Question

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \((x-2)^{2} y^{\prime \prime}+5(x-2) y^{\prime}+8 y=0\)

Short answer

Question: Find the general solution of the given differential equation \((x-2)^2y'' + 5(x-2)y' + 8y = 0\). Answer: The general solution of the given differential equation is \(y(x) = A (x-2)^{-2+2i} + B (x-2)^{-2-2i}\), where A and B are arbitrary constants.

Step-by-step solution

Introduce a substitution

First, let us introduce a substitution to simplify the differential equation. Let \(z = x - 2\). Then, our equation becomes: \((z)^{2} y^{\prime \prime}+5(z) y^{\prime}+8 y=0\)

Standard form of Equation

Now, let's rewrite the differential equation in the standard form. Divide the equation by \(z^2\) to get: \(y^{\prime \prime}+\frac{5}{z} y^{\prime}+\frac{8}{z^2} y=0\)

Solution Method

We can now see that this is a Cauchy-Euler differential equation. To solve this, we propose a solution of the form \(y(z) = z^r\), and find the corresponding values of \(r\).

Plug in the proposed solution

Replace the function \(y(z)\) and its derivatives into the differential equation: \(y'(z) = rz^{r-1}\) \(y''(z) = r(r-1)z^{r-2}\) Plug these into the differential equation: \(r(r-1)z^{r-2}+\frac{5}{z} rz^{r-1}+\frac{8}{z^2} z^r=0\)

Simplify and find characteristic equation

Notice that all terms have at least \(z^{r-2}\) as a factor. We can divide by \(z^{r-2}\) and simplify: \(r(r-1)+5r+8=0\) This equation is the characteristic equation. Solve for r: \(r^2 - r + 5r + 8 = 0\) \(r^2 + 4r + 8 = 0\)

Solve the characteristic equation

To find the roots of the quadratic equation, we can use the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) \(r = \frac{-4 \pm \sqrt{(-4)^2 - 4(1)(8)}}{2(1)}\) \(r = \frac{-4 \pm \sqrt{16 - 32}}{2}\) Since \(16-32=-16\), we have a complex solution: \(r = \frac{-4 \pm \sqrt{-16}}{2} = -2 \pm 2i\)

General solution

As the roots are complex, the general solution is of the form: \(y(z) = A z^{r_1} + B z^{r_2}\) \(y(z) = A z^{-2+2i} + B z^{-2-2i}\)

Transform back to x

Now let's revert back to the variable \(x\) using the substitution \(z = x - 2\): \(y(x) = A (x-2)^{-2+2i} + B (x-2)^{-2-2i}\) This is the general solution of the given differential equation that is valid in any interval not including the singular point.

Key concepts

Singular Point

In the context of a Cauchy-Euler differential equation, a singular point is a value of the independent variable where the coefficients of the equation become undefined or infinite. In simpler terms, it is a particular value that makes the original differential equation problematic or irregular. Identifying these points is crucial because solutions behave differently when crossing them.

In the exercise, we start with the differential equation:

• \((x-2)^{2} y^{\prime \prime}+5(x-2) y^{\prime}+8 y=0\)

The singular point here is clearly at \(x = 2\) because at this point the equation ceases to be defined. It’s needed to find solutions that are valid on intervals which do not include the singular point. By sidestepping these points, we ensure the solutions remain well-behaved and continuous.

Characteristic Equation

The characteristic equation is essentially a tool for finding solutions to a Cauchy-Euler differential equation. When we simplify the equation into its standard form, we are looking for a polynomial whose roots correspond to the values that solve the differential equation.

In the exercise provided, after substituting \(z = x-2\) and rewriting the equation in a more manageable form, we found the characteristic equation:

• \(r^2 + 4r + 8 = 0\)

This quadratic equation helps us determine possible solutions through its roots. The challenge often lies in solving this polynomial, which can sometimes yield complex roots.

Complex Roots

Complex roots arise when the discriminant in the quadratic formula is negative. This indicates that the solution involves imaginary numbers, which adds an oscillatory component to the general solution.

In the given exercise, solving the characteristic equation \(r^2 + 4r + 8 = 0\) using the quadratic formula, we find:

• \(r = \frac{-4 \pm \sqrt{-16}}{2} = -2 \pm 2i\)

These roots, \(-2+2i\) and \(-2-2i\), are complex, and require us to utilize a specific form for the general solution. This generally suggests a solution that involves exponential growth combined with sinusoidal functions, albeit transformed for the context of a Cauchy-Euler equation.

General Solution

The general solution for a Cauchy-Euler differential equation with complex roots incorporates both root components in its expression. This typically involves terms that use exponentials of the roots, capturing the oscillatory nature introduced by the imaginary part.

For our exercise scenario, with complex roots \(-2+2i\) and \(-2-2i\), the general solution takes the form:

• \(y(z) = A z^{-2+2i} + B z^{-2-2i}\)

This reflects the harmonic nature of the solution. We then revert back to the original variable using the transformation \(z = x-2\), to express it in terms of \(x\):

• \(y(x) = A (x-2)^{-2+2i} + B (x-2)^{-2-2i}\)

This general form encompasses all particular solutions and is valid across intervals excluding the singular point \(x = 2\).