Question

\(3 x y^{\prime \prime}+2 y^{\prime}+12 y=0\)

Short answer

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

Step-by-step solution

Identify the Type of Differential Equation

Observe the given differential equation, which is a second-order linear homogeneous differential equation with constant coefficients: \[ 3 x y^{\prime \prime} + 2 y^{\prime} + 12 y = 0 \]

Use the Characteristic Equation Method

For a second-order linear homogeneous differential equation of the form \[ a y^{\prime \prime} + b y^{\prime} + c y = 0 \] we assume a solution of the form \(y = e^{r x}\). Substitute \(y = e^{r x}\) into the equation to get the characteristic equation.

Set Up the Characteristic Equation

For \(y = e^{r x}\), the derivatives are \(y^{\prime} = r e^{r x}\) and \(y^{\prime \prime} = r^2 e^{r x}\). Substituting these into the original equation: \[ 3 x (r^2 e^{r x}) + 2 r e^{r x} + 12 e^{r x} = 0 \] Since \(e^{r x} eq 0\), we can divide the entire equation by \(e^{r x}\): \[ 3 r^2 + 2 r + 12 = 0 \]

Solve the Characteristic Equation

Solve the quadratic equation \(3 r^2 + 2 r + 12 = 0\) to find the roots. This can be done using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = 2\), and \(c = 12\). \[ r = \frac{-2 \pm \sqrt{2^2 - 4(3)(12)}}{2(3)} = \frac{-2 \pm \sqrt{4 - 144}}{6} = \frac{-2 \pm \sqrt{-140}}{6} = \frac{-2 \pm i \sqrt{140}}{6} = \frac{-1 \pm \frac{i \sqrt{35}}{3}}{3} \]

Express the General Solution

Given the complex roots \(r = -\frac{1}{3} \pm i \frac{\sqrt{35}}{3}\), the general solution for the differential equation is: \[ y(x) = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x)) \] where \(\alpha = -\frac{1}{3}\) and \(\beta = \frac{\sqrt{35}}{3}\). Therefore, the general solution is: \[ y(x) = e^{-\frac{1}{3} x} \left( C_1 \cos \left( \frac{\sqrt{35}}{3} x \right) + C_2 \sin \left( \frac{\sqrt{35}}{3} x \right) \right) \]

Key concepts

Characteristic Equation

The characteristic equation is essential in solving second-order linear homogeneous differential equations. When dealing with an equation like \[ ay'' + by' + cy = 0 \], we assume a solution of the form \( y = e^{rx} \).
By substituting \( y = e^{rx} \), the differential equation transforms into a polynomial, which is the characteristic equation.
This is done by calculating the derivatives:

• \( y' = r e^{rx} \)
• \( y'' = r^2 e^{rx} \)

After substituting these derivatives into the original equation, we get:
\[ 3 r^2 + 2 r + 12 = 0 \]This quadratic equation is what we call the characteristic equation. Solving this equation helps us determine the nature of the roots, which can be real and distinct, real and repeated, or complex. Each scenario has different methods for finding the general solution.

Homogeneous Differential Equation

A homogeneous differential equation has every term dependent on the function and its derivatives, with no standalone constants. In our example, \[ ay'' + by' + cy = 0 \], each term includes the dependent variable \( y \) or its derivatives.
Homogeneous differential equations are significant because they exhibit properties allowing for specific solution methods, especially through the characteristic equation.
For the given second-order linear homogeneous differential equation:\[ 3x y'' + 2y' + 12y = 0 \]
we see that each term involves \( y \) or its derivatives, ensuring the equation is homogeneous. Solving it involves finding the characteristic equation and subsequently deriving the roots, which then guide us to the general solution.

Complex Roots

Complex roots arise when the discriminant in the characteristic equation is negative. This happens in the quadratic equation \[ ar^2 + br + c = 0 \] when \[ b^2 - 4ac < 0 \].
In our example:
\[ 3 r^2 + 2 r + 12 = 0 \], \[ a = 3, b = 2, \] and \[ c = 12 \].
Calculating the discriminant gives us:\[ 2^2 - 4(3)(12) = 4 - 144 = -140 \]
Since the discriminant is negative, the roots are complex. They are found using the formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]\[ r = \frac{-2 \pm i \sqrt{140}}{6} \]The complex roots will then be in the form \[ r = \alpha \pm i \beta \].For complex roots, the general solution of the differential equation is given by:\[ y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \]In our example, with \( \alpha = -\frac{1}{3} \) and \( \beta = \frac{\sqrt{35}}{3} \), the final solution is:\[ y(x) = e^{-\frac{1}{3} x} \left( C_1 \cos \left( \frac{\sqrt{35}}{3} x \right) + C_2 \sin \left( \frac{\sqrt{35}}{3} x \right) \right) \]