Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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pls answer the exercises in the 2nd picture and I provide some example on guide you on how to do it thankyou!(pls answer 2-5)
Transcribed Image Text:Solutions of General Linear Homogeneous Equations
Solution: The differential equation is translated in terms of the operation D as
ye = Cie-2r+ e2x (C2+ xC3)
ye = Cie-2x+ C262r+ xCae2x or
D'y – D'y - 2Dy = 0 or (D3- D2 – 2D) y = 0
Example 3: Solve the equation (4D- 4D3- 23D?+12D+36) y=0
Solutions of General Linear Homogeneous Equations
The auxiliary equation is f(m) = m3 – m2 - 2m = m (m2- m - 2) = m (m - 2) (m + 1) = 0
Solution: The auxiliary equation is
The form of a general linear homogeneous differential equation with constant
whose roots are 0, 2. -1. Then the general solution is
f(m) = 4m - 4m - 23m2+ 12m + 36 = 0
coefficients is given by
ye= Cieor+ Cze2x+ C3e-1x
To find the roots, we use synthetic division,
d"y
dn-ly
dy
+*** ....tan-1 + any =
ye = Ci+ C2e2x + Cie-x
2
4
-4
-23
12
36
XP op
8
This form may also be written in a compact form as f(D)y = 0 where f(D) is a linear
4- 13+ 6y = 0
8
-30
-36
Example 4: Solve the equation
4
4
-15
-18
differential operation.
Solution: The differential equation is translated in terms of the operation D as
Theorem 1. If m is a root of the equation f(m)=0, then
Thus, m=2 is a root. Again by synthetic division;
4Day – 13Dy + 6y = 0 or (4D3 – 13D + 6) y =0
f(D)y = f(D)em = 0.
2|
-15
4
4
-18
Tne auxiliary equation is f(m) = 4m3 – 13m +6=0.
8.
24
18
The above theorem asserts that y = ex is a solution of f(D)y = 0. In general, we
%3=
By synthetic division, we have:
4
12
have the solution as y = Cems, C is constant. We define f(m) = 0 as the auxiliary
-2
4
-13
6
Thus, m=2 is another root.
equation.
-8
16
-6
Thus,
Distinct Real Roots of Auxiliary Equation
4
-8
3
f(m) = 4m – 4m – 23m2+ 12m + 36 = 0
Let the roots of the auxiliary equation by mı, mz, .., me. Then these will produce k
Thus, 4m3 – 13m + 6 = (m + 2) (4m2 - 8m + 3) = (m+ 2) (2m - 3) (2m – 1) = 0.
solutions as follows:
The roots of the auxiliary equation are 2 The general solution is
(m- 2) (m – 2) (4m2+ 12m + 9) = 0
(m – 2) (m - 2) (2m + 3) (2m + 3) = 0
-3 -3
3x
yi= emix, y2 = em2x, ... yk = emix
Y. = Ce-2x + Cze + Cze?
Thus, the roots of the auxiliary equation are 2.77. The solution is
The general solution, denoted by y. of the general linear homogeneous differential
-3x
equation with mı, m2,..., ma roots of the auxiliary equation is
Repeated Real Roots of Auxiliary Equation
Ye = e2*(C, + xC2) +eT (C3 + xC4)
If the auxiliary equation has a repeated root mı, then the general solution is
yc= Ciemix+ Czemaxt.. +Ckemix
Example 4: Solve the equation (D++ 3D3- 6D2– 28D – 24) y=0
yc= Ciemix+ xC26mix or
ye = emix (C1+ xC2)
Example 1: Solve the differential equation y+y-2y=0.
Solution: The auxiliary equation is f(m) = mt + 3m – 6m2 - 28m – 24 = 0
If the auxiliary equation has roots mi that are repeated 3 times, then the general solution is
Solution: The differential equation is translated in terms of the operation D as
To find the roots, we use synthetic division,
ye = Ciemix+ xC2emix+ x2C36mix
ye = 6mix (C1+ xC2+ x2Ca)
or
D'y + Dy - 2y = 0 or (D2+D- 2) y = 0
-2
1
3
-6
-28
-24
If the auxiliary equation has roots mi that are repeated n times, then the general solution is
-2
1
The auxiliary equation is f(m) = m2+ m – 2= (m +2) (m – 1) = 0.
-2
16
24
yc= Ciemix + xC2emix+ x2Caemixt... +xn-1Cnemix
y = emlx (Ci+ xC2+ xCs+... +x-1Cn)
1
-8
-12
The roots of f(m) = 0 are mi=-2, mz= 1.
Thus, m=-2 is a root. Again by synthetic division;
Thus, the general solution is
Example 1: Solve the differential equation y- 2y+y=0.
Solution: The differential equation is translated in terms of the operation D as
3
1
1
-8
-12
ye = Cie-2x + Czex
3
12
12
Example 2: Solve the equation (D2+2D - 1) y=0.
D'y - 2Dy + y= 0 or (D2- 2D + 1) y =0
1
4
4
The auxiliary equation is f(m) = m? - 2m + 1= (m – 1) (m - 1) = 0.
Thus, m=3 is another root.
Solution: The auxiliary equation is f(m) = m2+ 2m –1=0. The roots of the auxiliary
The roots of f(m) = 0 are mi= 1, mz= 1. The roots are repeated. Thus, the general solution is
Thus,
equation are obtained by quadratic formula,
ye= Cier+xCzex
f(m) = m+ 3m - 6m2 - 28m – 24 = 0
-2 ± /2² – 4(1)(-1)_ -2± v®_ -2 ± 2v2
= -1t v2
m%3D
2(1)
Example 2: Solve (D²– 4) (D– 2) y=0.
(m + 2) (m – 3) (m2 + 4m + 4) = 0
2
2
Thus, the solution is given by
Solution: The auxiliary equation is
(т + 2) (m - з) (m+2) (m +2) %3D0
f(m) = (m2 – 4) (m – 2) = 0
Thus, the roots of the auxiliary equation are -2, 3, -2, -2. The solution is
y = Cie (-1+v2)z+ Cre (-1-v2)
(т + 2) (m - 2) (т - 2) %3D0
ye = e-2x (C1+ xC2 + x2C3) + C4e3x
Example 3: Solve the equation y"-y-2y=0.
The roots are -2, 2, 2. Thus, the general solution is
![Imaginary Roots of Auxiliary Equations
ye= Ci cos 2x + C2 sin 2x
So we have distinct real root (3), repeated real roots (-2, -2). imaginary roots (+)where a=0 and
Consider the differential equation (boD"+ bıD" 14...+b 1D +ba) y=0
Example 2: Solve the equation (D-1) y=0
b=1.
Thus, the general solution is Ye = Ce3* + Cze 2x + xCze-2x + e0x (C, cos x + C, sin x
Ye = C,e + Cze=2* + xCze=z* + C4 cos x + Cs sin x )
Or the more compact form f(D)y=0
Solution: The auxiliary equation is
f(m) = m - 1 = 0
Supposed the root of the auxiliary equation is imaginary, say a + bi. It has to be noted
(m? – 1) (m²+ 1) = 0
that the conjugate a - bi is also a root of f(D)y = 0. Then the solution is given by the
(m + 1) (m – 1) (m? + 1) = 0
Repeated Imaginary Roots of Auxiliary Equations
following computations:
Repeated imaginary roots lead to solutions analogous to those of repeated real roots.
For example, if the roots m = a t bi are repeated twice then, the solution is given by
Y. = eax (C, cos bx + Cz sin bx) + xe®*(Cz cos bx + C4 sin bx) or
ye= Cse(a+b() x+ Cse(a-b) z
The roots of the auxiliary equation are
-1, 1, ti
ye = Caeaxebix + Cseaxe-bix
So we have distinct real roots of -1 and 1 and imaginary roots of ti with a=0 and b=1.
Y = (eax cos bx)(C, + xC2) + (ea* sin bx)(C3 + xC4)
From calculus, ei is defined by
Thus, the general solution is
eil = cos 0 + i sin 6
If the roots m = a ± biare repeated thrice then, the solution is given by
ye= Cie*+ Cze" + eor (C3 cos x + Ci sin x)
The above is called the Euler Equation.
Ye = (eax cos bx)(C, + xC2 + x?C3) + (eax sin bx)(C, + xC5 + x?C6)
ye= Cie+ Cze + Ca cos x + Ca sin x
Example 1: Solve the equation (D3 + D)?y = 0
Solution: The auxiliary equation is f(m) = (m³ + m)² = 0
The following illustrate the Euler Equation:
Example 3: Solve the equation (D3+3D?+ 3D+ 2) y =0
1. e = cos 8 + i sin 6
Solution: The auxiliary equation is
[m(m? + 1)]? = 0
2. eiar = cos ax +i sinax
f(m) = m + 3m? + 3m +2 = 0
m?(m? + 1)? = 0
3. e-ia = cos(-a) +i sin(-a) = cos a - i sin a
Solving for m we use synthetic equation,
4. e(a+bl) x= eax(ebix) = eax (cos bx +i sinbx)
The roots of the auxiliary equation are m = 0,0, ti, ti
-2|
1
3
2
With a=0, b=1, then the solution of the given differential equation is
5. ele-bi) *= 6a" (e-ber) = e*{cos(-bx) +i sin(-bx)} = 6* (cos bx – i sin bx}
-2
-2
-2
Y. = C,e0x + xCze0x + (e0x cos x)(C, + xC4) + (e0x sin x)(C+ xCg)
1
1
1
Thus, the solution of f(D)y = 0 with imaginary roots for its auxiliary equation is now given by
e = C, + xC2 + (cos x)(C3 + xC4) + (sin x)(Cs + xC6)
ye= Caeuxebix + Cieaxe-bix
Thus, m=-2 is a root. To find the root of m²+m +1=0, Use quadratic formula,
-1+y12 - 4(1)(1) -1+v-3 -1tiv3
y= Cser (cos bx +i sin bx) + Cie" (cos bx – i sin bx}
2nd Semester of A. Y. 2020-2021
EXERCISE NO. 06: Solutions of General Linear Homogeneous Equations
2(1)
2
2
y= Caer (cos bx) + Cae* (i sin bx) + Cier (cos bx) – Cie* (i sin bx)
-1t3
Thus, the roots of auxiliary equation are -2,- with a =,b=
Simplify,
Thus, the general solution is
I. Exercise
ye = er (cos bx) (C3+ Ca) + e (i sinbx) (Ca- Ca)
Ye = C,e-2x + e C2 cos
V3
-x)+ C3 sin
If we let C3+ C= Cı and i (Cs- Ca) = C2, where Ci and C2 are new arbitrary constants,
Direction: Solve the following equations with complete solution. Box your final answer.
Make your solutions easy to read.
then the general solution of f(D)y = 0 is
Example 4: Solve the equation (D5+ D*- 7D3–11D²- 8D – 12) y=0
y. = Cs" (cos bx) + C:e(sinbx) or y.= er (Ci cos bx + C2 sin bx)
Solution: The auxiliary equation is f(m) = m5+ m- 7m3– 11m? - 8m – 12 = 0
Thus, to determine the solution of a homogeneous differential equation with
imaginary roots of the auxiliary equation, we need to find the values of an and
b, and substitute it in the above general solution.
2. y" - y" + 9y' – 9y = 0
3. (D* + 4D²)y = 0
4. (D4 – 6D³ + 13D² – 12D + 4)y = 0
5. (D6 + 9D* + 24D² + 16)y = 0
Solving for m, we use synthetic equation,
3
-7
1
1
-11
-8
-12
12
15
12
12
Example 1: Solve the equation (D+4) y=0
1
4
4
4
Solution: The auxiliary equation is S(m) = m² + 4 = 0
-2
-2
-4
-4
m? = -4
-2|
1
2
1
2
-2
-2
m = tv-4
1
1
m = £21
Thus, we get
m5 + m* – 7m3 – 11m² – 8m – 12 = 0
The roots of the auxiliary equation are imaginary. From which a=0 and b=2. Thus, the
(m - 3)(m + 2)(m + 2)(m? + 1) = 0
general solution is
y = e* (Ci cos 2x + Cz sin 2x)
The roots of /(m) = 0 are 3,-2, -2, t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59e37926-5a74-4df7-8f60-ab86931c1501%2F469c73cb-8d1c-40a2-af49-a8ee83315541%2Fzvbn36f_processed.png&w=3840&q=75)
Transcribed Image Text:Imaginary Roots of Auxiliary Equations
ye= Ci cos 2x + C2 sin 2x
So we have distinct real root (3), repeated real roots (-2, -2). imaginary roots (+)where a=0 and
Consider the differential equation (boD"+ bıD" 14...+b 1D +ba) y=0
Example 2: Solve the equation (D-1) y=0
b=1.
Thus, the general solution is Ye = Ce3* + Cze 2x + xCze-2x + e0x (C, cos x + C, sin x
Ye = C,e + Cze=2* + xCze=z* + C4 cos x + Cs sin x )
Or the more compact form f(D)y=0
Solution: The auxiliary equation is
f(m) = m - 1 = 0
Supposed the root of the auxiliary equation is imaginary, say a + bi. It has to be noted
(m? – 1) (m²+ 1) = 0
that the conjugate a - bi is also a root of f(D)y = 0. Then the solution is given by the
(m + 1) (m – 1) (m? + 1) = 0
Repeated Imaginary Roots of Auxiliary Equations
following computations:
Repeated imaginary roots lead to solutions analogous to those of repeated real roots.
For example, if the roots m = a t bi are repeated twice then, the solution is given by
Y. = eax (C, cos bx + Cz sin bx) + xe®*(Cz cos bx + C4 sin bx) or
ye= Cse(a+b() x+ Cse(a-b) z
The roots of the auxiliary equation are
-1, 1, ti
ye = Caeaxebix + Cseaxe-bix
So we have distinct real roots of -1 and 1 and imaginary roots of ti with a=0 and b=1.
Y = (eax cos bx)(C, + xC2) + (ea* sin bx)(C3 + xC4)
From calculus, ei is defined by
Thus, the general solution is
eil = cos 0 + i sin 6
If the roots m = a ± biare repeated thrice then, the solution is given by
ye= Cie*+ Cze" + eor (C3 cos x + Ci sin x)
The above is called the Euler Equation.
Ye = (eax cos bx)(C, + xC2 + x?C3) + (eax sin bx)(C, + xC5 + x?C6)
ye= Cie+ Cze + Ca cos x + Ca sin x
Example 1: Solve the equation (D3 + D)?y = 0
Solution: The auxiliary equation is f(m) = (m³ + m)² = 0
The following illustrate the Euler Equation:
Example 3: Solve the equation (D3+3D?+ 3D+ 2) y =0
1. e = cos 8 + i sin 6
Solution: The auxiliary equation is
[m(m? + 1)]? = 0
2. eiar = cos ax +i sinax
f(m) = m + 3m? + 3m +2 = 0
m?(m? + 1)? = 0
3. e-ia = cos(-a) +i sin(-a) = cos a - i sin a
Solving for m we use synthetic equation,
4. e(a+bl) x= eax(ebix) = eax (cos bx +i sinbx)
The roots of the auxiliary equation are m = 0,0, ti, ti
-2|
1
3
2
With a=0, b=1, then the solution of the given differential equation is
5. ele-bi) *= 6a" (e-ber) = e*{cos(-bx) +i sin(-bx)} = 6* (cos bx – i sin bx}
-2
-2
-2
Y. = C,e0x + xCze0x + (e0x cos x)(C, + xC4) + (e0x sin x)(C+ xCg)
1
1
1
Thus, the solution of f(D)y = 0 with imaginary roots for its auxiliary equation is now given by
e = C, + xC2 + (cos x)(C3 + xC4) + (sin x)(Cs + xC6)
ye= Caeuxebix + Cieaxe-bix
Thus, m=-2 is a root. To find the root of m²+m +1=0, Use quadratic formula,
-1+y12 - 4(1)(1) -1+v-3 -1tiv3
y= Cser (cos bx +i sin bx) + Cie" (cos bx – i sin bx}
2nd Semester of A. Y. 2020-2021
EXERCISE NO. 06: Solutions of General Linear Homogeneous Equations
2(1)
2
2
y= Caer (cos bx) + Cae* (i sin bx) + Cier (cos bx) – Cie* (i sin bx)
-1t3
Thus, the roots of auxiliary equation are -2,- with a =,b=
Simplify,
Thus, the general solution is
I. Exercise
ye = er (cos bx) (C3+ Ca) + e (i sinbx) (Ca- Ca)
Ye = C,e-2x + e C2 cos
V3
-x)+ C3 sin
If we let C3+ C= Cı and i (Cs- Ca) = C2, where Ci and C2 are new arbitrary constants,
Direction: Solve the following equations with complete solution. Box your final answer.
Make your solutions easy to read.
then the general solution of f(D)y = 0 is
Example 4: Solve the equation (D5+ D*- 7D3–11D²- 8D – 12) y=0
y. = Cs" (cos bx) + C:e(sinbx) or y.= er (Ci cos bx + C2 sin bx)
Solution: The auxiliary equation is f(m) = m5+ m- 7m3– 11m? - 8m – 12 = 0
Thus, to determine the solution of a homogeneous differential equation with
imaginary roots of the auxiliary equation, we need to find the values of an and
b, and substitute it in the above general solution.
2. y" - y" + 9y' – 9y = 0
3. (D* + 4D²)y = 0
4. (D4 – 6D³ + 13D² – 12D + 4)y = 0
5. (D6 + 9D* + 24D² + 16)y = 0
Solving for m, we use synthetic equation,
3
-7
1
1
-11
-8
-12
12
15
12
12
Example 1: Solve the equation (D+4) y=0
1
4
4
4
Solution: The auxiliary equation is S(m) = m² + 4 = 0
-2
-2
-4
-4
m? = -4
-2|
1
2
1
2
-2
-2
m = tv-4
1
1
m = £21
Thus, we get
m5 + m* – 7m3 – 11m² – 8m – 12 = 0
The roots of the auxiliary equation are imaginary. From which a=0 and b=2. Thus, the
(m - 3)(m + 2)(m + 2)(m? + 1) = 0
general solution is
y = e* (Ci cos 2x + Cz sin 2x)
The roots of /(m) = 0 are 3,-2, -2, t
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