 0.
𝑡
We make the substitution 𝑢 = 𝑣 ′ . This gives the equation 𝑢 ′ − 2𝑡 𝑢 = 0, which is a first order linear
equation which we can solve by the method of the integrating factor. The integrating factor is
∫
𝑒
− 2𝑡 𝑑𝑡
−2 )
= 𝑒−2 ln 𝑡 = 𝑒ln(𝑡
= 𝑡−2 =
1
.
𝑡2
Multiplying through the equation 𝑢 ′ − 2𝑡 𝑢 = 0 by the integrating factor
1
𝑡2
gives
1 ′ 2
𝑢 − 3 𝑢 = 0.
𝑡2
𝑡
( )
( )
The left side is 𝑡12 𝑢 ′ , by the product rule.
So∫ we have 𝑡12 𝑢 ′ = 0, which implies
∫
𝑢 = 𝐶1 𝑡2 . Then since 𝑢 = 𝑣 ′ , we get 𝑣 = 𝑢 = 𝐶1 𝑡2 𝑑𝑡 = 𝐶1 𝑡3 /3 + 𝐶2 . Then
𝑦2 = 𝑣𝑒𝑡 = (𝐶1
1
𝑢
𝑡2
= 𝐶1 , or
𝑡3
𝐶1 3 𝑡
+ 𝐶2 )𝑒𝑡 =
𝑡 𝑒 + 𝐶2 𝑒 𝑡 .
3
3
Since we are just looking for one 𝑦2 , we can take 𝐶1 = 3 and 𝐶2 = 0 to obtain 𝑦2 = 𝑡3 𝑒𝑡 . Then the
general solution is 𝑦 = 𝐶1 𝑦1 + 𝐶2 𝑦2 , or
𝑦 = 𝐶1 𝑒𝑡 + 𝐶2 𝑡3 𝑒𝑡 ,
for arbitrary constants 𝐶1 and 𝐶2 .
Math 231.04, Spring 2010: Practice Exam 3 Solutions
3
3.) (10 points) For the equations below, write down the general form of the particular solution 𝑦𝑝
that you would use in the method of undetermined coefficients. Just write down the form; do not
attempt to solve for the coefficients.
a.) (5 points) 𝑦 ′′ − 9𝑦 = 𝑡2 sin 4𝑡.
We first find 𝑦ℎ , the solution to the homogeneous equation 𝑦 ′′ − 9𝑦 = 0. For this we guess
𝑦ℎ = 𝑒𝑟𝑡 , which leads to the characteristic equation 𝑟2 − 9 = 0. Factoring, we get (𝑟 + 3)(𝑟 − 3) = 0,
so 𝑟1 = −3 and 𝑟2 = +3 are the roots of the characteristic equation. Thus the general form of 𝑦ℎ is
𝑦ℎ = 𝐶1 𝑒−3𝑡 + 𝐶2 𝑒3𝑡 .
Now we look for a particular solution 𝑦𝑝 . The right side of the equation is 𝑡2 sin 4𝑡 we guess
𝑦𝑝 = (𝐴𝑡2 +𝐵𝑡+𝐶) sin 4𝑡+(𝐷𝑡2 +𝐸𝑡+𝐹 ) cos 4𝑡. Since none of these terms, after you multiply them
out, has any term in which is a constant multiple of any term in 𝑦ℎ , this form of 𝑦𝑝 is sufficient.
Hence
𝑦𝑝 = (𝐴𝑡2 + 𝐵𝑡 + 𝐶) sin 4𝑡 + (𝐷𝑡2 + 𝐸𝑡 + 𝐹 ) cos 4𝑡.
b.) (5 points) 𝑦 ′′ − 9𝑦 = 𝑡𝑒3𝑡 .
The left side of the equation is the same as for part a., so we have the same homogeneous
solution 𝑦ℎ = 𝐶1 𝑒−3𝑡 + 𝐶2 𝑒3𝑡 . Now we look for a particular solution 𝑦𝑝 . The right side of the
equation is of the form 𝑡𝑒3𝑡 , we try 𝑦𝑝 = (𝐴𝑡 + 𝐵)𝑒3𝑡 = 𝐴𝑡𝑒3𝑡 + 𝐵𝑒3𝑡 . However, the term 𝐴𝑒3𝑡 is a
constant multiple of the term 𝐶2 𝑒3𝑡 in 𝑦ℎ , so we need to modify 𝑦𝑝 by multiplying by 𝑡 to obtain
𝑦𝑝 = 𝑡(𝐴𝑡 + 𝐵)𝑒3𝑡 = 𝐴𝑡2 𝑒3𝑡 + 𝐵𝑡𝑒3𝑡 . This no longer has any overlap with 𝑦ℎ , so this form of 𝑦𝑝 is
sufficient. So
𝑦𝑝 = 𝐴𝑡2 𝑒3𝑡 + 𝐵𝑡𝑒3𝑡 .
4.) (10 points) Solve the Cauchy-Euler equation 𝑡2 𝑦 ′′ + 7𝑡𝑦 ′ + 9𝑦 = 0, for 𝑡 > 0.
Because this is a Cauchy-Euler equation, we try 𝑦 = 𝑡𝑟 as a solution. Then 𝑦 ′ = 𝑟𝑡𝑟−1 and
𝑦 ′′ = 𝑟(𝑟 − 1)𝑡𝑟−2 . Substituting into the equation gives
𝑡2 𝑟(𝑟 − 1)𝑡𝑟−2 + 7𝑡𝑟𝑡𝑟−1 + 9𝑡𝑟 = 0,
or
𝑟(𝑟 − 1)𝑡𝑟 + 7𝑟𝑡𝑟 + 9𝑡𝑟 = 0.
Cancelling 𝑡𝑟 gives 𝑟(𝑟 − 1) + 7𝑟 + 9 = 0, or
0 = 𝑟2 + 6𝑟 + 9 = (𝑟 + 3)2 .
Hence 𝑟 = −3 is a double root, and we get solutions 𝑦1 = 𝑡−3 and 𝑦2 = 𝑡−3 ln 𝑡. The general solution
𝑦 is a linear combination of 𝑦1 and 𝑦2 ; that is,
𝑦 = 𝐶1 𝑡−3 + 𝐶2 𝑡−3 ln 𝑡,
for arbitrary constants 𝐶1 and 𝐶2 .
Math 231.04, Spring 2010: Practice Exam 3 Solutions
4
5.) (a) (5 points) Check that the functions 𝑦1 = 𝑡 and 𝑦2 = 𝑡3 are solutions of
3
3
𝑦 ′′ − 𝑦 ′ + 2 𝑦 = 0,
𝑡
𝑡
𝑡 > 0.
Since 𝑦1 = 𝑡, we get 𝑦1 ′ = 1 and 𝑦1 ′′ = 0. Therefore
3
3
3
3 3
3
𝑦1 ′′ − 𝑦1 ′ + 2 𝑦1 = 0 − + 2 ⋅ 𝑡 = − + = 0,
𝑡
𝑡
𝑡 𝑡
𝑡
𝑡
hence 𝑦1 is a solution.
For 𝑦2 = 𝑡3 we get 𝑦2 ′ = 3𝑡2 and 𝑦2 ′′ = 6𝑡. Hence
3
3
3
3
𝑦2 ′′ − 𝑦2 ′ + 2 𝑦2 = 6𝑡 − 3𝑡2 + 2 𝑡3 = 6𝑡 − 9𝑡 + 3𝑡 = 0,
𝑡
𝑡
𝑡
𝑡
so 𝑦2 is a solution.
(b) (15 points) Find the general solution of
3
3
𝑦 ′′ − 𝑦 ′ + 2 𝑦 = 𝑡2 ,
𝑡
𝑡
𝑡 > 0.
To find the general solution, we need to find a particular solution 𝑦𝑝 . Using the method of
variation of parameters, we guess a solution of the form 𝑦𝑝 = 𝑣1 𝑦1 +𝑣2 𝑦2 . The equations determining
𝑣1′ and 𝑣2′ are 𝑦1 𝑣1′ + 𝑦2 𝑣2′ = 0 and 𝑦1′ 𝑣1′ + 𝑦2′ 𝑣2′ = 𝑔(𝑡), where 𝑔(𝑡) is the right side of the equation.
Using our values of 𝑦1 , 𝑦2 and 𝑔, we get
𝑡𝑣1′ + 𝑡3 𝑣2′ = 0
1 ⋅ 𝑣1′ + 3𝑡2 𝑣2′ = 𝑡2 .
To solve these equations simultaneously for 𝑣1′ and 𝑣2′ , it is easiest to first cancel a factor of 𝑡 from the
first equation to get 𝑣1′ + 𝑡2 𝑣2′ = 0. Then when we subtract this equation from the second equation
𝑣1′ + 3𝑡2 𝑣2′ = 𝑡2 to get 2𝑡2 𝑣2 ′ = 𝑡2 . Hence 𝑣2′ = 1/2. Substituting 𝑣2′ = 1/2 into 𝑣1′ + 𝑡2 𝑣2′ = 0 gives
𝑣1′ + 21 𝑡2 = 0 or 𝑣1′ = − 21 𝑡2 .
3
Integrating the equation 𝑣1′ = − 21 𝑡2 gives 𝑣1 = − 21 𝑡3 = − 16 𝑡3 while integrating 𝑣2′ = 1/2 gives
𝑣2 = 21 𝑡 (the constants of integration are not needed here). Hence
(
)
(
)
1 3
1
1 1 4
1 3 4 2 4 1 4
3
𝑦𝑝 = 𝑣1 𝑦1 + 𝑣2 𝑦2 = − 𝑡 ⋅ 𝑡 + 𝑡 ⋅ 𝑡 = − +
𝑡 = − +
𝑡 = 𝑡 = 𝑡.
6
2
6 2
6 6
6
3
Then the general solution is 𝑦 = 𝑦ℎ + 𝑦𝑝 , or
1
𝑦 = 𝐶1 𝑡 + 𝐶2 𝑡3 + 𝑡4 ,
3
for arbitrary constants 𝐶1 , 𝐶2 , and 𝐶3 .
Math 231.04, Spring 2010: Practice Exam 3 Solutions
5
6.) (10 points) A spring has mass 2 𝑘𝑔, damping constant 8 𝑁 -𝑠𝑒𝑐/𝑚, and spring constant (stiffness)
26 𝑁/𝑚. There is no external force driving the spring. Find the general formula for the displacement
from equilibrium position of the spring at time 𝑡 seconds, and determine whether this spring is
undamped, underdamped, critically damped, or overdamped.
The general equation for the displacement from equilibrium 𝑦 of the spring is 𝑚𝑦 ′′ +𝑏𝑦 ′ +𝑘𝑦 = 0,
where 𝑚 is the mass, 𝑏 is the stiffness, and 𝑘 is the spring constant. Here we get 2𝑦 ′′ +8𝑦 ′ +26𝑦 = 0,
or, after dividing by 2,
𝑦 ′′ + 4𝑦 ′ + 13𝑦 = 0.
The characteristic equation is 𝑟2 + 4𝑟 + 13 = 0. We cannot factor this quadratic, so using the
quadratic formula we get
√ √
√
√
−4 ± −36
−4 ± 36 −1
−4 ± 6𝑖
−4 ± 16 − 4 ⋅ 13
=
=
=
= −2 ± 3𝑖.
𝑟=
2
2
2
2
This implies that the general solution is
𝑦 = 𝐶1 𝑒−2𝑡 cos 3𝑡 + 𝐶2 𝑒−2𝑡 sin 3𝑡,
for arbitrary constants 𝐶1 and 𝐶2 . This spring is underdamped.
7.) (10 points) Solve 𝑦 ′′′ − 4𝑦 ′′ = 48𝑡.
We first find 𝑦ℎ , the solution to the homogeneous equation 𝑦 ′′′ − 4𝑦 ′′ = 0. The characteristic
equation is 𝑟3 − 4𝑟2 = 0, or 𝑟2 (𝑟 − 4) = 0. Hence 𝑟 = 4 is a root and 𝑟 = 0 is a double root.
Therefore
𝑦ℎ = 𝐶1 𝑒4𝑡 + 𝐶2 𝑒0𝑡 + 𝐶3 𝑡𝑒0𝑡 = 𝐶1 𝑒4𝑡 + 𝐶2 + 𝐶3 𝑡,
for arbitraty constants 𝐶1 , 𝐶2 and 𝐶3 .
Next we find a particular solution 𝑦𝑝 of 𝑦 ′′′ − 4𝑦 ′′ = 48𝑡. Because the inhomogeneous part on
the right hand side of the equation is 48𝑡, a polynomial of degree 1, the usual guess for 𝑦𝑝 is 𝐴𝑡 + 𝐵,
the general polynomial of degree 1. However, 𝐴𝑡 is a constant multiple of the term 𝐶3 𝑡 in 𝑦ℎ (and
𝐵 is a multiple of the term 𝐶2 ), so we multiply by 𝑡 and try 𝑡(𝐴𝑡 + 𝐵) = 𝐴𝑡2 + 𝐵𝑡. However, 𝐵𝑡 is
a constant multiple of the term 𝐶3 𝑡 in 𝑦ℎ , so we multiply by 𝑡 again, to arrive at
𝑦𝑝 = 𝑡(𝐴𝑡2 + 𝐵) = 𝐴𝑡3 + 𝐵𝑡2 .
This is not a multiple of any term in 𝑦ℎ , so it should work. We calculate
𝑦𝑝′ = 3𝐴𝑡2 + 2𝐵𝑡,
𝑦𝑝′′ = 6𝐴𝑡 + 2𝐵, and 𝑦𝑝′′′ = 6𝐴.
Hence
𝑦𝑝′′′ − 4𝑦𝑝′′ = 6𝐴 − 4(6𝐴𝑡 + 2𝐵) = 6𝐴 − 24𝐴𝑡 − 8𝐵 = −24𝐴𝑡 + 6𝐴 − 8𝐵.
Therefore 𝑦 ′′′ − 4𝑦 ′′ = 48𝑡 holds if −24𝐴 = 48 and 6𝐴 − 8𝐵 = 0. Therefore 𝐴 = −2 and hence
8𝐵 = 6𝐴 = 6(−2) = −12, or 𝐵 = −12/8 = −3/2. Therefore
3
𝑦𝑝 = −2𝑡3 − 𝑡2 .
2
′′′
′′
(c) Finally, the general solution of 𝑦 − 4𝑦 = 48𝑡 is 𝑦 = 𝑦𝑝 + 𝑦ℎ , or
3
𝑦 = −2𝑡3 − 𝑡2 + 𝐶1 𝑒4𝑡 + 𝐶2 + 𝐶3 𝑡,
2
for arbitrary constants 𝐶1 , 𝐶2 and 𝐶3 .