extat + b
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
How to convert the found solution in the first image to the form in the second image
Transcribed Image Text:ett
at + b
ct + d
![Then
((x+3)=8x-5-1 (-x²+ 3x)
⇒ ½ (-x+ 3x¹) = 8x− 1 2 (-5x+15x)
4
⇒ −x+3x² = 16X+5x²-15x
⇒ x²+2x²+x=0.
Auxiliary equation of becomes
m²+2m+1=0
⇒ [m+1)² = 0
⇒m=-1,-1
• The general solution of is
x (t) = (₁+c₂t)ēt; C₁₂C₂ are
arbitrary constants
and x²(t)=-(4+C₂t)ēt+çēt
NOW, 2ylt)=-x²(t) + 3x lt)
1) = (a + c₂₁t) ē+ - c₂ē ² + 3 (G₁+q₂t) ēt
⇒ 2y(t) = 4(₁+₂t)ēt - sēt
→y(t) = 2 (₁+²₂t)ēt - /12/2 ś czēt
NOW, x(0) = 2 and y(0) = 2
⇒ C₁=2
⇒26₁-1/2²₂ =2
⇒4 - 12/2C₂₁₂=2
⇒1/12²2₂=4-2=2
1 = €₂=4
Solution
Hence, the solution of IVP gives
xlt) = (2+ 4t) et = 2ēt + ute
and y(t) = 2 (2+4t)et-zet = 2ět+8tēt
• The solution of the given system is
·t.
- x(+) = [2]ē* + [²] tēt
4
st
8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65851872-5405-4226-8ece-14f44369e59f%2F16234892-bbe5-40a2-a5e5-35253a3735e8%2F1bb90uo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Then
((x+3)=8x-5-1 (-x²+ 3x)
⇒ ½ (-x+ 3x¹) = 8x− 1 2 (-5x+15x)
4
⇒ −x+3x² = 16X+5x²-15x
⇒ x²+2x²+x=0.
Auxiliary equation of becomes
m²+2m+1=0
⇒ [m+1)² = 0
⇒m=-1,-1
• The general solution of is
x (t) = (₁+c₂t)ēt; C₁₂C₂ are
arbitrary constants
and x²(t)=-(4+C₂t)ēt+çēt
NOW, 2ylt)=-x²(t) + 3x lt)
1) = (a + c₂₁t) ē+ - c₂ē ² + 3 (G₁+q₂t) ēt
⇒ 2y(t) = 4(₁+₂t)ēt - sēt
→y(t) = 2 (₁+²₂t)ēt - /12/2 ś czēt
NOW, x(0) = 2 and y(0) = 2
⇒ C₁=2
⇒26₁-1/2²₂ =2
⇒4 - 12/2C₂₁₂=2
⇒1/12²2₂=4-2=2
1 = €₂=4
Solution
Hence, the solution of IVP gives
xlt) = (2+ 4t) et = 2ēt + ute
and y(t) = 2 (2+4t)et-zet = 2ět+8tēt
• The solution of the given system is
·t.
- x(+) = [2]ē* + [²] tēt
4
st
8
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