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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![**Problem:** Find the GS of \( y'' - y' - 12y = 0 \)
This is a second-order linear homogeneous differential equation with constant coefficients. To solve it, we find the characteristic equation:
\[ r^2 - r - 12 = 0 \]
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), \( c = -12 \), we calculate:
\[ r = \frac{1 \pm \sqrt{1 + 48}}{2} \]
\[ r = \frac{1 \pm \sqrt{49}}{2} \]
\[ r = \frac{1 \pm 7}{2} \]
This gives roots \( r_1 = 4 \) and \( r_2 = -3 \).
The general solution (GS) is:
\[ y(t) = C_1 e^{4t} + C_2 e^{-3t} \]
where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F093e7116-3e01-49ad-9157-7a22a5caeb16%2F5516b923-6c67-4abf-8da4-da4de4fb82cd%2Fx7uqp2b_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:** Find the GS of \( y'' - y' - 12y = 0 \)
This is a second-order linear homogeneous differential equation with constant coefficients. To solve it, we find the characteristic equation:
\[ r^2 - r - 12 = 0 \]
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), \( c = -12 \), we calculate:
\[ r = \frac{1 \pm \sqrt{1 + 48}}{2} \]
\[ r = \frac{1 \pm \sqrt{49}}{2} \]
\[ r = \frac{1 \pm 7}{2} \]
This gives roots \( r_1 = 4 \) and \( r_2 = -3 \).
The general solution (GS) is:
\[ y(t) = C_1 e^{4t} + C_2 e^{-3t} \]
where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.
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