Question

\(3 x y^{\prime \prime}+y^{\prime}+12 y=0\)

Short answer

The general solution is \[ y = x^{-(1/3)}(C_1\cos(\frac{\sqrt{35}}{3}\ln(x)) + C_2\sin(\frac{\sqrt{35}}{3}\ln(x))) \].

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \[ 3xy'' + y' + 12y = 0 \]

Convert to Standard Form

Write the differential equation in standard form by isolating the highest-order derivative: \[ y'' + \frac{1}{3x}y' + \frac{4}{x}y = 0 \]

Assume a Solution Form

Assume a solution of the form \( y = x^r \). Compute the derivatives: \[ y' = rx^{r-1} \] and \[ y'' = r(r-1)x^{r-2} \].

Substitute and Simplify

Substitute \( y = x^r \), \( y' \), and \( y'' \) into the standard form equation: \[ r(r-1)x^{r-2} + \frac{1}{3x}(rx^{r-1}) + \frac{4}{x}(x^r) = 0 \]. Simplify to get: \[ r(r-1)x^{r-2} + \frac{r}{3}x^{r-2} + 4x^{r-1} = 0 \]. Combine like terms: \[ (r(r-1) + \frac{r}{3} + 4)x^{r-2} = 0 \]

Solve the Characteristic Equation

Solve the characteristic equation: \[ r(r-1) + \frac{r}{3} + 4 = 0 \]which simplifies to: \[ r^2 - r + \frac{r}{3} + 4 = 0 \] Combine terms: \[ r^2 + \frac{2r}{3} + 4 = 0 \] Multiply through by 3 to clear fractions: \[ 3r^2 + 2r + 12 = 0 \].

Find the Roots of the Quadratic Equation

Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = 3\), \(b = 2\), and \(c = 12\): \[ r = \frac{-2 \pm \sqrt{4 - 144}}{6} \] Simplify under the square root: \[ r = \frac{-2 \pm \sqrt{-140}}{6} \] \[ r = \frac{-2 \pm i\sqrt{140}}{6} \] \[ r = \frac{-1}{3} \pm \frac{i\sqrt{35}}{3} \]. The roots are complex.

Write the General Solution

Given the complex roots of the form \( \alpha \pm \beta i \), write the general solution: \[ y = x^{\alpha}(C_1\cos(\beta\ln(x)) + C_2\sin(\beta\ln(x))) \] Here, \( \alpha = -\frac{1}{3} \) and \( \beta = \frac{\sqrt{35}}{3} \): \[ y = x^{-(1/3)}(C_1\cos(\frac{\sqrt{35}}{3}\ln(x)) + C_2\sin(\frac{\sqrt{35}}{3}\ln(x))) \].

Key concepts

Homogeneous Differential Equation

A differential equation is called homogeneous if it can be written so that every term involves either the unknown function or its derivatives. In this case, the given second-order linear differential equation is homogeneous because each term involves the function \(y\), its first derivative \(y'\), or its second derivative \(y''\).
The standard form of a second-order homogeneous linear differential equation with constant coefficients is:
\[ ay'' + by' + cy = 0 \]
In our case, we start with:
\[ 3xy'' + y' + 12y = 0 \]. Notice how we have terms involving \( y \), \( y' \), and \( y''\), making it homogeneous.

Characteristic Equation

To solve a second-order homogeneous linear differential equation, we often convert it to an algebraic equation called the characteristic equation. This is done by assuming the solutions have the form \( y = x^r \) and then substituting this guess into the differential equation.
For our example, we write the differential equation in standard form: \[ y'' + \frac{1}{3x} y' + \frac{4}{x} y = 0 \].
Assuming \( y = x^r \), we get the derivatives \(y' = r x^{r-1}\) and \( y'' = r(r-1) x^{r-2}\).
Substituting these into the standard form, we get the characteristic equation:
\[ r(r-1) x^{r-2} + \frac{1}{3x} (rx^{r-1}) + \frac{4}{x} (x^r) = 0 \].
Simplifying this, we combine like terms to obtain:
\[ r^2 + \frac{2r}{3} + 4 = 0 \].

Complex Roots

When solving the characteristic equation, we encounter different types of roots, which determine the form of the solution. In our case, the characteristic equation is \( 3r^2 + 2r + 12 = 0 \).
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \(a = 3\), \(b = 2\), and \(c = 12\), we have:
\[ r = \frac{-2 \pm \sqrt{4 - 144}}{6} \]
Upon simplifying, we find:
\[ r = \frac{-2 \pm i\sqrt{140}}{6} \], yielding the roots:
\( r = \frac{-1}{3} \pm \frac{ i \sqrt{35}}{3} \).
These roots are complex and will influence the form of our general solution.

General Solution

The general solution of a second-order linear differential equation with complex roots is formed using sine and cosine terms.
Given complex roots \( \alpha \pm \beta i \), the general solution can be written as:
\[ y = x^{\alpha} ( C_1 \cos ( \beta \ln x) + C_2 \sin ( \beta \ln x)) \].
For the characteristic equation from our example, we have: \( \alpha = -\frac{1}{3} \) and \( \beta = \frac{ \sqrt{35}}{3} \). So, the general solution becomes:
\[ y = x^{-1/3}( C_1 \cos ( \frac{ \sqrt{35}}{3} \ln x ) + C_2 \sin ( \frac{ \sqrt{35}}{3} \ln x )) \]
This complex solution fits the form required for our second-order linear differential equation.

Quadratic Formula

The quadratic formula is a standard method for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It is used extensively in solving characteristic equations in second-order differential equations.
The formula is:

• \( r = \frac{-b \pm \sqrt{ b^2 - 4ac }}{ 2a } \)

In our problem, to solve the characteristic equation \(3r^2 + 2r + 12 = 0\), we substitute:
\(a = 3\), \(b = 2\), and \(c = 12 \).
Using the quadratic formula, we compute:
\( r = \frac{-2 \pm \sqrt{4 - 144}}{6} \)
This results in:
\(r = \frac{-2 \pm i \sqrt{ 140 }}{6} \), giving the complex roots \( r = \frac{-1}{3} \pm \frac{ i\sqrt{35}}{3} \).
Understanding the quadratic formula helps in identifying the nature of the roots (real or complex) and constructing the corresponding solution.