Direction: Solve the following equations with complete solution. Box your final answer. Make your solufions easy to read. 5. (D* + 9D* + 24D² + 16)y = 0
Direction: Solve the following equations with complete solution. Box your final answer. Make your solufions easy to read. 5. (D* + 9D* + 24D² + 16)y = 0
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!
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+ Cz 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+ bıDn 14...+b 1D+ba) y=0
Example 2: Solve the equation (D- 1) y=0
b=1.
Or the more compact form f(D)y=0
Thus, the general solution is Ye = Ce3x + Cze-2x + xCze-2x + e0x (C, cos x + C, sin x
Ye = Ce + C2e 2x + xCze2x + C4 cos x + C sin x
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
(m2 – 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) (m2 + 1) = 0
Repeated Imaginary Roots of AUxiliary Equations
following computations:
The roots of the auxiliary equation are
Repeated imaginary roots lead to solutions analogous to those of repeated real roots.
For example, if the roots m = a ± bi gre repeated twice then, the solution is given by
Ye = eax (C, cos bx + Cz sin bx) + xeax (C; cos bx + C4 sin bx) or
ye = Cse(a+b() x+ Cie(a-b) x
-1, 1, ti
y= Caeaxebix + Cieuxe-bix
So we have distinct real roots of -1 and 1 and imaginary roots of ti with a=0 and b=1.
From calculus, ei is defined by
Ye = (eax cos bx)(G + xC2) + (ea* sin bx)(C3 + xC4)
Thus, the general solution is
eii = cos 8 + i sin e
If the roots m= at bi are repeated thrice then, the solution is given by
Y. = (eax cos bx)(C, + xCz + x?C3) + (eax sin bx)(C, + xC; + x?C,)
ye= Cie+ Cze + sx (Ca cos x+ Cı sin x)
The above is called the Euler Equation.
ye = Cie-+ Cze+ Cs cos x + Casin x
Example 1: Solve the equation (D3 + D)?y = 0
The following illustrate the Euler Equation:
1. si = cos 6 +i sin 6
Example 3: Solve the equation (D3+3D2+3D+2) y=0
Solution: The auxiliary equation is
f(m) = (m + m)? = 0
Solution: The auxiliary equation is
[m(m? + 1)] = 0
2. eiax= cos ax +i sinax
f(m) = ma + 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+bi) x= eux(ebix) = eax (cos bx +i sinbx)
The roots of the auxiliary equation are m = 0,0,±i, ti
-2
1
3
2
With a=0, b=1, then the solution of the given differential equation is
5. ele-b) x= gar(e-bis) = gar{cos(-bx) +i sin(-bx)} = eux {cos bx – i sin bx}
-2
1
-2
1
-2
Y. = C,e0x + xC2e0x + (e0* cos x)(C, + xC4) + (e0* sin x)(C,+ xC6)
1
Thus, the solution of f(D)y = 0 with imaginary roots for its auxiliary equation is now given by
Ye = C + xC2 + (cos x)(C, + xC,) + (sin x)(C; + xC6)
y = Cieaxebix + Cseuxe-bix
Thus, m=-2 is a root. To find the root of m? + m +1= 0, use quadratic formula,
-1±/1 - 4(1)(1) -1+v-3 -1tiv3
ye= C3ear (cos bx +i sin bx) + Ciear (cos bx – i sin bx}
m =
2(1)
2
2
ye = Cae (cos bx) + Csear (i sin bx) + Csear (cos bx) – Ciear (i sin bx)
-2,- with a =b =
2nd Semester of A.Y. 2020-2021
EXERCISE NO. 06: Solutions of General Linear Homogeneous Equations
Thus, the roots of auxiliary equation are
Simplify,
Thus, the general solution is
V3
x) + C3 sin-
y= eaz (cos bx) (Cs+ Ca) + ear (i sinbx) (Ca- C4)
Ye = C,e=2* + ec,
Cz cos
I Exercise
If we let C3+ C = Cı and i (C3- C:) = C2, where Ci and C2 are new arbitrary constants,
then the general solution of f(D)y = 0 is
Example 4: Solve the equation (D5+ D-7D3- 11D2- 8D – 12) y=0
Direction: Solve the following equations with complete solution. Box your final answer.
y = Ciear (cos bx) + Czea (sinbx) or ye= ear (C1cos bx + C2 sin bx)
Make your solutions easy to read.
Solution: The auxiliary equation isf(m) = m5+ mt- 7m3- 11m2- 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.
Solving for m, we use synthetic equation,
3
|
-7
-11
-8
-12
1
12
15
12
12
Example 1: Solve the equation (D2+4) y=0
-2|
1
4
4
4
Solution: The auxiliary equation is f(m) = m? + 4 = 0
-2
-4
1
-2
2
-4
-2
5. (D6 + 9D* + 24D² + 16)y = 0
m2 = -4
1
-2
-2
m = tv-4
1
1
m = +21
Thus, we get
m5 + m* - 7m - 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) = o
general solution is
y = el (Ci cos 2x + C2 sin 2x)
The roots of /f (m) = 0 are 3, -2,-2, ti](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59e37926-5a74-4df7-8f60-ab86931c1501%2F1dea695b-b5c5-4769-8e52-67b75e9ed2ab%2F8ztxp1a_processed.png&w=3840&q=75)
Transcribed Image Text:Imaginary Roots of Auxiliary Equations
ye= Ci cos 2x+ Cz 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+ bıDn 14...+b 1D+ba) y=0
Example 2: Solve the equation (D- 1) y=0
b=1.
Or the more compact form f(D)y=0
Thus, the general solution is Ye = Ce3x + Cze-2x + xCze-2x + e0x (C, cos x + C, sin x
Ye = Ce + C2e 2x + xCze2x + C4 cos x + C sin x
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
(m2 – 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) (m2 + 1) = 0
Repeated Imaginary Roots of AUxiliary Equations
following computations:
The roots of the auxiliary equation are
Repeated imaginary roots lead to solutions analogous to those of repeated real roots.
For example, if the roots m = a ± bi gre repeated twice then, the solution is given by
Ye = eax (C, cos bx + Cz sin bx) + xeax (C; cos bx + C4 sin bx) or
ye = Cse(a+b() x+ Cie(a-b) x
-1, 1, ti
y= Caeaxebix + Cieuxe-bix
So we have distinct real roots of -1 and 1 and imaginary roots of ti with a=0 and b=1.
From calculus, ei is defined by
Ye = (eax cos bx)(G + xC2) + (ea* sin bx)(C3 + xC4)
Thus, the general solution is
eii = cos 8 + i sin e
If the roots m= at bi are repeated thrice then, the solution is given by
Y. = (eax cos bx)(C, + xCz + x?C3) + (eax sin bx)(C, + xC; + x?C,)
ye= Cie+ Cze + sx (Ca cos x+ Cı sin x)
The above is called the Euler Equation.
ye = Cie-+ Cze+ Cs cos x + Casin x
Example 1: Solve the equation (D3 + D)?y = 0
The following illustrate the Euler Equation:
1. si = cos 6 +i sin 6
Example 3: Solve the equation (D3+3D2+3D+2) y=0
Solution: The auxiliary equation is
f(m) = (m + m)? = 0
Solution: The auxiliary equation is
[m(m? + 1)] = 0
2. eiax= cos ax +i sinax
f(m) = ma + 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+bi) x= eux(ebix) = eax (cos bx +i sinbx)
The roots of the auxiliary equation are m = 0,0,±i, ti
-2
1
3
2
With a=0, b=1, then the solution of the given differential equation is
5. ele-b) x= gar(e-bis) = gar{cos(-bx) +i sin(-bx)} = eux {cos bx – i sin bx}
-2
1
-2
1
-2
Y. = C,e0x + xC2e0x + (e0* cos x)(C, + xC4) + (e0* sin x)(C,+ xC6)
1
Thus, the solution of f(D)y = 0 with imaginary roots for its auxiliary equation is now given by
Ye = C + xC2 + (cos x)(C, + xC,) + (sin x)(C; + xC6)
y = Cieaxebix + Cseuxe-bix
Thus, m=-2 is a root. To find the root of m? + m +1= 0, use quadratic formula,
-1±/1 - 4(1)(1) -1+v-3 -1tiv3
ye= C3ear (cos bx +i sin bx) + Ciear (cos bx – i sin bx}
m =
2(1)
2
2
ye = Cae (cos bx) + Csear (i sin bx) + Csear (cos bx) – Ciear (i sin bx)
-2,- with a =b =
2nd Semester of A.Y. 2020-2021
EXERCISE NO. 06: Solutions of General Linear Homogeneous Equations
Thus, the roots of auxiliary equation are
Simplify,
Thus, the general solution is
V3
x) + C3 sin-
y= eaz (cos bx) (Cs+ Ca) + ear (i sinbx) (Ca- C4)
Ye = C,e=2* + ec,
Cz cos
I Exercise
If we let C3+ C = Cı and i (C3- C:) = C2, where Ci and C2 are new arbitrary constants,
then the general solution of f(D)y = 0 is
Example 4: Solve the equation (D5+ D-7D3- 11D2- 8D – 12) y=0
Direction: Solve the following equations with complete solution. Box your final answer.
y = Ciear (cos bx) + Czea (sinbx) or ye= ear (C1cos bx + C2 sin bx)
Make your solutions easy to read.
Solution: The auxiliary equation isf(m) = m5+ mt- 7m3- 11m2- 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.
Solving for m, we use synthetic equation,
3
|
-7
-11
-8
-12
1
12
15
12
12
Example 1: Solve the equation (D2+4) y=0
-2|
1
4
4
4
Solution: The auxiliary equation is f(m) = m? + 4 = 0
-2
-4
1
-2
2
-4
-2
5. (D6 + 9D* + 24D² + 16)y = 0
m2 = -4
1
-2
-2
m = tv-4
1
1
m = +21
Thus, we get
m5 + m* - 7m - 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) = o
general solution is
y = el (Ci cos 2x + C2 sin 2x)
The roots of /f (m) = 0 are 3, -2,-2, ti
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