Section 17.2 Nonhomogeneous Linear Equations

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Section 17.2 Nonhomogeneous Linear Equations

Review(p.1221-1222)

7. Solve the differential equation. ^d_dx²^y2 − 2^dy_dx+ y = x cos x.

Solution:

738 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS

17 Review

1. True. See Theorem 17.1.3.

2. False. The differential equation is not homogeneous.

3. True. cosh  and sinh  are linearly independent solutions of this linear homogeneous equation.

4. False.  = ^is a solution of the complementary equation, so we have to take () = ^.

1. The auxiliary equation is 4²− 1 = 0 ⇒ (2 + 1)(2 − 1) = 0 ⇒  = ±¹2. Then the general solution is  = 1^2+ 2^−2.

2. The auxiliary equation is ²− 2 + 10 = 0 ⇒  = 1 ± 3, so  = ^(1cos 3 + 2sin 3).

3. The auxiliary equation is ²+ 3 = 0 ⇒  = ±√3 . Then the general solution is  = 1cos√

3 

+ 2sin√

3 .

4. The auxiliary equation is ²+ 8 + 16 = 0 ⇒ ( + 4)² = 0 ⇒  = −4, so the general solution is

 = 1^−4+ 2^−4.

5. ²− 4 + 5 = 0 ⇒  = 2 ± , so ^() = ^2(1cos  + 2sin ). Try () = ^2 ⇒ ⁰^= 2^2

and ⁰⁰= 4^2. Substitution into the differential equation gives 4^2− 8^2+ 5^2= ^2 ⇒  = 1 and the general solution is () = ^2(1cos  + 2sin ) + ^2.

6. ²+  − 2 = 0 ⇒  = 1,  = −2 and ^() = 1^+ 2^−2. Try () = ²+  +  ⇒ ^⁰ = 2 +  and ⁰⁰= 2. Substitution gives 2 + 2 +  − 2²− 2 − 2 = ² ⇒  =  = −¹2,  = −³₄ so the general solution is () = 1^+ 2^−2− ¹2²− ¹2 −³4.

7. ²− 2 + 1 = 0 ⇒  = 1 and ^() = 1^+ 2^. Try () = ( + ) cos  + ( + ) sin  ⇒

⁰ = ( −  − ) sin  + ( +  + ) cos  and ⁰⁰^ = (2 −  − ) cos  + (−2 −  − ) sin . Substitution gives (−2 + 2 − 2 − 2) cos  + (2 − 2 + 2 − 2) sin  =  cos  ⇒  = 0,  =  =  = −¹₂. The general solution is () = 1^+ 2^−¹2cos  −¹2( + 1) sin .

8. ²+ 4 = 0 ⇒  = ±2 and ^() = 1cos 2 + 2sin 2. Try () =  cos 2 +  sin 2so that no term of is a solution of the complementary equation. Then ⁰= ( + 2) cos 2 + ( − 2) sin 2 and

⁰⁰= (4 − 4) cos 2 + (−4 − 4) sin 2. Substitution gives 4 cos 2 − 4 sin 2 = sin 2 ⇒

 = −¹4 and  = 0. The general solution is () = 1cos 2 + 2sin 2 −¹4 cos 2.

8. Solve the differential equation. ^d_dx²^y2 + 4y = sin 2x.

Solution:

17 Review

1. True. See Theorem 17.1.3.

2. False. The differential equation is not homogeneous.

3. True. cosh  and sinh  are linearly independent solutions of this linear homogeneous equation.

4. False.  = ^is a solution of the complementary equation, so we have to take () = ^.

1. The auxiliary equation is 4²− 1 = 0 ⇒ (2 + 1)(2 − 1) = 0 ⇒  = ±¹2. Then the general solution is  = 1^2+ 2^−2.

2. The auxiliary equation is ²− 2 + 10 = 0 ⇒  = 1 ± 3, so  = ^(1cos 3 + 2sin 3).

3. The auxiliary equation is ²+ 3 = 0 ⇒  = ±√

3 . Then the general solution is  = 1cos√

3 

+ 2sin√

3  .

4. The auxiliary equation is ²+ 8 + 16 = 0 ⇒ ( + 4)² = 0 ⇒  = −4, so the general solution is

 = 1^−4+ 2^−4.

5. ²− 4 + 5 = 0 ⇒  = 2 ± , so ^() = ^2(1cos  + 2sin ). Try () = ^2 ⇒ ⁰^= 2^2

and ⁰⁰= 4^2. Substitution into the differential equation gives 4^2− 8^2+ 5^2= ^2 ⇒  = 1 and the general solution is () = ^2(1cos  + 2sin ) + ^2.

6. ²+  − 2 = 0 ⇒  = 1,  = −2 and ^() = 1^+ 2^−2. Try () = ²+  +  ⇒ ^⁰ = 2 +  and ⁰⁰= 2. Substitution gives 2 + 2 +  − 2²− 2 − 2 = ² ⇒  =  = −¹2,  = −³4 so the general solution is () = 1^+ 2^−2− ¹2²− ¹2 −³4.

7. ²− 2 + 1 = 0 ⇒  = 1 and ^() = 1^+ 2^. Try () = ( + ) cos  + ( + ) sin  ⇒

⁰ = ( −  − ) sin  + ( +  + ) cos  and ⁰⁰^ = (2 −  − ) cos  + (−2 −  − ) sin . Substitution gives (−2 + 2 − 2 − 2) cos  + (2 − 2 + 2 − 2) sin  =  cos  ⇒  = 0,  =  =  = −¹₂. The general solution is () = 1^+ 2^−¹2cos  −¹2( + 1) sin .

8. ²+ 4 = 0 ⇒  = ±2 and ^() = 1cos 2 + 2sin 2. Try () =  cos 2 +  sin 2so that no term of is a solution of the complementary equation. Then ⁰= ( + 2) cos 2 + ( − 2) sin 2 and

_⁰⁰= (4 − 4) cos 2 + (−4 − 4) sin 2. Substitution gives 4 cos 2 − 4 sin 2 = sin 2 ⇒

 = −¹4 and  = 0. The general solution is () = 1cos 2 + 2sin 2 −¹4 cos 2.

21. Assume that the earth is a solid sphere of uniform density with mass M and radius R = 3960 mi. For a particle of mass m within the earth at a distance r from the earths center, the gravitational force attracting the particle to the center is

F_r= −GMrm r²

where G is the gravitational constant and Mr is the mass of the earth within the sphere of radius r.

(a) Show that Fr= ^{−GM m}_R3 r.

(b) Suppose a hole is drilled through the earth along a diameter. Show that if a particle of mass m is dropped from rest at the surface, into the hole, then the distance y = y(t) of the particle from the center of the earth at time t is given by

y⁰⁰(t) = −k²y(t) where k²= GM/R³= g/R.

(c) Conclude from part (b) that the particle undergoes simple harmonic motion. Find the period T . (d) With what speed does the particle pass through the center of the earth?

Solution:

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7= (−1)³

3 · 5 · 7 = (−1)³2³3!

7! ,    , 2+1= (−1)^2^!

(2 + 1)! for  = 0 1 2    . Thus the solution to the initial-value problem is () = ^∞

=0

^= ^∞

=0

(−1)^2^!

(2 + 1)! ^2+1.

18. Let () = ^∞

=0

^. Then ⁰⁰() = ^∞

=0 ( − 1) ^^⁻²= ^∞

=0

( + 2)( + 1)+2^and the differential equation

becomes ^∞

=0

[( + 2)( + 1)+2− ( + 2)^]^= 0. Thus the recursion relation is +2= 

 + 1 for

 = 0, 1, 2,    . Given 0and 1, we have 2 = 0

1, 4 = 2

3 = 0

1 · 3, 6 = 4

5 = 0

1 · 3 · 5,    ,

2= 0

1 · 3 · 5 · · · (2 − 1) = 0

2^⁻¹( − 1)!

(2 − 1)! . Similarly 3 = 1

2, 5= 3

4 = 1

2 · 4,

7 = 5

6 = 1

2 · 4 · 6,    , 2+1= 1

2 · 4 · 6 · · · 2 = 1

2^!. Thus the general solution is

() = ^∞

=0

^= 0+ 0

∞

=1

2^⁻¹( − 1)! ^2

(2 − 1)! +  ^∞

=0

^2+1

2^!. But ^∞

=0

^2+1

2^! =  ^∞

=0

₁

2²

! = ^²^2, so () = 1^²^2+ 0+ 0

∞

=1

2^⁻¹( − 1)! ^2

(2 − 1)! .

19. Here the initial-value problem is 2⁰⁰+ 40⁰+ 400 = 12,  (0) = 001, ⁰(0) = 0. Then

() = ^−10(1cos 10 + 2sin 10)and we try () = . Thus the general solution is

() = ^−10(1cos 10 + 2sin 10) +₁₀₀³ . But 001 = ⁰(0) = 1+ 003and 0 = ⁰⁰(0) = −10¹+ 102, so 1= −002 = ². Hence the charge is given by () = −002^−10(cos 10 + sin 10) + 003.

20. By Hooke’s Law the spring constant is  = 64 and the initial-value problem is 2⁰⁰+ 16⁰+ 64 = 0, (0) = 0,

⁰(0) = 24. Thus the general solution is () = ^−4(1cos 4 + 2sin 4). But 0 = (0) = 1and

24 = ⁰(0) = −4¹+ 42 ⇒ ¹= 0, 2= 06. Thus the position of the mass is given by () = 06^−4sin 4.

21. (a) Since we are assuming that the earth is a solid sphere of uniform density, we can calculate the density  as follows:

 = mass of earth

volume of earth= 

4

3³. If is the volume of the portion of the earth which lies within a distance  of the center, then = ⁴₃³and = =  ³

³ . Thus = −

² = − 

³ .

(b) The particle is acted upon by a varying gravitational force during its motion. By Newton’s Second Law of Motion,

²

² = = − 

³ , so ⁰⁰() = −² ()where ²= 

³ . At the surface, − = ^= − 

² , so

 = 

² . Therefore ²= 

.

CHAPTER 17 REVIEW ¤ 741

(c) The differential equation ⁰⁰+ ² = 0has auxiliary equation ²+ ²= 0. (This is the  of Section 17.1,

not the  measuring distance from the earth’s center.) The roots of the auxiliary equation are ±, so the general solution of our differential equation for  is () = 1cos  + 2sin . It follows that ⁰() = −¹ sin  + 2 cos . Now

 (0) = and ⁰(0) = 0, so 1= and 2 = 0. Thus () =  cos  and ⁰() = − sin . This is simple harmonic motion (see Section 17.3) with amplitude , frequency , and phase angle 0. The period is  = 2.

 ≈ 6370 km = 6370 × 10⁶mand  = 98 ms², so  =

 ≈ 124 × 10⁻³s⁻¹and

 = 2 ≈ 5079 s ≈ 85 min.

(d) () = 0 ⇔ cos  = 0 ⇔  = ^₂ + for some integer  ⇒ ⁰() = − sin 2 + 

= ±.

Thus the particle passes through the center of the earth with speed  ≈ 7899 kms ≈ 28,400 kmh.

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