Chapter 9: Problem 15
\((x+1) y^{\prime \prime}+(x-1) y^{\prime}-2 y=0\)
Short Answer
Expert verified
y = C_1 e^x + C_2 e^{-2x}
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \( (x+1) y^{\prime \prime}+(x-1) y^{\prime}-2 y=0 \), which is a second-order linear ordinary differential equation.
02
Find the Characteristic Equation
To solve this, assume a solution of the form \( y = e^{rx} \). Plug this into the differential equation. This gives the characteristic equation: \[ r^2 (x+1) + r (x-1) - 2 = 0 \].
03
Simplify the Characteristic Equation
Distribute and combine like terms in the characteristic equation: \[ r^2 x + r^2 + r x - r - 2 = 0 \]. Set it to zero to find roots: \[ r^2 x + r^2 + r x - r - 2 = 0 \].
04
Solve for Values of r
Solve the characteristic equation for \( r \): \[ r^2 + r - 2 = 0 \]. Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), substitute a=1, b=1, and c=-2. This gives: \[ r = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \]. So, \( r_1 = 1 \) and \( r_2 = -2 \).
05
Write the General Solution
Use the roots to write the general solution of the differential equation: \[ y = C_1 e^{r_1 x} + C_2 e^{r_2 x} \]. Substituting the roots, \( y = C_1 e^{x} + C_2 e^{-2x} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
To solve a second-order linear ordinary differential equation, we start by finding the characteristic equation. Assume a solution of the form \( y = e^{rx} \). Substituting this into the differential equation \( (x+1) y^{\prime \prime}+(x-1) y^{\prime}-2 y=0 \) transforms it into an algebraic equation in terms of \( r \). This is our characteristic equation, which can help us find the values of \( r \) that satisfy the differential equation. To break it down further, plug in the assumed solution: \( (x+1) r^2 e^{rx} + (x-1) r e^{rx} - 2 e^{rx} = 0 \). Factor out \( e^{rx} \) (since it can't be zero), which simplifies to: \( r^2 (x+1) + r (x-1) - 2 = 0 \). This is our characteristic equation.
Quadratic Formula
Once we have the characteristic equation, we often get a quadratic equation like \( r^2 + r - 2 = 0 \). To find the values of \( r \) (roots), we can use the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula is crucial for solving any quadratic equation. For our characteristic equation, set \( a = 1 \), \( b = 1 \), and \( c = -2 \). Plug these into the quadratic formula: \( r = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \).We get two roots: \( r_1 = 1 \) and \( r_2 = -2 \). These roots allow us to move to the next step: writing the general solution.
General Solution
With the roots \( r_1 \) and \( r_2 \) from the quadratic formula, we can write the general solution to our differential equation. For a second-order linear differential equation, the general solution is of the form: \( y = C_1 e^{r_1 x} + C_2 e^{r_2 x} \).Here, \( C_1 \) and \( C_2 \) are constants that can be determined if we have initial conditions or additional information. Substituting our specific roots \( r_1 = 1 \) and \( r_2 = -2 \) into the general form gives us: \( y = C_1 e^{x} + C_2 e^{-2x} \).This is the complete general solution to the original differential equation \( (x+1) y^{\prime \prime}+(x-1) y^{\prime}-2 y=0 \).Understanding how to derive this general solution is critical for solving similar types of differential equations in the future.