Question

Identify the singular point. Find the general solution that is valid for values of \(t\) on either side of the singular point. $$ t^{2} y^{\prime \prime}+t y^{\prime}+9 y=0 $$

Short answer

Question: Determine the singular point of the given second-order ODE and find the general solution that is valid for values of \(t\) on either side of the singular point: \(t^2 y'' + t y' + 9y = 0\). Answer: The singular point is \(t=0\), and the general solution is \(y(t) = t^{\frac{1}{2}}\left[ C_1 \cos{(3\ln |t|)} + C_2 \sin{(3\ln |t|)} \right]\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

1. Identifying the Singular Point

To find the singular point, we need to locate the value of \(t\) at which the leading coefficient of the highest derivative becomes zero. In our case, the coefficient of the highest derivative (which is \(y''\)) is \(t^2\). Setting this equal to zero, we get: $$ t^2 = 0 $$ Solving for \(t\), we get \(t=0\) as the singular point.

2. Formulating the equation with the leading coefficient as 1

To find the general solution of the ODE, we start by dividing the equation through by the leading coefficient \(t^2\) to make the coefficient of the highest derivative equal to 1: $$ y'' + \frac{1}{t}y' + \frac{9}{t^2}y = 0 $$

3. Identifying the equation type and its solution method

The ODE we have now is of the form $$ y'' + \frac{P(t)}{t}y' + \frac{Q(t)}{t^2}y = 0 $$ which is a Cauchy-Euler equation. To solve a Cauchy-Euler equation, we try a solution of the form \(y(t) = t^r\) where \(r\) is a constant to be determined.

4. Substituting the proposed solution into the ODE

We differentiate the proposed solution \(y(t) = t^r\) with respect to \(t\): $$ y'(t) = r t^{r-1} $$ and again to get the second derivative: $$ y''(t) = r(r-1)t^{r-2} $$ Now we substitute \(y(t)\), \(y'(t)\), and \(y''(t)\) into the ODE: $$ r(r-1)t^{r-2} + \frac{1}{t}\cdot r t^{r-1} + \frac{9}{t^2} \cdot t^{r} = 0 $$ which simplifies to $$ r(r-1)t^{r} + rt^{r} + 9t^{r} = 0 $$

5. Solving the characteristic equation

Since \(t^{r}\) cannot be equal to zero, we can divide both sides by \(t^{r}\): $$ r(r-1) + r + 9 = 0 $$ which gives the quadratic equation $$ r^2 - r + 9 = 0 $$ The roots of this equation are complex conjugates: $$ r_1 = \frac{1}{2} + 3i, \quad r_2 = \frac{1}{2} - 3i $$

6. Writing down the general solution

Since the roots are complex conjugates, the general solution for the Cauchy-Euler ODE can be expressed as the linear combination of the real part and the imaginary part: $$ y(t) = t^{\frac{1}{2}}\left[ C_1 \cos{(3\ln |t|)} + C_2 \sin{(3\ln |t|)} \right] $$ where \(C_1\) and \(C_2\) are arbitrary constants. This general solution is valid for values of \(t\) on either side of the singular point \(t=0\).

Key concepts

Singular Point

In the context of Ordinary Differential Equations (ODEs), a singular point is a value where the function behaves "differently" than expected, often where the function may not be defined in the same manner. Specifically for Cauchy-Euler equations, singular points are where the leading coefficient of the highest-order derivative becomes zero. To find a singular point, examine the leading term, and set its coefficient equal to zero. For the given problem, the coefficient of the highest derivative is \( t^2 \). By setting \( t^2 = 0 \), we determine that \( t = 0 \) is the singular point. Identifying singular points is crucial, as they may imply special behavior or restrictions for the solution of a differential equation.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving a function and its derivatives. ODEs model a wide range of phenomena in physics, engineering, and other sciences.When working with ODEs, it's common to simplify and solve them by reducing the equation to a standard form. For example, by dividing through by the leading coefficient, we streamline the equation of interest. In the provided example, after dividing the entire equation by \( t^2 \), we obtained:\[y'' + \frac{1}{t}y' + \frac{9}{t^2}y = 0\]This transformation helps set the stage for methods specific to the form of the ODE, like the Cauchy-Euler method used here.

Complex Roots

When solving differential equations, you may encounter equations that have characteristic polynomials with complex roots. These arise, particularly in linear differential equations, often signaling oscillatory solutions.In the given problem, the characteristic equation derived was \( r^2 - r + 9 = 0 \). Solving this polynomial, we found two complex conjugate roots: - \( r_1 = \frac{1}{2} + 3i \)- \( r_2 = \frac{1}{2} - 3i \)Complex roots prompt us to use Euler's formula to express solutions involving sinusoidal functions. This is visible in the use of sine and cosine in the final general solution.

General Solution

The ultimate goal in solving an ODE is often to determine the general solution. This solution describes all possible behaviors governed by the differential equation.For our Cauchy-Euler equation, the general solution was derived by using the roots of the characteristic equation. Once determining the roots as complex, we used the relation to describe the solution in terms of real functions:\[y(t) = t^{\frac{1}{2}}\left[ C_1 \cos{(3\ln |t|)} + C_2 \sin{(3\ln |t|)} \right]\]Here, \( C_1 \) and \( C_2 \) are constants determined by initial conditions, allowing for a flexible solution that fits specific boundary conditions. This general solution is valid across intervals, excluding the singular point, in this case, \( t=0 \). Understanding the form and derivation of a general solution is essential for applying mathematical and theoretical models to real-world scenarios.