Sketch the graph of the solution and describe its behavior as t increases. 9/144) y" +y' – 2y = 0, y(0) = 1, y'(0) = 1 =(y" + y' – 2y) = 0 =(b? +b – 2) = 0 =b? + 2b – b - 2 = b(b + 2) – 1(b + 2) = 0 =(b – 1)(b + 2) = 0 =1, = 1,22 = -2 Solution:y(x) = c,e-A;* + Cze¬h2* → y(x) = c,e* + cze²2* → 1 =y(0) = c,e(0) +cze(0) → 1 = c1+c2 =y'(x) = ce*x(-1) + 2czez* → y'(0) = c;(-1) + 2c, →1 = -c, + 2c2 €s+ C2 = 1 &s+ 2c2 = 1- 3c, = q + c2 = 1+ C = 1- c, → c; = 1- -y(x) = e* + y= ;le*+ e²*] 11/144) 6y" - 5y' +y=0, (0) = 4, y'(0) = 0 (6b2 – 5b + 1) = 0 → (622 – 5A + 1) = 0 → 6A2 – 32 – 21 + 1 = 0 - 31(21 – 1) – 1(21 – 1) = 0 → (31 – 1)(21 – 1) = 0 11 3'2 y(x) = c,e¬h,* + c,e¬* → y(x) = c,e³+c,e?* 4 = c, + c; and -= 0 3 2 0 = 2c, + 3c2 8 = 2€5+ 2c2 0 = 2€ş+ 3c, → C2 = -8 =2c, - 24 = 0 → c = 12 Now: y(x) = 12e+(-8)e* → y = 12e - 8e*
Sketch the graph of the solution and describe its behavior as t increases. 9/144) y" +y' – 2y = 0, y(0) = 1, y'(0) = 1 =(y" + y' – 2y) = 0 =(b? +b – 2) = 0 =b? + 2b – b - 2 = b(b + 2) – 1(b + 2) = 0 =(b – 1)(b + 2) = 0 =1, = 1,22 = -2 Solution:y(x) = c,e-A;* + Cze¬h2* → y(x) = c,e* + cze²2* → 1 =y(0) = c,e(0) +cze(0) → 1 = c1+c2 =y'(x) = ce*x(-1) + 2czez* → y'(0) = c;(-1) + 2c, →1 = -c, + 2c2 €s+ C2 = 1 &s+ 2c2 = 1- 3c, = q + c2 = 1+ C = 1- c, → c; = 1- -y(x) = e* + y= ;le*+ e²*] 11/144) 6y" - 5y' +y=0, (0) = 4, y'(0) = 0 (6b2 – 5b + 1) = 0 → (622 – 5A + 1) = 0 → 6A2 – 32 – 21 + 1 = 0 - 31(21 – 1) – 1(21 – 1) = 0 → (31 – 1)(21 – 1) = 0 11 3'2 y(x) = c,e¬h,* + c,e¬* → y(x) = c,e³+c,e?* 4 = c, + c; and -= 0 3 2 0 = 2c, + 3c2 8 = 2€5+ 2c2 0 = 2€ş+ 3c, → C2 = -8 =2c, - 24 = 0 → c = 12 Now: y(x) = 12e+(-8)e* → y = 12e - 8e*
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Sketch The Graph Of The Solution And Describe Its ... | Chegg.com
Transcribed Image Text:Sketch the graph of the solution and describe its behavior as t increases.
9/144) y" + y' – 2y = 0, y(0) = 1, y'(0) = 1
=(y" + y' – 2y) = 0
=(b? + b – 2) = 0
=b? + 2b – b – 2 = b(b + 2) – 1(b + 2) = 0
=(b – 1)(b + 2) = 0
=1, = 1,22 = -2
Solution:y(x) = cze-ds* + cze¬h2* → y(x) = c,e¬* + cze2x → 1
=y(0) = c,e0) + cze(0) → 1 = c1 + c2
=y'(x) = ce¯*x(-1) + 2c,e²* → y'(0) = c,(-1) + 2c2 →1 = -c, + 2c2
&s+ cz = 1
& + 2c, = 1- 3c2 =
1
2
C + c2 = 1- c, = 1– c2 → C = 1–
=y(x) = ;e* +e* -y = le* + e
11/144) 6y" – 5y' +y=0, ¥(0) = 4, y'(0) = 0
(6b? – 5b + 1) = 0 → (6x² – 51 + 1) = 0 → 6A² – 3A – 21 + 1 = 0
— зА(2л — 1)- 1(2n — 1) %3D о— (3л — 1) (24 — 1) %3D 0
1 1
3'2
y(x) = c,e¬^s* + cze¬hz* → y(x) = c,e3+cze7*
4 = c, + Cz and -= 0
3
0 = 2c, + 3c2
8 = 2€5+ 2c2
0 = 2€z+ 3c, → c2 = -8
=2c, - 24 = 0 → cq = 12
Now: y(x) = 12e i+(-8)e* → y = 12e – 8e *
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