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.