Find the values of the constants C and L such that the function y(x) = Cel satisfies to ODE y + y =y".

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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**Problem Statement:**

Find the values of the constants \( C \) and \( L \) such that the function \( y(x) = Ce^{Lx} \) satisfies the ordinary differential equation (ODE) \( y + y' = y'' \).

**Options:**

- \( \circ \) Only two solutions with \( L = \frac{1 \pm \sqrt{5}}{2} \) and \( C = 1 \).

- \( \circ \) Two possible solutions with \( L = \frac{1 - i\sqrt{5}}{2} \) and with \( L = \frac{1 + i\sqrt{5}}{2} \).

- \( \circ \) Two possible solutions with \( L = \frac{1 - \sqrt{5}}{2} \) and with \( L = \frac{1 + \sqrt{5}}{2} \), and \( C \) can be arbitrary.

- \( \circ \) Only one solution with \( L = \frac{1 - \sqrt{5}}{2} \) and \( C \) can be arbitrary.
Transcribed Image Text:**Problem Statement:** Find the values of the constants \( C \) and \( L \) such that the function \( y(x) = Ce^{Lx} \) satisfies the ordinary differential equation (ODE) \( y + y' = y'' \). **Options:** - \( \circ \) Only two solutions with \( L = \frac{1 \pm \sqrt{5}}{2} \) and \( C = 1 \). - \( \circ \) Two possible solutions with \( L = \frac{1 - i\sqrt{5}}{2} \) and with \( L = \frac{1 + i\sqrt{5}}{2} \). - \( \circ \) Two possible solutions with \( L = \frac{1 - \sqrt{5}}{2} \) and with \( L = \frac{1 + \sqrt{5}}{2} \), and \( C \) can be arbitrary. - \( \circ \) Only one solution with \( L = \frac{1 - \sqrt{5}}{2} \) and \( C \) can be arbitrary.
Expert Solution
Step 1: Solution

Given differential equation: y plus y apostrophe equals y apostrophe apostrophe

We have to find the value of constant C and L such that the function y left parenthesis x right parenthesis equals C e to the power of L x end exponent satisfies to ODE 

y plus y apostrophe equals y apostrophe apostrophe

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