Problem 54 - Differential Equations and Linear Algebra

Question

Find the general solution to the given differential equation on the interval \((0, \infty).\) $$x^{2} y^{\prime \prime}+9 x y^{\prime}+15 y=0.$$

Step-by-Step Solution

Answer

The general solution of the given Cauchy-Euler differential equation \(x^{2} y^{\prime \prime}+9 x y^{\prime}+15 y=0\) on the interval \((0, \infty)\) is \(y(x) = C_1x^{-3} + C_2x^{-5}\), where \(C_1\) and \(C_2\) are constants.

Step 1: Write the characteristic equation

The characteristic equation for a Cauchy-Euler differential equation is: \(m(m-1) + am + b = 0\) For our given differential equation, \(a = 9\) and \(b = 15\). So we have: \(m(m-1) + 9m + 15 = 0\)

Step 2: Solve the characteristic equation

Now we need to solve for the roots of the characteristic equation: \(m(m-1) + 9m + 15 = 0\) We can rewrite this as a quadratic equation in terms of m: \(m^2 - m + 9m + 15 = 0\) Simplifying, we get: \(m^2 + 8m + 15 = 0\) Now we can use the quadratic formula to solve for the roots: \(m = \frac{-B \pm \sqrt{B^2-4AC}}{2A}\) In this case, \(A = 1\), \(B = 8\), and \(C = 15\). Plugging these values into the quadratic formula, we get: \(m = \frac{-8 \pm \sqrt{8^2-4(1)(15)}}{2(1)}\) Simplifying, we obtain: \(m = \frac{-8 \pm \sqrt{64-60}}{2}\) So the roots are: \(m = \frac{-8 \pm \sqrt{4}}{2}\) Which gives us \(r_1 = -3\) and \(r_2 = -5\).

Step 3: Write the general solution

Now that we have the roots of the characteristic equation, we can write the general solution for the given differential equation: \(y(x) = x^m (C_1x^{r_1} + C_2x^{r_2})\) Plugging in the roots we found, \(r_1 = -3\) and \(r_2 = -5\), we obtain the general solution: \(y(x) = x^{m}(C_1x^{-3} + C_2x^{-5})\) Since we're solving on the interval (0, ∞), our final general solution is: \(y(x) = C_1x^{-3} + C_2x^{-5}\)

Key Concepts

Characteristic Equation

The Cauchy-Euler differential equation is a special type of linear differential equation. It can generally be written in the form \[ a_n x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \ + \ \ldots + a_1 x y' + a_0 y = 0, \] where each term involves products of powers of x.
The characteristic equation is a key mathematical tool used for solving these differential equations. To find the characteristic equation, we substitute a solution of the form \(y = x^m\).
For the particular equation given in the exercise, \[x^2 y'' + 9xy' + 15y = 0,\] the characteristic equation becomes \[ m(m-1) + am + b = 0,\] where for this problem \(a = 9\) and \(b = 15\).
By substituting these values, the characteristic equation simplifies to \[ m(m-1) + 9m + 15 = 0.\]
This step is crucial as it transforms a differential equation into a manageable algebraic problem. Once we have the characteristic equation, we can solve it for the roots.

General Solution

After finding the roots of the characteristic equation, which are often denoted as \(r_1\) and \(r_2\), we can construct the general solution for the differential equation.
In our exercise, solving the characteristic equation \[ m^2 + 8m + 15 = 0 \] led to the roots \(r_1 = -3\) and \(r_2 = -5\). The general solution for a Cauchy-Euler differential equation on the interval \((0, \infty)\) is given by:
\[ y(x) = C_1 x^{r_1} + C_2 x^{r_2}, \] where \(C_1\) and \(C_2\) are arbitrary constants.
Substituting the roots, the solution becomes: \[ y(x) = C_1 x^{-3} + C_2 x^{-5}. \]
This form of the solution leverages the unique power-law behavior of Euler-Cauchy equations, and these solutions are valid across the specified interval.

Quadratic Formula

The quadratic formula is a classic mathematical tool used for solving quadratic equations of the form \(ax^2 + bx + c = 0\).
It's given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] This formula provides a systematic way to find the roots of the quadratic equation.
In our step-by-step solution, we used the quadratic formula to solve the characteristic equation \[ m^2 + 8m + 15 = 0. \] Here, \(A = 1\), \(B = 8\), and \(C = 15\).
Plugging these values into the quadratic formula, we have: \[ m = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 15}}{2 \cdot 1} \] which simplifies to \[ m = \frac{-8 \pm \sqrt{64 - 60}}{2}. \]
The final result gives us the real and distinct roots \(r_1 = -3\) and \(r_2 = -5\), which are crucial for writing the general solution. Using the quadratic formula helps ensure accuracy when dealing with complex equations and is especially helpful in algebraic problems like these.