Apply the method of reduction of order (not it's formula) to find the second solution if 6y "+ y'- y = 0 y, = e3

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Problem Statement:**

Apply the method of reduction of order (not its formula) to find the second solution of the differential equation given by:

\[ 6y'' + y' - y = 0 \]

where the first solution is:

\[ y_1 = e^{\frac{x}{3}} \]

**Explanation for Educational Website:**

In this problem, we are tasked with finding the second solution to a second-order linear homogeneous differential equation using the method of reduction of order. The differential equation provided is:

\[ 6y'' + y' - y = 0 \]

We are given that \( y_1 = e^{\frac{x}{3}} \) is one solution to the equation.

To apply the reduction of order method, one typically assumes the second solution \( y_2 \) takes the form:

\[ y_2 = v(x) y_1 \]

where \( v(x) \) is a function to be determined. By substituting \( y_2 \) and its derivatives into the original differential equation, and using the known solution \( y_1 \), you can solve for \( v(x) \).

This method eliminates certain terms, simplifying the problem and allowing you to find a particular expression or differential equation for \( v(x) \). Solving this will yield the unknown second solution \( y_2 \).
Transcribed Image Text:**Problem Statement:** Apply the method of reduction of order (not its formula) to find the second solution of the differential equation given by: \[ 6y'' + y' - y = 0 \] where the first solution is: \[ y_1 = e^{\frac{x}{3}} \] **Explanation for Educational Website:** In this problem, we are tasked with finding the second solution to a second-order linear homogeneous differential equation using the method of reduction of order. The differential equation provided is: \[ 6y'' + y' - y = 0 \] We are given that \( y_1 = e^{\frac{x}{3}} \) is one solution to the equation. To apply the reduction of order method, one typically assumes the second solution \( y_2 \) takes the form: \[ y_2 = v(x) y_1 \] where \( v(x) \) is a function to be determined. By substituting \( y_2 \) and its derivatives into the original differential equation, and using the known solution \( y_1 \), you can solve for \( v(x) \). This method eliminates certain terms, simplifying the problem and allowing you to find a particular expression or differential equation for \( v(x) \). Solving this will yield the unknown second solution \( y_2 \).
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