Chapter 17: Problem 28
Find the general solution of the given equation. $$4 \frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+y=0$$
Short Answer
Expert verified
The general solution is \(y(x) = (C_1 + C_2 x) e^{\frac{1}{2}x}\).
Step by step solution
01
Identify the Differential Equation Type
The given equation is \(4 \frac{d^{2} y}{d x^{2}} - 4 \frac{d y}{d x} + y = 0\). This is a second-order linear homogeneous differential equation with constant coefficients.
02
Write the Characteristic Equation
For the given differential equation, we assume a solution of the form \(y = e^{rt}\). Substituting this into the differential equation, we obtain the auxiliary (or characteristic) equation: \[ 4r^2 - 4r + 1 = 0. \]
03
Solve the Characteristic Equation
To solve \(4r^2 - 4r + 1 = 0\), we use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4, b = -4, c = 1\). This gives: \[ r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{4 \pm 0}{8} = \frac{1}{2}. \] Thus, we have a repeated root \(r = \frac{1}{2}\).
04
Write the General Solution
Since the characteristic equation has a repeated root \(r = \frac{1}{2}\), the general solution to the differential equation is \[ y(x) = (C_1 + C_2 x) e^{\frac{1}{2}x}, \] where \(C_1\) and \(C_2\) are arbitrary constants.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
A characteristic equation is a crucial concept when solving differential equations, especially second-order linear ones like the given example. This equation often emerges from a method that proposes solutions of the form \( y = e^{rt} \). By substituting this into the differential equation, we transform the problem of solving a differential equation into solving a polynomial equation, specifically a quadratic equation for second-order cases.
This polynomial, known as the characteristic equation, is found by replacing derivatives in the differential equation with powers of \( r \), as was done in the problem: \( 4r^2 - 4r + 1 = 0 \).
This polynomial, known as the characteristic equation, is found by replacing derivatives in the differential equation with powers of \( r \), as was done in the problem: \( 4r^2 - 4r + 1 = 0 \).
- The coefficients in the differential equation determine the coefficients of the characteristic equation.
- Solving the characteristic equation yields the roots \( r \), which are used to construct the general solution of the differential equation.
Repeated Roots
Repeated roots occur when solving the characteristic equation yields the same value for \( r \) more than once. In the context of second-order linear differential equations, a repeated root presents a unique scenario.
In this exercise, the characteristic equation \( 4r^2 - 4r + 1 = 0 \) simplifies to a single repeated root, \( r = \frac{1}{2} \).
In this exercise, the characteristic equation \( 4r^2 - 4r + 1 = 0 \) simplifies to a single repeated root, \( r = \frac{1}{2} \).
- Repeated roots occur when the discriminant \( b^2 - 4ac \) of the quadratic formula is zero, leading to a single solution being repeated.
- This means instead of two distinct solutions, we obtain one solution repeated, which requires special handling in terms of forming the general solution.
General Solution
The general solution to a differential equation encompasses all possible solutions. For a second-order linear homogeneous differential equation with repeated roots, this solution takes a distinctive form.
In cases like \( r = \frac{1}{2} \) repeated, the solution is not just \( e^{rac{1}{2}x} \). We use the formula for repeated roots, which adds a term involving \( x \):
\[ y(x) = (C_1 + C_2 x) e^{rac{1}{2}x} \]
In cases like \( r = \frac{1}{2} \) repeated, the solution is not just \( e^{rac{1}{2}x} \). We use the formula for repeated roots, which adds a term involving \( x \):
\[ y(x) = (C_1 + C_2 x) e^{rac{1}{2}x} \]
- \( C_1 \) and \( C_2 \) are arbitrary constants, showing the infinite nature of solutions as \( x \) varies.
- The term \( C_2 x \) accounts for the repeated nature of the root, ensuring completeness in capturing all potential solutions.