Question

Find the general solution of the given differential equation. $$ 6 y^{\prime \prime}-y^{\prime}-y=0 $$

Short answer

Answer: The general solution for the given differential equation is $$y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-\frac{1}{3}x}$$, where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Write down the given differential equation

First, let's rewrite the given second-order linear differential equation: $$ 6y^{\prime\prime}(x) - y^{\prime}(x) - y(x) = 0 $$

Write the characteristic equation

The characteristic equation for the given differential equation is found by replacing \(y^{\prime\prime}\) with \(m^2\), \(y^{\prime}\) with \(m\), and \(y\) with \(1\). So the characteristic equation becomes: $$ 6m^2 - m - 1 = 0 $$

Solve the characteristic equation for roots

To solve the quadratic equation, we can use the quadratic formula: $$ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where \(a=6\), \(b=-1\), and \(c=-1\). Plugging in these values, we get: $$ m = \frac{1 \pm \sqrt{(-1)^2 - 4(6)(-1)}}{2(6)} = \frac{1 \pm \sqrt{25}}{12} $$ The roots m are: $$ m_1 = \frac{1 + 5}{12} = \frac{1}{2}\\ m_2 = \frac{1 - 5}{12} = -\frac{1}{3} $$

Write the fundamental solutions

Using the roots \(m_1\) and \(m_2\), we can write the fundamental solutions as: $$ y_1(x) = e^{\frac{1}{2}x}\\ y_2(x) = e^{-\frac{1}{3}x} $$

Write the general solution

The general solution of the given differential equation is the linear combination of the fundamental solutions with constants \(C_1\) and \(C_2\): $$ y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-\frac{1}{3}x} $$ This is the general solution for the given second-order linear differential equation.

Key concepts

Second-Order Linear Differential Equations

Second-order linear differential equations have the standard form: \[ a y'' + b y' + c y = g(t) \] where \( a \), \( b \), and \( c \) are constants, and \( g(t) \) is a given function. These equations involve the second derivative of the function, which means we are dealing with how the rate of change itself changes over time. This type of equation often models physical systems like oscillations in springs or electrical circuits. The equations can be:

• Homogeneous if \( g(t) = 0 \)
• Non-homogeneous if \( g(t) eq 0 \)

In our example, the equation is homogeneous, thus simplifying our approach to finding solutions by focusing on the characteristic part.

Characteristic Equation

To solve second-order linear homogeneous differential equations, we use the characteristic equation. This technique transforms the original differential equation into an algebraic equation, which is often simpler to solve. Given the differential equation: \[ a y'' + b y' + c y = 0 \] we replace \( y'' \) with \( m^2 \), \( y' \) with \( m \), and \( y \) with \( 1 \). This yields the characteristic equation: \[ a m^2 + b m + c = 0 \] This equation is a quadratic equation in terms of \( m \), and finding the roots of this equation helps us determine the behavior of the solutions to the differential equation. It is crucial as it directly leads to finding the fundamental solutions.

Fundamental Solutions

Fundamental solutions are the basic building blocks for constructing a general solution to a second-order differential equation. Once we have the roots of the characteristic equation, the nature of these roots tells us the form of the fundamental solutions.

• Real and distinct roots: If the roots \( m_1 \) and \( m_2 \) are real and different, the fundamental solutions are \( e^{m_1 x} \) and \( e^{m_2 x} \).
• Real and repeated roots: If there is one repeated root \( m \), the solutions are \( e^{mx} \) and \( x e^{mx} \).
• Complex roots: If the roots are complex, \( m = ext{α} \pm ext{β}i \), use \( e^{ ext{α}x}\cos( ext{β}x) \) and \( e^{ ext{α}x}\sin( ext{β}x) \).

In our exercise, with real and distinct roots \( \frac{1}{2} \) and \( -\frac{1}{3} \), the fundamental solutions become \( e^{\frac{1}{2}x} \) and \( e^{-\frac{1}{3}x} \).

General Solution

The general solution is a comprehensive expression that includes all possible solutions for a given differential equation. It is formed by taking a linear combination of the fundamental solutions obtained. For the differential equation at hand, the general solution is: \[ y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-\frac{1}{3}x} \] Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial or boundary conditions if provided. This presentation of the solution captures the general behavior of the system described by the differential equation and can be tailored to specific cases with additional constraints. Thus, the general solution allows for adaptability, covering everything from initial conditions to more complex scenarios.