Question

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \((x+1)^{2} y^{\prime \prime}+3(x+1) y^{\prime}+0.75 y=0\)

Short answer

Answer: The general solution of the given differential equation is \(y(x) = C_{1}(x+1)^{-1 + \frac{1}{2}\sqrt{2}} + C_{2}(x+1)^{-1 - \frac{1}{2}\sqrt{2}}\), in any interval not containing the singular point \(x = -1\).

Step-by-step solution

Identify the Singular Point

First, consider the equation given: \((x+1)^2y'' + 3(x+1)y' + 0.75y = 0\). The singular point occurs when the coefficient of the highest-order derivative, \((x+1)^2\), is equal to zero. Let's find the singular point: \((x+1)^2 = 0\) \(x+1 = 0\) \(x = -1\). So, the singular point is \(x = -1\).

Use the Change of Variables and Transformation

To obtain a simpler differential equation, we perform a change of variables and transformation. Set \(u = x + 1\), and express the given equation in terms of u. The derivatives also need to be transformed: \(y'(x) = y'(u - 1) = y'(u)\frac{du}{dx}\), \(y''(x) = y''(u - 1)\frac{du^2}{dx^2}\). Now, express the given equation in terms of u and remove the factors \((u-1)\): \(u^{2}y'' + 3uy' + 0.75y = 0\).

Applying Reduction of Order

The differential equation can be rewritten as: \(u^{2}y'' + 3uy' + 0.75y = 0\). Now, we need to apply reduction of order by using a substitution method. We are going to assume that \(y(u) = u^r\). Derive \(y(u)\) w.r.t \(u\) twice: \(y'(u) = ru^{r-1}\), \(y''(u) = r(r-1)u^{r-2}\). Now we plug \(y(u)\), \(y'(u)\), and \(y''(u)\) into the given equation: \(u^2(r(r-1)u^{r-2}) + 3u(ru^{r-1}) + 0.75(u^r) = 0\). Simplify the equation by dividing by \(u^{r-2}\): \(r(r-1)u^2 + 3ru + 0.75u^{r+2} = 0\).

Solve for the Characteristic Equation

To find the general solution, solve the characteristic equation: \(r(r-1) + 3r + 0.75 = 0\) \(r^2 - 1r + 3r + 0.75 = 0\) \(r^2 + 2r + 0.75 = 0\). Now, we solve this quadratic equation for 'r': \(r_{1,2} = \frac{-2 \pm \sqrt{(-2)^2 - 4 (1)(0.75)}}{2(1)}\) \(r_{1,2} = -1 \pm \frac{1}{2}\sqrt{2}\). Thus, we have two different real solutions for \(r\): \(r_1 = -1 + \frac{1}{2}\sqrt{2}\) and \(r_2 = -1 - \frac{1}{2}\sqrt{2}\).

Find the General Solution

Contrast the general solution using the found r-values: \(y(u) = C_{1}u^{r_1} + C_{2}u^{r_2}\) \(y(u) = C_{1}u^{-1 + \frac{1}{2}\sqrt{2}}+ C_{2}u^{-1 - \frac{1}{2}\sqrt{2}}\). Now, convert back to \(x\) variable: \(y(x) = C_{1}(x+1)^{-1 + \frac{1}{2}\sqrt{2}}+ C_{2}(x+1)^{-1 - \frac{1}{2}\sqrt{2}}\). Thus, we have found the general solution of the differential equation that is valid in any interval not including the singular point \(x = -1\): \(y(x) = C_{1}(x+1)^{-1 + \frac{1}{2}\sqrt{2}}+ C_{2}(x+1)^{-1 - \frac{1}{2}\sqrt{2}}\).

Key concepts

Singular Point

In the study of differential equations, a singular point is a value for which the equation may not behave 'nicely', often leading to solutions that can't be expressed in a standard form. Specifically, it's a point where the coefficient of the highest derivative is zero, which can create complications in finding solutions.

For our example, \( (x+1)^2 y'' + 3(x+1) y' + 0.75 y = 0 \), we determine the singular point by setting the highest-order coefficient \( (x+1)^2 \) to zero and solving for \( x \). The result is \( x = -1 \), indicating that the solutions may be problematic at this point, and any analysis or solution we generate should avoid including the singular point in the interval.

Reduction of Order

When dealing with second-order differential equations, reduction of order is a valuable technique, especially when one solution is known and we need to find another. That being said, it can also be useful in other contexts as a way to simplify the process of finding the solution to a differential equation by reducing its order.

In our example, the substitution \( y(u) = u^r \) helps to reduce the original equation into a form where the unknown function appears with only its derivatives, significantly simplifying the process. Once the appropriate derivatives of \( y(u) \) are substituted into the original equation, we can solve for \( r \) to find the general form of the solution.

Characteristic Equation

The characteristic equation is a cornerstone when solving linear homogeneous differential equations with constant coefficients. It's derived from the original equation by assuming solutions of the exponential form and leads to an algebraic equation whose roots are vital in finding the general solution.

In the process of applying the reduction of order, as seen in our example, we eventually arrive at an algebraic equation dependent on \( r \) which is the characteristic equation. Solving this allows us to get the specific values for \( r \) necessary to construct the general solution. For our differential equation, the characteristic equation \( r^2 + 2r + 0.75 = 0 \) results in two distinct real roots, which paves the way to construct the general solution.

General Solution of Differential Equation

The general solution of a differential equation represents the set of all possible solutions to the equation. It incorporates arbitrary constants to account for the various initial conditions the equation might be subjected to.

In the context of second-order linear homogeneous differential equations, once we have the solutions to the characteristic equation, we can write the general solution as a linear combination of the functions represented by these roots. In our exercise, after determining the roots \( r_1 \) and \( r_2 \), the general solution is expressed as \( y(x) = C_{1}(x+1)^{r_1} + C_{2}(x+1)^{r_2} \) where \( C_{1} \) and \( C_{2} \) are the arbitrary constants. The final step involves translating that solution back in terms of the original variable \( x \) to complete the process.