Download Math 2142 Homework 4 Part 1: Due Friday March 11

Math 2142 Homework 4 Part 1: Due Friday March 11
Problem 1. For the following homogeneous second order differential equations, give the
general solution and the particular solution satisfying the given initial conditions.
d2 y
dy
+ 2 − 3y = 0 with y(0) = 6 and y 0 (0) = −2
2
dx
dx
2
dy
dy
+ 4 + 20y = 0 with y(0) = 2 and y 0 (0) = −8
2
dx
dx
d2 y
dy
− 4 + 4y = 0 with y(0) = y 0 (0) = 1
2
dx
dx
Solution. For the first equation,
d2 y
dy
+ 2 − 3y = 0 with y(0) = 6 and y 0 (0) = −2
2
dx
dx
the characteristic polynomial is r2 + 2r − 3 = (r + 3)(r − 1) with roots are r = −3 and r = 1.
Therefore, the general solution is
y(x) = c1 e−3x + c2 ex
To find the particular solution, we use y 0 (x) = −3c1 e−3x + c2 ex . Therefore, the initial conditions give
y(0) = 6 ⇒ 6 = c1 + c2
y (0) = −2 ⇒ −2 = −3c1 + c2
0
Subtracting these equations gives 8 = 4c1 and so c1 = 2. Therefore, c2 = 4 and the particular
solution is
y(x) = 2e−3x + 4ex
For the second equation,
dy
d2 y
+ 4 + 20y = 0 with y(0) = 2 and y 0 (0) = −8
2
dx
dx
the characteristic polynomial is r2 + 4r + 20. The roots are given by
p
√
−4 ± 42 − 4(1)(20)
−4 ± −64
r=
=
= −2 ± 4i
2(1)
2
Therefore, the general solution is
y(x) = c1 e−2x cos 4x + c2 e−2x sin 4x
To find the particular solution, we plug in y(0) = 2 to get 2 = c1 . To use the second condition
we differentiate
y(x) = 2e−2x cos 4x + c2 e−2x sin 4x
y 0 (x) = −4e−2x cos 4x − 8e−2x sin 4x − 2c2 e−2x sin 4x + 4c2 e−2x cos 4x
and therefore
y 0 (0) = −4 + 4c2
so −8 = −4 + 4c2 and hence c2 = −1. The particular solution is
y(x) = 2e−2x cos 4x − e−2x sin 4x
For the last equation
dy
d2 y
− 4 + 4y = 0 with y(0) = y 0 (0) = 1
2
dx
dx
the characteristic polynomial is r2 − 4r + 4 = (r − 2)2 and therefore there is a single (repeated)
real root r = 2. The general solution is
y(x) = c1 e2x + c2 xe2x
The initial condition y(0) = 1 gives us that 1 = c1 . Differentiating gives
y(x) = e2x + c2 xe2x
y 0 (x) = 2e2x + c2 e2x + 2c2 xe2x
The condition y 0 (0) = 1 gives us 1 = 2 + c2 and hence c2 = −1. The particular solution is
y(x) = e2x − xe2x
Problem 2. Find the general solutions for the following differential equations.
dy
d2 y
− 5 + 4y = e4x
2
dx
dx
d2 y
dy
+
3
+ 2y = x2
dx2
dx
Solution. To solve the first equation, we consider the homogeneous differential equation
y 00 − 5y 0 + 4y = 0. The characteristic polynomial is r2 − 5r + 4 = (r − 4)(r − 1) with roots
r = 4 and r = 1. Therefore, the general solution to the homogeneous equation is c1 ex + c2 e4x .
To guess a solution to the non-homogeneous equation, we try y = Axe4x (since we know
y = Ae4x is a solution to the homogeneous equation). Calculating the derivatives, we have
y 0 = Ae4x + 4Axe4x
y 00 = 4Ae4x + 4Ae4x + 16Axe4x = 8Ae4x + 16Axe4x
Plugging these values into the non-homogeneous equation, we have
y 00 − 5y 0 + 4y = e4x
(8Ae4x + 16Axe4x ) − 5(Ae4x + 4Axe4x ) + 4(Axe4x ) = e4x
(8A − 5A)e4x + (16A − 20A + 4A)xe4x = e4x
3Ae4x = e4x
and so A = 1/3. Therefore, one solution to the non-homogeneous equation is y = 1/3 xe4x
and the general solution to the non-homogeneous equation is
1
y = xe4x + c1 ex + c2 e4x
3
For the second equation, we consider the homogeneous differential equation y 00 +3y 0 +2y =
0. The characteristic polynomial is r2 + 3r + 2 = (r + 1)(r + 2) with roots r = −1 and r = −2.
Therefore, the general solution to the homogeneous equation is c1 e−x + c2 e−2x .
To guess a solution to the non-homogeneous equation, we try y = Ax2 + Bx + C. Calculating the derivatives, we have
y 0 = 2Ax + B
y 00 = 2A
Plugging these values into the non-homogeneous equation gives us
y 00 + 3y 0 + 2y = x2
2A + 3(2Ax + B) + 2(Ax2 + Bx + C) = x2
2Ax2 + (6A + 2B)x + (2A + 3B + 2C) = x2
Comparing the coefficients for the powers of x2 , we know 2A = 1, so A = 1/2. Comparing
the coefficients of x, we know 6A + 2B = 0. Since A = 1/2, this equation tells us 3 + 2B = 0
so B = −3/2. Finally, comparing the constant terms, we know 2A + 3B + 2C = 0. Since
A = 1/2 and B = −3/2, this equation tells us 1 − 9/2 + 2C = 0 which means C = 7/4.
Therefore, y = x2 /2 − 3x/2 + 7/4 is a single solution to the non-homogeneous equation and
the general solution to the non-homogeneous equation is
y = x2 /2 − 3x/2 + 7/4 + c1 e−x + c2 e−2x
Problem 3. Solve the following initial value problems.
d2 y
dy
+ 4 + 13y = −3e−2x with y(0) = y 0 (0) = 0
2
dx
dx
d2 y
dy
+ 6 + 8y = cos(x) with y(0) = y 0 (0) = 0
2
dx
dx
Solution. For the first equation, we consider the homogeneous equation y 00 + 4y 0 + 13y =
0. The characteristic polynomial is r2 + 4r + 13 which (using the quadratic formula) has
roots −2 ± 3i. Using r = −2 + 3i, we know that a complex solution to the homogeneous
equation is given by y = e−2x cos 3x + ie−2x sin 3x and therefore, the general (real) solution is
c1 e−2x cos 3x + c2 e−2x sin 3x.
To guess a particular solution to the non-homogeneous equation, we try y = Ae−2x .
Calculating the derivatives, we have
y 0 = −2Ae−2x
y 00 = 4Ae−2x
Plugging these values into the non-homogeneous equation gives us
4Ae−2x + 4(−2Ae−2x ) + 13Ae−2x = −3e−2x
which means 9A = −3, so A = −1/3. Therefore, y = −1/3e−2x is a solution to the nonhomogeneous equation and the general solution to the non-homogeneous equation is
1
y = − e−2x + c1 e−2x cos 3x + c2 e−2x sin 3x
3
To find the values of c1 and c2 corresponding to our initial conditions, we calculate (using the
product and chain rules)
2
y 0 = e−2x − 2c1 e−2x cos 3x − 3c1 e−2x sin 3x − 2c2 e−2x sin 3x + 3c2 e−2x cos 3x
3
2
y 0 = e−2x + (−2c1 + 3c2 )e−2x cos 3x + (−3c1 − 2c2 )e−2x sin 3x
3
To use y(0) = 0, we plug x = 0 into our general equation for y to get
1
− e0 + c1 e0 cos 0 + c2 e0 sin 0 = 0
3
which tells us that c1 = 1/3. To use y 0 (0) = 0, we plug x = 0 into our general equation for y 0
to get
2 0
e + (−2c1 + 3c2 )e0 cos 0 + (−3c1 − 2c2 )e0 sin 0 = 0
3
which tells us that −2c1 + 3c2 = −2/3. Since c1 = 1/3, we have 3c2 = 0 and so c2 = 0.
Therefore, the particular solution is
1
1
y = − e−2x + e−2x cos 3x
3
3
For the second equation, we consider the homogeneous equation y 00 + 6y 0 + 8y = 0. The
characteristic equation is r2 + 6r + 8 = (r + 2)(r + 4) which has roots r = −2 and r = −4.
Therefore, the general solution to the homogeneous equation is y = c1 e−4x + c2 e−2x .
To find a single solution to the non-homogeneous equation, we guess y = A sin x + B cos x.
Calculating derivatives gives us
y 0 = A cos x − B sin x
y 00 = −A sin x − B cos x
Plugging these values into the non-homogeneous equation gives
y 00 + 6y 0 + 8y = cos x
−A sin x − B cos x + 6(A cos x − B sin x) + 8(A sin x + B cos x) = cos x
(−A − 6B + 8A) sin x + (−B + 6A + 8B) cos x = cos x
(7A − 6B) sin x + (6A + 7B) cos x = cos x
Setting the coefficients of cos x and sin x equal from the opposite sides of the equation, we
have 7A − 6B = 0 and 6A + 7B = 1. Solving this system of equations yields A = 6/85 and
B = 7/85. Therefore, the general solution to the non-homogeneous equation is
y=
6
7
sin x +
cos x + c1 e−4x + c2 e−2x
85
85
To find the specific solution, we use y(0) = 0 to get
7
6
sin 0 +
cos 0 + c1 e0 + c2 e0 = 0
85
85
which means c1 + c2 = −7/85. To use y 0 (0) = 0, we first take the derivative
y0 =
6
7
cos x −
sin x − 4c1 e−4x − 2c2 e−2x
85
85
and then use y 0 (0) = 0 by plugging in x = 0 to get
7
6
cos 0 −
sin 0 − 4c1 e0 − 2c2 e0 = 0
85
85
to get −4c1 − 2c2 = −6/85. Solving c1 + c2 = −7/85 and −4c1 − 2c2 = −6/85 gives c1 = 2/17
and c2 = −1/5. Therefore, the specific solution to the non-homogeneous equation is
y=
6
7
2
1
sin x +
cos x + e−4x − e−2x
85
85
17
5
Problem 4. Consider a pair of linear second order differential equations with the same left
side.
d2 y
dy
+ P1 (x)
+ P2 (x) y = Q1 (x)
2
dx
dx
d2 y
dy
+ P1 (x)
+ P2 (x) y = Q2 (x)
2
dx
dx
Prove that if y1 (x) is a solution to the top equation and y2 (x) is a solution to the bottom
equation, then y1 (x) + y2 (x) is a solution to
d2 y
dy
+ P2 (x) y = Q1 (x) + Q2 (x)
+ P1 (x)
2
dx
dx
Hint. I would let L(y) = y 00 + P1 (x)y 0 + P2 (x)y and use the fact that L(y) is a linear operator
to write a very short proof for this problem.
Solution. Let L(y) = y 00 + P1 (x)y 0 + P2 (x)y. From our work in class, we know that L(y) is
a linear operator. The assumptions of this problem tell us that L(y1 ) = Q1 (x) and L(y2 ) =
Q2 (x). Using the linearity of L(y), we know L(y1 + y2 ) = L(y1 ) + L(y2 ) = Q1 (x) + Q2 (x),
which means that y1 + y2 is a solution to y 00 + P1 (x)y 0 + P2 (x)y = Q1 (x) + Q2 (x).
Alternately, you can solve this problem directly. We are given that
y100 + P1 (x) y10 + P2 (x) y1 = Q1 (x)
y200 + P1 (x) y20 + P2 (x) y2 = Q2 (x)
We calculate
d
d2
(y1 + y2 ) + P1 (x) (y1 + y2 ) + P2 (x)(y1 + y2 )
2
dx
dx
= (y100 + y200 ) + P1 (x)(y10 + y20 ) + P2 (x)(y1 + y2 )
= (y100 + P1 (x)y10 + P2 (x)y1 ) + (y200 + P1 (x)y20 + P2 (x)y2 )
= Q1 (x) + Q2 (x)
and hence y1 + y2 satisfies the required equation.
Problem 5(a). Find the general solution for
d2 y
+ y = e x + x3 + x + 2
2
dx
Hint. By Problem 4, you can separate this problem into three steps. First, find the general
solution to the homogeneous equation. Second, find a particular solution to y 00 + y = ex .
Third, find a particular solution to y 00 + y = x3 + x + 2. The sum of these three parts will be
the general solution.
5(b). Find the particular solution to the equation in 3(a) satisfying y(0) = 2 and y 0 (0) = 0.
Solution. For 5(a), we break the problem into three pieces. For the first piece, we solve the
homogeneous equation y 00 + y = 0. The characteristic polynomial is r2 + 1 which has roots
r = ±i. Using r = i = 0 + i, a complex solution to the homogeneous equation is
y = e0x cos x + ie0x sin x = cos x + i sin x
and therefore the general (real) solution is y = c1 cos x + c2 sin x.
For the second piece, we find a single solution to the non-homogeneous equation y 00 +y = ex
by guessing y = Aex . Taking derivatives, we have y 0 = y 00 = Aex . Plugging into the nonhomogeneous equation gives
y 00 + y = ex
Aex + Aex = ex
2Aex = ex
and so A = 1/2. This tells us that one solution to y 00 + y = ex is y = 1/2 ex .
For the third piece, we find a single solution to the non-homogeneous equation y 00 + y =
3
x + x + 2 by guessing y = Ax3 + Bx2 + Cx + D. Taking derivatives gives us
y 0 = 3Ax2 + 2Bx + C
y 00 = 6Ax + 2B
Plugging into the non-homogeneous equation gives us
y 00 + y = x3 + x + 2
(6Ax + 2B) + (Ax3 + Bx2 + Cx + D) = x3 + x + 2
Ax3 + Bx2 + (6A + C)x + (2B + D) = x3 + x + 2
Comparing the coefficients on each side of the equation gives A = 1, B = 0, 6A + C = 1 and
2B + D = 2. Solving for C and D gives C = −5 and D = 2. Therefore, a single solution to
this non-homogeneous equation is y = x3 − 5x + 2.
Putting the three pieces together, the general solution for the original non-homogeneous
equation is
1
y = c1 cos x + c2 sin x + ex + x3 − 5x + 2
2
To find the specific solution for 5(b), we first use y(0) = 2 to get
1
c1 cos 0 + c2 sin 0 + e0 + 03 − 5(0) + 2 = 2
2
which means c1 + 1/2 + 2 = 2 so c1 = −1/2. Our specific solution now has the form
1
1
y = − cos x + c2 sin x + ex + x3 − 5x + 2
2
2
Taking a derivative gives us
y0 =
1
1
sin x − c2 cos x + ex + 3x2 − 5
2
2
Plugging in y 0 (0) = 0 we get
1
1
sin 0 − c2 cos 0 + e0 + 3(02 ) − 5 = 0
2
2
which means c2 = 9/2. Therefore, the specific solution is
1
9
1
y = − cos x + sin x + ex + x3 − 5x + 2
2
2
2
Problem 6. Do the following problems in Exercises 10.4 on page 382: 1, 4, 6, 7, 13.
Solution to Exercise 1. Replacing n by the real variable x, we calculate
x+1
x
−
=1−1=0
x→∞ x + 1
x
lim
where in the second step we use L’Hospital’s Rule on each of the fractions.
Solution to Exercise 4. Replacing n by the real variable x, we use L’Hospital’s Rule to
calculate
x2 + 3x − 2
1
lim
=
2
x→∞
5x
5
Solution to Exercise 6. The terms in this sequence alternate between 0 and 2, so the
sequence diverges.
Solution to Exercise 7. The numerators of the terms in this sequence alternate between 0
and 2. Therefore, for any n, we have
0≤
2
1 + (−1)n
≤
n
n
Since limn→∞ 2/n = 0, we have limn→∞ (1 + (−1)n )/n = 0 by the Squeeze Theorem.
Solution to Exercise 13. Replacing n by the real variable x, we use our limit tricks to
calculate the limit as follows.
√
√
√
√
√
√
x+1+ x
lim x + 1 − x = lim ( x + 1 − x) · √
√
x→∞
x→∞
x+1+ x
(x + 1) − x
= lim √
√
x→∞
x+1+ x
1
= lim √
√
x→∞
x+1+ x
= 0
The last equality holds because the numerator is constant but the denominator goes to ∞.