Homework6 Solutions

§3.7In Problem 1 through 4 use the method of variation of parameters to find a particular solutionof the given diÆerential equation. Then check your answer by using the method of undeterminedcoe±ents.

1. y00 ° 5y0 + 6y = 2et

Solution: Characteristic equation is r2°5r+6 = 0, therefore r1 = 2, r2 = 3, and y1(t) = e2t,

y2(t) = e3t. Wronskian

W (y1, y2)(t) =

ØØØØe2t 2e2t

e3t 3e3t

ØØØØ = e5t

Using the formula given in the text, we have

Y (t) = °e2t

Ze3t

· 2et

e5tdt + e3t

Ze2t

· 2et

e5tdt

= °2e2t

Ze°t dt + 2e3t

Ze°2t dt = et .

2. y00 ° y ° 2y = 2e°t

Solution: Characteristic equation is r2°r°2 = 0, therefore r1 = °1, r2 = 2, and y1(t) = e°t,

y2(t) = e2t. Wronskian

W (y1, y2)(t) =

ØØØØe°t

°e°t

e2t 2e2t

ØØØØ = 3et

Using the formula given in the text, we have

Y (t) = °e°t

Ze2t

· 2e°t

2etdt + e2t

Ze°t

· 2e°t

3etdt

= °2

3te°t

°

2

9e°t

We can choose Y (t) = °23te

°t.

In problem 5–12 find the general solution of the given equation.

5. y00 + y = tan t, 0 < t < º/2.

Solution: Characteristic equation is r2 + 1 = 0, therefore r1 = i, r2 = °i, and y1(t) = cos t,y2(t) = sin t. Wronskian

W (y1, y2)(t) =

ØØØØcos t ° sin tsin t cos t

ØØØØ = 1 .

1

Using the formula given in the text, we have

Y (t) = ° cos t

Zsin t tan t dt + sin t

Zcos t tan t dt

= ° cos t

Zsin2 t

cos tdt + sin t

Zsin t dt

= ° cos t

Zsin2 t

1° sin2 td(sin t)° sin t cos t

= ° cos t

Z(

1

1° sin2 t° 1) d(sin t)° sin t cos t

= ° cos t(° sin t +1

2ln

1 + sin t

1° sin t)° sin t cos t

= °1

2cos t ln

1 + sin t

1° sin t= ° cos t ln

1 + sin t

cos t,

or Y (t) = ° cos t ln(sec t + tan t). Note the range for t guarantees the function inside ln ispositive.

10. y00 ° 2y0 + y = et/(1 + t2)

Solution: Characteristic equation is r2° 2r + 1 = 0, therefore r1 = r2 = 1, and y1(t) = et,

y2(t) = tet. Wronskian

W (y1, y2)(t) =

ØØØØet et

tet et + tet

ØØØØ = e2t

Using the formula given in the text, we have

Y (t) = °et

Zt

1 + t2dt + tet

Z1

1 + t2dt

= °1

2et ln(1 + t2) + tet arctan t .

In problems 13–20 verify that the given functions y1 and y2 are solutions to the correspondinghomogeneous equation; then find a particular solution of the given nonhomogeneous equation.

13. t2y00 ° 2y = 3t2 ° 1, t > 0, y1 = t2, y2 = t°1

Solution: It’s easy to verify that the given functions are solutions to the homogeneousequation. We will concentrate on finding a particular solution. First rewrite the equationso that the leading coe±cient is 1.

y00 °2

t2y = 3°

1

t2.

The Wronskian is

W (y1, y2)(t) =

ØØØØt2 2tt°1

°1/t2

ØØØØ = °3

2

and

Y (t) = °t2Z

t°1(3° 1/t2)

°3dt + t°1

Zt2(3° 1/t2)

°3dt

=t2

3(3 ln t +

1

2t2)°

1

3t(t3 ° t)

= t2 ln t°t2

3+

1

2.

15. ty00 ° (1 + t)y0 + y = t2e2t, t > 0, y1 = 1 + t, y2 = et

Solution: Rewrite the equation so that the leading coe±cient is 1.

y00 °1 + t

ty0 +

1

ty = te2t .

The Wronskian is

W (y1, y2)(t) =

ØØØØ1 + t 1et et

ØØØØ = tet

and

Y (t) = °(1 + t)

Zet(te2t)

tetdt + et

Z(1 + t)te2t

tetdt

= °(1 + t)

Ze2t dt + et

Z(1 + t)et dt

= °1 + t

2e2t + et((1 + t)et

° et)

= °1

2e2t +

1

2te2t =

1

2e2t(t° 1) .

§3.8In problems 1–4 determine !0, R, and ± so as to write the given expression in the form u =R cos(!0t° ±).

1. u = 3 cos 2t + 4 sin 2t

Solution:

u = 5(3

5cos 2t +

4

5sin 2t) = 5 cos(2t° arccos

3

5)

Therefore R = 5, !0 = 2, ± = arccos(3/5).

2. u = ° cos t +p

3 sin t

Solution:

u = 2(°1

2cos t +

p

3

2sin t) = 2 cos(t°

2º

3) .

Therefore R = 2, !0 = 1, ± = 2º/3.

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6. A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibriumposition with a downward velocity of 10 cm/sec, and if there is no damping, determine theposition u of the mass at any time t. When does the mass first return to its equilibriumposition?

Solution: m = 100 = 0.1 kg, L = 5 = 0.05 m, u(0) = 0 m, u0(0) = 10 = 0.1 m/sec.Therefore k = 0.1£ 9.8/0.05 = 98/5 N/m. The equation for the system is

1

10u00 +

98

5u = 0

Solving the characteristic equation we get r = ±14i. Therefore the general solution is

u(t) = C1 cos 14t + C2 sin 14t .

Using the initail condition we get C1 = 0, C2 = 1/140. Thus

u(t) =1

140sin 14t .

Setting u(t) = 0 we get t = º/14 sec as the time when the mass first return to its equilibriumposition.

13. A certain vibrating system satisfies the equation u00 + ∞u0 + u = 0. Find the value of thedamping coe±cient ∞ for which the quasi period of the damped motion is 50% geater thanthe period of the corresponding undamped motion.

Solution: We can solve the equation to find the quasi period directly. But since there is aformula in the text for the ratio of the quasi period to the period of the undamped systemwe simply use that one. By this formula (28) in the text, and notice that m = 1, k = 1, wehave

Td

T=

µ1°

∞2

4km

∂°1/2

=

µ1°

∞2

4

∂°1/2

= 150% =3

2,

solving for ∞ we get ∞ = 23

p

5.

17. A mass weighing 8 lb stretches a sping 1.5 in. The mass is also attached to a damper withcoe±cient ∞. Determine the value of ∞ for which the system is critically damped; be sure togive the units for ∞.

Solution: m = 8/32 = 1/4 lb·sec2/ft, L = 1.5/12 = 1/8 ft. Thus k = 8/(1/8) = 64 lb/ft.Thus the equation is

1

4u00 + ∞u0 + 64u = 0 .

So in order for the system to be critically damped we need

∞2 = 4mk = 64

or ∞ = 8 lb·sec/ft.

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19. Assume the system decribed by the equation mu00 + ∞u0 + ku = 0 is critically damped oroverdamped. Show that the mass can pass through the equilibrium position at most once,regardless of the initial condition.

Solution: First assume the system is critically damped, then the general solution to theequation is u = C1ert + C2tert where r is the double root to the characteristic equation, andr is negative. Set u = 0 and we get

ert(C1 + C2t) = 0 .

Clearly this equation has no solution if C2 = 0 and exactly one solution if C2 6= 0.

For the overdamped case the general solution is u = C1er1t +C2er2t and the proof is similar.

§3.9

6. A mass of 6 kg stretches a spring 10 cm. The mass is acted on by an external force of10 sin(t/2) N and moves in a medium that imparts a viscous force of 2 N when the speedof the mass is 4 cm/sec. If the mass is set in mostion from its equilibrium position with aninitial velocity of 3 cm/sec, formulate the initial value problem describing the motion of themass.

Solution: m = 5 kg, L = 0.1m, ∞ = 2/4 N·sec/cm = 50 N·sec/m, k = 5 £ 9.8/0.1 = 490N/m. Therefore the equation is

5u00 + 50u0 + 490u = 10 sin t/2 .

Initial condition is u(0) = 0, u0(0) = 3/100 m/sec.

8. (a) Find the solution of the initial value problem in Problem 6.

(b) Identify the transient and steady-state parts of the solution.

(c) Plot the graph of the steady-state solution.

(d) If the given external force is replaced by a force 2 cos !t of frequency !, find the valueof ! for which the amplitude of the forced response is maximum.

Solution

(a) Solving the characteristic equation we get r = °5 ±

p

73i, so the general solution tothe homogeneous equation is

e°5t(C1 cosp

73t + C2 sinp

73t) .

One way to find a particular solution is using undetermined coe±cients, but that’s abit tedious. What we will do here is to use equation (8) and (9) in the text. Let’s firstrewrite the equation as u00 + 10u0 + 98u = 2 sin t/2. Thus m = 1, ∞ = 10, !0 =

p

73,! = 1/2, F0 = 2. Then according to equation (9)

¢ =q

m2(!20 ° !2)2 + ∞2!2 = 72.92 ,

5

and R = F0/¢ = 0.02743, ± = cos°1 m(!20°!2)¢ = cos°1 72.75

72.92 = 0.06830. However wecannot say u(t) = R cos(!t ° ±) is a particular solution to our equation since that’s asolution to equation (1) in the text in which the right hand side is F0 cos !t, while inour equation the right hand side is F0 sin !t. To remedy the situation we consider thefunction u(t) = R sin(!t ° ±). It’s easy to check that this is a particular solution toour equation. Therefore

u(t) = e°5t(C1 cosp

73t + C2 sinp

73t) + 0.02743 sin(0.5t° 0.06830)

is the general solution. Using the initial condition we get C1 = 0.00187 m = 0.187 cm,C2 = 0.0062 m = 0.62 cm.

(b) The transient part is u(t) = e°5t(0.187 cosp

73t + 0.62 sinp

73t) cm, steady-state partis u(t) = 2.743 sin(0.5t° 0.06830) cm.

(c) Use your calculator (or computer :).

(d) The amplitude of the forced response R is given as F0/¢, therefore in order for it toreach maximum ¢ should reach the minimum. Note ¢ =

pm2(!2

0 ° !2)2 + ∞2!2. Tofind the minimum we take the derivative of ¢2 with respect to ! and set it to 0. Thus

2m2(!20 ° !2)(°2!) + 2∞2! = 0

! =

r!2

0 °∞2

2m2=p

23 .

9. If an undamped spring-mass system with a mass that weighs 6 lb and a spring constant 1lb/in. is suddenly set in motion at t = 0 by an external force of 4 cos 7t lb, determine theposition of the mass at any time and draw a graph of the displacement versus t.

Solution: m = 6/32 = 3/16 lb·sec2/ft, k = 12 lb/ft, thus the equation is 316u

00+12u = 4 cos 7twith initial condition u(0) = 0, u0(0) = 0. Rewrite the equation as u00 + 64u = 64

3 cos 7t we

see that !0 =p

64 = 8. Using equation (5) in the text we have the solution

u =F0

m(!20 ° !2)

(cos !t° cos !0t)

=64

45(cos 7t° cos 8t) .

18. Consider the forced but undamped system described by the initial value problem

u00 + u = 3 cos !t , u(0) = 0, u0(0) = 0 .

(a) Find the solution u(t) for ! 6= 1.

(b) Plot the solution u(t) versus t for ! = 0.7, ! = 0.8 and ! = 0.9. Describe how theresponse u(t) changes as ! varies in this interval. What happens as ! takes on valuescloser and closer to 1? Note that the natural frequency of the unforced system is !0 = 1.

Solution:

(a) Again using equation (5) we get the solution

u(t) =3

1° !2(cos !t° cos t) .

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(b) We will not show the graph here, but we can describe what happens as ! gets closerto 1. Rewrite the solution as

u(t) =3

2(1° !2)sin

1° !

2t sin

1 + !

2t

We may view this as a vibration with a varying amplitude 32(1°!2) sin 1°!

2 t. As ! gets

closer to 1 the term 32(1°!2) becomes very large, which means the maximal amplitude

is very big and also the period according to which the amplitude changes gets very bigtoo since it’s equal to 4º/(1° !).

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