Math 2280 - Assignment 6 · Math 2280 - Assignment 6 Dylan Zwick Summer 2013 Section 3.7 - 1, 5,...
Transcript of Math 2280 - Assignment 6 · Math 2280 - Assignment 6 Dylan Zwick Summer 2013 Section 3.7 - 1, 5,...
Math 2280 - Assignment 6
Dylan Zwick
Summer 2013
Section 3.7 - 1, 5, 10, 17, 19
Section 3.8 - 1, 3, 5, 8, 13
Section 4.1 - 1, 2, 13, 15, 22
Section 4.2 - 1, 10, 19, 28
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Section 3.7 - Electrical Circuits
3.7.1 This problem deals with the RL circuit pictured below. It is a series circuit containing an inductor with an inductance of L henries,a resistor with a resistance of R ohms, and a source of electromotiveforce (emf), but no capacitor. In this case the equation governing oursystem is the first-order equation
LI’ + RI = E(t).
C
Suppose that L = 511, R = 25Q, and the source E of emf is a batterysupplying 1OOl to the circuit. Suppose also that the switch has beenin position 1 for a long time, so that a steady current of 4A is flowingin the circuit. At time t = 0, the switch is thrown to position 2, sothatl(0)=r4andE=Ofort>0. FindI(t).
Solution - The ODE that describes this system is:
We can rewrite this as:
51’ + 251 = 0.
I’ + SI = 0.
2
This is a first-order linear ODE, and the corresponding integrating
factor is ρ(t) = eR
5dt = e5t. Multiplying both sides of the ODE by thisintegrating factor we get:
e5tI ′ + 5e5tI = 0 ⇒ d
dt
(
e5tI)
= 0.
Integrating both sides of the above equation we get:
e5tI = C ⇒ I(t) = Ce−5t.
If we plug in the initial condition I(0) = 4 = C we get the finalsolution:
I(t) = 4e−5t.
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3.7.5 - In the circuit from Problem 3.7.1, with the switch in position 1, sup-pose that E(t) = 100e−10t cos 60t, R = 20, L = 2, and I(0) = 0. FindI(t).
Solution - The differential equation governing this system will be:
2I ′ + 20I = 100e−10t cos (60t).
If we divide both sides by 2 we get the ODE:
I ′ + 10I = 50e−10t cos (60t).
The integrating factor for this first-order linear ODE will be ρ(t) =e
R
10dt = e10t. Multiplying both sides of our equation by this integrat-ing factor we get:
d
dt
(
e10tI)
= 50 cos (60t).
Integrating both sides of this equation gives us:
e10tI =5
6sin (60t) + C
⇒ I(t) =5
6e−10t sin (60t) + Ce−10t.
If we plug in the initial condition I(0) = 0 we get C = 0, and oursolution is:
I(t) =5
6e−10t sin (60t).
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3.7.10 - This problem deals with an RC circuit pictured below, containinga resistor (R ohms), a capacitor (C farads), a switch, a source of emf,but no inductor. This system is governed by the linear first-orderdifferential equation
dQ 1R-- + = E(t).
for the charge Q = Q(t) on the capacitor at time t. Note that 1(t)Q’(t).
C
Suppose an emf of voltage E(t) = B0 cos wt is applied to the RCcircuit at time t = 0 (with the switch closed), and Q(0) = 0. SubstituteQ5(t) = A cos wt + B sin wt in the differential equation to show thatthe steady periodic charge on the capacitor is
E0CQ5 (t) =
__________
cos (wt—
3)p
where /3 = tan’ (wRC).
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Solution - Our steady periodic charge will be of the form:
Qsp = A cos (ωt) + B sin (ωt).
Its derivative will is:
Q′sp = −ωA sin (ωt) + ωB cos (ωt).
If we plug these values into the ODE:
RQ′sp +
1
CQsp = E0 cos (ωt),
we get:
−ARω sin (ωt) + BRω cos (ωt) +A
Ccos (ωt) +
B
Csin (ωt) = E0 cos (ωt).
Grouping like terms together and multiplying both sides by C wehave:
(B − AωRC) sin (ωt) + (A + BωRC) cos (ωt) = CE0 cos (ωt).
Equating coefficients we get B − AωRC = 0, which means B =AωRC. Plugging these into the above equation gives us:
A(1 + ω2R2C2) cos (ωt) = CE0 cos (ωt).
So, A =E0C
1 + ω2R2C2, and B =
E0C2ωR
1 + ω2R2C2. From this we get:
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Qsp =E0C
1 + ω2R2C2(cos (ωt) + ωRC sin (ωt)).
If we define α = tan−1(ωRC) then
sin α =ωRC√
1 + ω2R2C2,
cos α =1√
1 + ω2R2C2.
Using these we can rewrite Qsp as:
Qsp =E0C√
1 + ω2R2C2(cos (α) cos (ωt) + sin (α) sin (ωt)).
If we use the trigonometric identity
cos (θ − φ) = cos (θ) cos (φ) + sin (θ) sin (φ)
we can rewrite the above equation as:
Qsp =E0C√
1 + ω2R2C2cos (ωt− α).
This is our desired form.
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3.7.17 For the RLC circuit pictured below find the current 1(t) using thegiven values of R, L, C and V(t), and the given initial values.
L
(7C
R = 16Q, L = 2H, C = .02F;
E(t) = 100V; 1(0) = 0, Q(0) 5.
Solution - The differential equation that governs this system is:
We can rewrite this as:
21’+161-i-50Q= 100.
I’ + 81 + 25Q 50.
The corresponding homogeneous equation is:
I’ + 81 + 25Q = 0.
The characteristic polynomial for this linear homogeneous ODE is:
r+8r+25,
8
which has roots:
−8 ±√
82 − 4(1)(25)
2= −4 ± 3i.
So, the corresponding homogeneous solution is:
Qh(t) = c1e−4t cos (3t) + c2e
−4t sin (3t).
As for a particular solution, we guess our particular solution is of theform Qp(t) = A. Plugging this into our ODE we get:
25A = 50,
so A = 2. Therefore the solution to our differential equation is:
Q(t) = c1e−4t cos (3t) + c2e
−4t sin (3t) + 2.
Its derivative is:
I(t) =−3c1e
−4t sin (3t) − 4c1e−4t cos (3t) + 3c2e
−4t cos (3t) − 4c2e−4t sin (3t).
Plugging in our initial conditions gives us:
Q(0) = c1 + 2 = 5,
I(0) = −4c1 + 3c2 = 0.
Solving this system of equations gives us c1 = 3, c2 = 4. So, oursolution is:
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Q(t) = 3e−4t cos (3t) + 4e−4t sin (3t) + 2,
and
I(t) = −25e−4t sin (3t).
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3.7.19 Same instructions as Problem 3.7.17, but with the values:
R = 60Ω, L = 2H , C = .0025F ;
E(t) = 100e−10tV ; I(0) = 0, Q(0) = 1.
Solution - The differential equation that governs this circuit is:
2I ′ + 60I + 400Q = 100e−10t.
We can rewrite this ODE as:
I ′ + 30I + 200Q = 50e−10t.
The corresponding homogeneous equation is:
I ′ + 30I + 200Q = 0.
This has characteristic polynomial:
r2 + 30r + 200 = (r + 20)(r + 10).
The roots of this polynomial are: r = −10,−20, and so the homoge-neous solution is:
Qh = c1e−20t + c2e
−10t.
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As for a particular solution, our first guess is the particular solutionwill be of the form Ae−10t, but this would not be linearly independentof our homogeneous solution, so we need to use the guess Ate−10t.With this guess we get:
Qp = Ate−10t,
Q′p = A(−10te−10t + e−10t),
Q′′p = A(100te−10t − 20e−10t).
Plugging these into the ODE we get:
100Ate−10t − 20Ae−10t − 300Ate−10t + 30Ae−10t + 200Ate−10t =10Ae−10t = 50e−10t
⇒ A = 5.
So, our solution is:
Q(t) = 5te−10t + c1e−20t + c2e
−10t.
Its derivative is:
Q′(t) = I(t) = −50te−10t + 5e−10t − 20c1e−20t − 10c2e
−10t.
Plugging in our initial conditions we get:
Q(0) = c1 + c2 = 1,
I(0) = 5 − 20c1 − 10c2 = 0.
Solving this system we get c1 =3
2, c2 = −1
2, and our current will be:
I(t) = 10e−20t − 10e−10t − 50te−10t.
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Section 3.8 - Endpoint Problems and Eigenvalues
3.8.1 For the eigenvalue problem
y′′ + λy = 0; y′(0) = 0, y(1) = 0,
first determine whether λ = 0 is an eigenvalue; then find the positiveeigenvalues and associated eigenfunctions.
Solution - First, if λ = 0 then the solution to the differential equation
y′′ = 0
is
y = Ax + B.
From this we get y′ = A, and so if y′(0) = 0 we must have A = 0. Thiswould mean y = B, and if y(1) = 0 then B = 0. So, only the trivialsolution A = B = 0 works, and therefore λ = 0 is not an eigenvalue.
For λ > 0 the characteristic polynomial for our linear differentialequation is:
r2 + λ = 0,
which has roots r = ±√−λ. The corresponding solution to our ODE
will be:
y = A cos (√
λx) + B sin (√
λx).
with derivative
y′ = −A√
λ sin (√
λx) + B√
λ cos (√
λx).
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So, y′(0) = B√
λ, and therefore if y′(0) = 0 then we must have B = 0,as λ > 0. So, our solution must be of the form:
y = A cos (√
λx).
If we plug in y(1) = 0 we get:
y(1) = A cos (√
λ) = 0.
If A 6= 0 we must have cos(√
λ) = 0, which is true only if√
λ =π
2+ nπ. So, the eigenvalues are:
λn =
(
π
(
1
2+ n
))2
, with n ∈ N,
and corresponding eigenfunctions
yn = cos((π
2+ nπ
)
x)
.
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3.8.3 Same instructions as Problem 3.8.1, but for the eigenvalue problem:
y′′ + λy = 0; y(−π) = 0, y(π) = 0.
Solution - If λ = 0 then, as in Problem 3.8.1, our solution will be ofthe form:
y = Ax + B.
This means y(π) = Aπ + B = 0, and y(−π) = −Aπ + B = 0. Addingthese two equations we get 2B = 0, which means B = 0. If B = 0then Aπ = 0, which means A = 0. So, the only solution is the trivialsolution A = B = 0, and therefore λ = 0 is not an eigenvalue.
Now if λ > 0 then again just as in Problem 3.8.1 we’ll have a solutionof the form:
y(x) = A cos (√
λx) + B sin (√
λx).
If we plug in our endpoint values we get:
y(π) = A cos (√
λπ) + B sin (√
λπ) = 0,
y(−π) = A cos (−√
λπ) + B sin (−√
λπ) =
A cos (√
λπ) − B sin (√
λπ) = 0,
where in the second line above we use that cos is an even function,while sin is odd.
If we add these two equations together we get:
2A cos (√
λπ) = 0.
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This is true if either A = 0 or√
λ =
(
1
2+ n
)
. If√
λ =
(
1
2+ n
)
then
y(π) = B sin
((
1
2+ n
)
π
)
= 0.
As sin
((
1
2+ n
)
π
)
= ±1 we must have B = 0.
On the other hand, if A = 0 above then we have:
y(π) = B sin (√
λπ).
If B 6= 0 then we must have√
λ = n. Combining our two results weget that the possible eigenvalues are:
λn =n2
4,
for n ∈ N, and n > 0, with corresponding eigenfunctions:
yn(x) =
cos(
n2x)
n oddsin
(
n2x)
n even
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3.8.5 Same instructions as Problem 3.8.1, but for the eigenvalue problem:
y′′ + λy = 0; y(−2) = 0, y′(2) = 0.
Solution - If λ = 0 then, just as in Problem 3.8.1, the solution to theODE will be:
y(x) = Ax + B,
y′(x) = A.
If we plug in our endpoint conditions we get y(−2) = −2A + B = 0and y′(2) = A = 0. These equations are satisfied if and only if A =B = 0, which is the trivial solution. So, λ = 0 is not an eigenvalue.
If λ > 0 then, just as in Problem 3.8.1, the solution to the ODE will beof the form:
y(x) = A cos (√
λx) + B sin (√
λx),
with
y′(x) = −A√
λ sin (√
λx) + B√
λ cos (√
λx).
Plugging in the endpoint conditions, and using that cos is even andsin is odd, we get:
y(−2) = A cos (−2√
λ) + B sin (−2√
λ) =
A cos (2√
λ) − B sin (2√
λ) = 0,
y′(2) = −A√
λ sin (2√
λ) + B√
λ cos (2√
λ) = 0.
If we divide both sides of the second equality by√
λ we get
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−A sin (2√
λ) + B cos (2√
λ) = 0.
From these equations we get:
A cos (2√
λ) = B sin (2√
λ) ⇒ A
B= tan (2
√λ),
B cos (2√
λ) = A sin (2√
λ) ⇒ B
A= tan (2
√λ).
So,
A
B=
B
A⇒ A2 = B2.
So, either A = B or A = −B.
If A = B then tan (2√
λ) = 1, which means 2√
λ =π
4+ nπ, and
therefore
λ =
((
1 + 4n
8
)
π
)2
.
If A = −B then tan (2√
λ) = −1, which means 2√
λ =3π
4+ nπ, and
therefore
λ =
((
3 + 4n
8
)
π
)2
.
So, the eigenvalues are:
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λn =
((
1 + 2n
8
)
π
)2
with n ∈ N and n > 0, with corresponding eigenfunctions:
yn =
cos((
1+2n8
)
πx)
+ sin((
1+2n8
)
πx)
n evencos
((
1+2n
8
)
πx)
− sin((
1+2n
8
)
πx)
n odd
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3.8.8 - Consider the eigenvalue problem
y′′ + λy = 0; y(0) = 0 y(1) = y′(1) (not a typo).;
all its eigenvalues are nonnegative.
(a) Show that λ = 0 is an eigenvalue with associated eigenfunctiony0(x) = x.
(b) Show that the remaining eigenfunctions are given by yn(x) =sin βnx, where βn is the nth positive root of the equation tan z =z. Draw a sketch showing these roots. Deduce from this sketchthat βn ≈ (2n + 1)π/2 when n is large.
Solution -
(a) - If λ = 0 then the solution to the ODE will be of the form:
y(x) = Ax + B,
with
y′(x) = A.
So, y(0) = B = 0, and y(1) = A = y′(1). So, any function of theform y(x) = Ax will work, and our eigenfunction for λ = 0 is:
y0 = x.
(b) - For λ > 0 the solutions will all be of the form:
y(x) = A cos (λx) + B sin (λx).
If we plug in y(0) = A = 0 we get the solutions are of the form:
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y(x) = Bsin(\/5x),
with
= Bvcos (\/I).
If we plug in the other endpoint values we get:
y(l) = Bsin(v) = BVcos(v’) = y’(l).
If B 0 then we must have:
tan(/)=
So, v”X works if it’s a root of the equation tan z = z, and if 8, is
the iith such root, then the associated eigenfunction is:
A sketchintersect:
sin (/3x).
of z and tan z are below. The roots are where they
As ii gets large it occurs at approximately
L
/277 + iN2
J7F.
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3.8.13 - Consider the eigenvalue problem
y′′ + 2y′ + λy = 0; y(0) = y(1) = 0.
(a) Show that λ = 1 is not an eigenvalue.
(b) Show that there is no eigenvalue λ such that λ < 1.
(c) Show that the nth positive eigenvalue is λn = n2π2 + 1, withassociated eigenfunction yn(x) = e−x sin (nπx).
Solution -
(a) - If λ = 1 then the characteristic polynomial is:
r2 + 2r + 1 = (r + 1)2,
which has roots r = −1,−1. So, −1 is a root with multiplicity 2. Thecorresponding solution to the ODE will be:
y(x) = Ae−x + Bxe−x.
If we plug in the endpoint values we get:
y(0) = A = 0,
y(1) = Ae−1 + Be−1 = Be−1 = 0.
From these we see the only solution is the trivial solution A = B = 0,so λ = 1 is not an eigenvalue.
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(b) - If λ < 1 then the characteristic polynomial will be:
r2 + 2r + λ,
which has roots
r =−2 ±
√
22 − 4(1)λ
2= −1 ±
√1 − λ.
If λ < 1 then√
1 − λ will be real, and the solution to our ODE will beof the form:
y(x) = Ae(−1+√
1−λ)x + Be(−1−√
1−λ)x.
Plugging in our endpoint values we get:
y(0) = A + B = 0,
y(1) = Ae−1+√
1−λ + Be−1−√
1−λ = 0.
From these we get, after a little algebra:
A(1 − e−2√
1−λ) = 0.
If λ < 1 then e−2√
1−λ < 1, and therefore 1 − e2√
1−λ > 0. So, for theabove equality to be true we must have A = 0, which means B = 0,and so the only solution is the trivial solution A = B = 0. Therefore,no value λ < 1 is an eigenvalue.
(c) - If λ > 1 then again using the roots from the quadratic equation inpart (b) we get that our solutions will be of the form:
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y(x) = Ae−x cos (√
λ − 1x) + Be−x sin (√
λ − 1x).
If we plug in the endpoint values we get:
y(0) = A = 0,
and so
y(x) = Be−x sin (√
λ − 1x).
If we plug in our other endpoint value we get:
y(1) = Be−1 sin (√
λ − 1) = 0.
If B 6= 0 then we must have sin (√
λ − 1) = 0, which is only possibleif
√λ − 1 = nπ,
⇒ λn = n2π2 + 1.
So, the eigenvalues are given above, and the corresponding eigen-functions are:
yn = e−x sin (nπx),
for n ∈ N, n > 0.
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Section 4.1 - First-Order Systems and Applications
4.1.1 - Transform the given differential equation into an equivalent systemof first-order differential equations.
x′′ + 3x′ + 7x = t2.
Solution - If we define x = x1 then define:
x′1 = x2,
x′2 = t2 − 3x2 − 7x1.
So, the system is:
x′1 = x2
x′2 = −7x1 − 3x2 + t2
.
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4.1.2 - Transform the given differential equation into an equivalent systemof first-order differential equations.
x(4) + 6x′′ − 3x′ + x = cos 3t.
Solution - Define x = x1. Then the equivalent system is:
x′1 = x2
x′2 = x3
x′3 = x4
x′4 = −6x3 + 3x2 − x1 + cos (3t)
.
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4.1.13 - Find the particular solution to the system of differential equationsbelow. Use a computer system or graphing calculator to construct adirection field and typical solution curves for the given system.
x′ = −2y, y′ = 2x; x(0) = 1, y(0) = 0.
Solution - If we differentiate y′ = 2x, we get y′′ = 2x′ = −4y. So, wehave the differential equation:
y′′ + 4y = 0.
The solution to this ODE is:
y(t) = A cos (2t) + B sin (2t).
Now,
x(t) =1
2y′ =
1
2(−2A sin (2t) + 2B cos (2t)) = −A sin (2t) + B cos (2t).
If we plug in x(0) = B = 1 and y(0) = A = 0 we get:
x(t) = cos (2t)
y(t) = sin (2t).
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More room, if necessary, for Problem 4.1.13.
Pirec
/‘4
o(
lviV
7’Ndfe; 5hoI) beCI(cIJ 40
A 1• Ibei
19
4.1.15 - Find the general solution to the system of differential equationsbelow. Use a computer system or graphing calculator to construct adirection field and typical solution curves for the given system.
x′ =1
2y, y′ = −8x.
Solution - If we differentiate y′ = −8x we get y′′ = −8x′ = −4y. So,our ODE is:
y′′ + 4y = 0.
The solution to this ODE is:
y(t) = A cos (2t) + B sin (2t).
The function x(t) is:
x(t) = −1
8y′(t) = −A
4sin (2t) +
B
4cos (2t).
So, the general solution to this system of ODEs is:
x(t) = −A
4sin (2t) +
B
4cos (2t)
y(t) = A cos (2t) + B sin (2t).
29
More room, if necessary, for Problem 4.1.15.
Pirecio
21
4.1.22 (a) - Beginning with the general solution of the system from Prob-lem 13, calculate x2 + y2 to show that the trajectories are circles.
(b) - Show similarly that the trajectories of the system from Problem15 are ellipses of the form 16x2 + y2 = C2.
(a) - The general solution to the system of ODEs from Problem 4.1.13 is:
x(t) = −A sin (2t) + B cos (2t)
y(t) = A cos (2t) + B sin (2t).
From these we get:
x(t)2 + y(t)2 = (−A sin (2t) + B cos (2t))2 + (A cos (2t) + B sin (2t))2
= A2 sin2 (2t) − 2AB sin (2t) cos (2t) + B2 cos2 (2t) + A2 cos2 (2t) +2AB sin (2t) cos (2t) + B2 sin2 (2t)
= A2 + B2.
So, circles.
(b) - The general solution to the system of ODEs from Problem 4.1.15 is:
x(t) = −A
4sin (2t) +
B
4cos (2t)
y(t) = A cos (2t) + B sin (2t).
So,
16x(t)2 = A2 sin2 (2t) − 2AB sin (2t) cos (2t) + B2 cos2 (2t),
y(t)2 = A2 cos2 (2t) + 2AB sin (2t) cos (2t) + B2 sin2 (2t).
Combining these we get 16x(t)2 + y(t)2 = A2 + B2 = C2. So, ellipses.
31
Section 4.2 - The Method of Elimination
4.2.1 - Find a general solution to the linear system below. Use a computersystem or graphing calculator to construct a direction field and typi-cal solution curves for the system.
x′ = −x + 3yy′ = 2y
Solution - The differential equation
y′ = 2y
has the solution
y(t) = Ae2t.
So,
x′ = −x + 3Ae2t ⇒ x′ + x = 3Ae2t.
This is a first-order linear ODE. Its integrating factor is:
ρ(t) = eR
1dt = et.
Multiplying both sides by this integrating factor our linear ODE be-comes:
d
dt
(
etx)
= 3Ae3t.
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Integrating both sides we get:
etx = Ae3t + B
x Ae2 + Be.
So, the general solution to this system is:
x(t) = Ae2t + Be_t,
j(t) = Ae2t.
We can write this in vector form as:
(I)4(i)2t+B(i)t
looks kind of like this:The direction field
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4.2.10 Find a particular solution to the given system of differential equa-tions that satisfies the given initial conditions.
x′ + 2y′ = 4x + 5y,
2x′ − y′ = 3x;
x(0) = 1, y(0) = −1.
Solution - If we add 2 times the second equation to the first we get:
5x′ = 10x + 5y.
If we subtract 2 times the first equation from the second we get:
−5y′ = −5x − 10y ⇒ 5y′ = 5x + 10y.
Differentiating 5x′ = 10x+5y and plugging in 5y′ = 5x+10y we get:
5x′′ = 10x′ + 5y′ = 10x′ + (5x + 10y)
⇒ 5x′′ = 10x′ + (5x + 10x′ − 20x)
⇒ 5x′′ = 20x′ − 15x
⇒ x′′ = 4x′ − 3x.
The linear homogeneous differential equation x′′ − 4x′ + 3x = 0 hascharacteristic equation:
r2 − 4r + 3 = (r − 3)(r − 1).
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So, the roots are r = 3, 1, and the general solution to the ODE is:
x(t) = c1e3t + c2e
t.
From the equation 5y′ = 5x + 10y we get y′ = x + 2y, and therefore:
y′ − 2y = c1e3t + c2e
t.
If we multiply both sides by the integrating factor e−2t we get:
d
dt
(
e−2ty)
= c1et + c2e
−t.
Integrating both sides we get:
e−2ty = c1et − c2e
−t + C,
and so:
y(t) = c1e3t − c2e
t + Ce2t.
Plugging this into any of the equations in our system gives us C = 0.So,
y(t) = c1e3t − c2e
t.
We can write this solution in matrix form as:
x(t) = c1
(
11
)
e3t + c2
(
1−1
)
et.
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If we plug in
x(0) =
(
1−1
)
= c1
(
11
)
+ c2
(
1−1
)
we can see immediately that c1 = 0 and c2 = 1. So, the solution toour initial value problem is:
x(t) =
(
1−1
)
et.
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4.2.19 Find a general solution to the given system of differential equations.
x′ = 4x − 2y,
y′ = −4x + 4y − 2z,
z′ = −4y + 4z.
Solution - If we differentiate the first equation we get:
x′′ = 4x′ − 2y′ = 4x′ − 2(−4x + 4y − 2z)
⇒ x′′ = 4x′ + 8x − 8y + 4z.
Differentiating again we get:
x(3) = 4x′′ + 8x′ − 8y′ + 4z′ = 4x′′ + 8x′ − 8y′ + 4(−4y + 4z)
⇒ x(3) = 4x′′ + 8x′ − 8y′ − 16y + 16z
⇒ x(3) = 4x′′ + 8x′ − 8y′ − 16y + 8(−y′ − 4x + 4y)
⇒ x(3) = 4x′′ + 8x′ − 16y′ + 16y − 32x
⇒ x(3) = 4x′′ + 8x′ − 8(4x′ − x′′) + 8(4x − x′) − 32x
⇒ x(3) = 12x′′ − 32x′ ⇒ x(3) − 12x′′ + 32x′ = 0.
The characteristic equation for this ODE is
r3 − 12r2 + 32r = r(r − 8)(r − 4).
So,
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x(t) = c1 + c2e8t + c3e
4t.
From this we get:
x′(t) = 8c2e8t + 4c3e
4t
and
y(t) = 2x − 1
2x′ = 2c1 − 2c2e
8t.
Finally,
z(t) = −2x′(t) + 2y(t) − 1
2y′(t) = 2c1 + 2c2e
8t − 2c3e4t.
So,
x(t) = c1 + c2e8t + c3e
4t,
y(t) = 2c1 − 2c2e8t,
z(t) = 2c1 + 2c2e8t − 2c3e
4t.
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4.2.28 For the system below first calculate the operational determinant todetermine how many arbitrary constants should appear in a generalsolution. Then attempt to solve the system explicitly so as to findsuch a general solution.
(D2 + D)x + D2y = 2e−t
(D2 − 1)x + (D2 − D)y = 0
Solution - The operational determinant of the system above is:
(D2 +D)(D2 −D)−D2(D2 − 1) = D4 −D3 +D3 −D2 −D4 +D2 = 0.
So, there are 0(!) arbitrary constants. How is this possible? Well, ifwe subtract the second relation from the first we get:
(D + 1)x + Dy = 2e−t
⇒ Dy = 2e−t − (D + 1)x
⇒ D2y = −2e−t − (D2 + D)x
⇒ (D2 + D)x + D2y = −2e−t.
However, this cannot be, as our first relation above is:
(D2 + D)x + D2y = 2e−t,
and 2e−t 6= −2e−t. So, there is no solution to the system.
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