Mathematics II 4.6 Higher order linear equation with constant coefficent *Homogenous Example 1//Solve the equation d³y d4y d³y d²y dx5 - 3. +3 dx4 = 0 dx3 dx² Solution/ (D5-3D+3D3-D²)=0 D² (D3-3D² + 3D - 1) = 0 r² (33²+3r - 1) = 0 either r²=0r1 = r2 = 0 or (33r2+3r - 1) = 0 (r− 1)3 = 0 → r3 = r4 = r5 = 1 y=(c₁ex + c2 xe*x + c3e** + c4xe** + c5x²e**) → y=(c1ex + c2 xex + c3e1x + c4xetx +c5x2e1x) Exercise: Solve the equation d³y d²y +3 +3 dx3 dx² + y = 0 dx
Mathematics II 4.6 Higher order linear equation with constant coefficent *Homogenous Example 1//Solve the equation d³y d4y d³y d²y dx5 - 3. +3 dx4 = 0 dx3 dx² Solution/ (D5-3D+3D3-D²)=0 D² (D3-3D² + 3D - 1) = 0 r² (33²+3r - 1) = 0 either r²=0r1 = r2 = 0 or (33r2+3r - 1) = 0 (r− 1)3 = 0 → r3 = r4 = r5 = 1 y=(c₁ex + c2 xe*x + c3e** + c4xe** + c5x²e**) → y=(c1ex + c2 xex + c3e1x + c4xetx +c5x2e1x) Exercise: Solve the equation d³y d²y +3 +3 dx3 dx² + y = 0 dx
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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Please (re)solve these examples in more detail as well.................
![Mathematics II
4.6 Higher order linear equation with constant
coefficent
*Homogenous
Example 1//Solve the equation
d³y
d4y d³y
d²y
dx5
- 3. +3
dx4
= 0
dx3
dx²
Solution/
(D5-3D+3D3-D²)=0
D² (D3-3D² + 3D - 1) = 0
r² (33²+3r - 1) = 0
either r²=0r1 = r2 = 0
or
(33r2+3r - 1) = 0
(r− 1)3 = 0 → r3 = r4 = r5 = 1
y=(c₁ex + c2 xe*x + c3e** + c4xe** + c5x²e**) →
y=(c1ex + c2 xex + c3e1x + c4xetx +c5x2e1x)
Exercise: Solve the equation
d³y d²y
+3 +3
dx3
dx²
+ y = 0
dx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1ecab767-ba67-485c-ab81-0a1604555596%2F385d22d3-eefb-4496-b7ce-bd926964fdf9%2Fxg1id49_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Mathematics II
4.6 Higher order linear equation with constant
coefficent
*Homogenous
Example 1//Solve the equation
d³y
d4y d³y
d²y
dx5
- 3. +3
dx4
= 0
dx3
dx²
Solution/
(D5-3D+3D3-D²)=0
D² (D3-3D² + 3D - 1) = 0
r² (33²+3r - 1) = 0
either r²=0r1 = r2 = 0
or
(33r2+3r - 1) = 0
(r− 1)3 = 0 → r3 = r4 = r5 = 1
y=(c₁ex + c2 xe*x + c3e** + c4xe** + c5x²e**) →
y=(c1ex + c2 xex + c3e1x + c4xetx +c5x2e1x)
Exercise: Solve the equation
d³y d²y
+3 +3
dx3
dx²
+ y = 0
dx
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