Question

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \((x-1)^{2} y^{\prime \prime}+8(x-1) y^{\prime}+12 y=0\)

Short answer

Answer: The general solution is \(y(x)= C_1 e^{r_1 (x - 1)} + C_2 e^{r_2 (x - 1)}\), where \(r_1\) and \(r_2\) are the roots of the characteristic equation.

Step-by-step solution

Perform substitution

Replace \(x-1\) with \(t\). So we have, \(t^{2} y^{\prime \prime}+8t y^{\prime}+12 y=0\).

Write the characteristic equation

Convert the given second-order linear differential equation to its corresponding characteristic equation. This is done by replacing \(y^{\prime \prime}\) with \(r^2\), \(y^{\prime}\) with \(r\), and \(y\) with 1 in the original differential equation to obtain: \(t^2r^2 + 8tr + 12 = 0\).

Solve for r

To find the roots of the characteristic equation, we first divide the equation by \(t^2\) to give: \(r^2 + 8r + \frac{12}{t^2} = 0\). Now, use the quadratic formula to find the roots of the equation: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b = 8\), and \(c = \frac{12}{t^2}\). So, we have: \(r = \frac{-8 \pm \sqrt{64 - \frac{48}{t^2}}}{2}\).

Find the general form of the solution

Since the differential equation has two different real roots, the general form of the solution is: \(y(t)=C_1 e^{r_1 t} + C_2 e^{r_2 t}\), where \(C_1\) and \(C_2\) are constants and \(r_1\) and \(r_2\) are the roots of the characteristic equation.

Substitute back to find the solution in terms of x

Recall that we have \(t = x - 1\). So we can write the general form of the solution in terms of x as follows: \(y(x)= C_1 e^{r_1 (x - 1)} + C_2 e^{r_2 (x - 1)}\), where \(r_1\) and \(r_2\) are the roots from Step 3. Thus, the general solution of the given differential equation that is valid in any interval not including the singular point is \(y(x)= C_1 e^{r_1 (x - 1)} + C_2 e^{r_2 (x - 1)}\).

Key concepts

Second-Order Linear Differential Equation

Understanding the second-order linear differential equation is fundamental for solving many problems in mathematics and physics. These equations are characterized by the highest derivative being second-order and the equation itself being linear in the unknown function and its derivatives.

In the given exercise, we have the second-order linear differential equation \( (x-1)^{2} y''+8(x-1) y'+12 y=0 \). The coefficients of \( y'' \) and \( y' \) are functions of \( x \) and are non-zero when \( x eq 1 \)—making \( x=1 \) a singular point. In intervals where \( x eq 1 \) the equation can be tackled effectively by simplification techniques such as substitution, which help transform it into a more standard form where known solution methods can be applied.

Characteristic Equation

The characteristic equation is a key aspect of solving linear differential equations with constant coefficients. It is an algebraic equation obtained by substituting the unknown function with an exponential trial solution.

In our exercise, once we perform the substitution suggested in the step-by-step solution (\( x-1 \) with \( t \)), we replace \( y'' \) with \( r^2 \) and \( y' \) with \( r \) for the corresponding characteristic equation: \(t^2r^2 + 8tr + 12 = 0\). This transformation paves the way for determining the roots which in turn define the general solution of the differential equation.

Roots of Characteristic Equation

Finding the roots of the characteristic equation is a pivotal step in solving second-order linear differential equations. These roots indicate the nature of the solutions to the differential equation. They can be real and distinct, real and repeated, or complex conjugates.

In our example, after simplifying the characteristic equation, the quadratic formula is used to find the roots \( r \). The solutions \( r_1 \) and \( r_2 \) are then used to construct the general solution of the original differential equation. The quadratic equation's discriminant (\( b^2 - 4ac \) in the quadratic formula) determines if the roots are real and distinct, or if they are complex.

Singular Points in Differential Equations

Singular points, or singularities, in differential equations are values of the independent variable at which the coefficients of the highest derivatives become either zero or undefined, leading to complications in finding a solution.

Our exercise highlights the singular point at \( x=1 \) because the coefficient \( (x-1)^2 \) of \( y'' \) would be zero, which invalidates the standard form of a second-order linear differential equation. It is crucial to exclude such singular points from the interval of solution as they are points where the behavior of the solution may change fundamentally. Near singular points, special techniques or a separate analysis may be needed to understand the solutions.