Question

A Euler equation \(\left(t-t_{0}\right)^{2} y^{\prime \prime}+\alpha\left(t-t_{0}\right) y^{\prime}+\beta y=0\) is known to have the given general solution. What are the constants \(t_{0}, \alpha\), and \(\beta\) ? $$ y(t)=c_{1}(t+2)+c_{2} \frac{1}{(t+2)^{2}}, \quad t \neq-2 $$

Short answer

Question: Determine the values of \(t_0\), \(\alpha\), and \(\beta\) in the Euler equation for the given general solution \(y(t) = c_1(t+2) + c_2 \frac{1}{(t+2)^{2}}, \quad t \neq -2\). Answer: The constants are \(t_0 = -2\), \(\alpha = 1\), and \(\beta = c_1\).

Step-by-step solution

Identifying the given general solution

The given general solution is: $$ y(t) = c_1(t+2) + c_2 \frac{1}{(t+2)^{2}}, \quad t \neq -2 $$

Differentiating the general solution

Let's find the first and second derivatives of \(y(t)\) with respect to \(t\). First derivative: $$ y'(t) = \frac{d}{dt}(c_1(t+2)+c_2\frac{1}{(t+2)^2}) = c_1 - 2c_2 \frac{1}{(t+2)^3} $$ Second derivative: $$ y''(t) = \frac{d^2}{dt^2}(y(t)) = \frac{d}{dt}(c_1 - 2c_2 \frac{1}{(t+2)^3}) = 6c_2 \frac{1}{(t+2)^4} $$

Substituting into the Euler equation

Now, substitute the given solution, \(y(t)\), and its derivatives, \(y'(t)\) and \(y''(t)\), into the Euler equation: $$ \left(t-t_0\right)^2 y''(t) + \alpha \left(t-t_0\right) y'(t) + \beta y(t) = 0 $$ Substitute: $$ \left(t-t_0\right)^2\left(6c_2 \frac{1}{(t+2)^4}\right) + \alpha \left(t-t_0\right) \left(c_1 - 2c_2\frac{1}{(t+2)^3}\right) + \beta \left(c_1(t+2)+c_2 \frac{1}{(t+2)^2}\right) = 0 $$

Comparing coefficients

From the given general solution, we can observe that: $$ \left(t+2\right)^{4} = (t-t_0)^2 $$ So, comparing exponents on both sides, we have \(t_0 = -2\). Now let's compare the coefficients: 1. For the \(y''(t)\) term: \((t-t_0)^2 = (t+2)^2\), so no additional constants. 2. For the \(y'(t)\) term: \(\alpha(t-t_0) = \alpha(t+2)\), so \(\alpha = 1\). 3. For the \(y(t)\) term: \(\beta (c_1(t+2)+c_2 \frac{1}{(t+2)^2}) = \beta (t+2)\), so \(\beta = c_1\). Thus, the constants are \(t_0 = -2\), \(\alpha = 1\), and \(\beta = c_1\).

Key concepts

General Solution

The general solution of a differential equation represents a family of solutions that encompasses all possible specific solutions. It generally contains arbitrary constants, which can then be adjusted to meet specific initial conditions or boundary conditions. In the given problem, the general solution for the Euler equation is given by:\[ y(t) = c_1(t+2) + c_2 \frac{1}{(t+2)^2}, \quad t eq -2 \]- **Arbitrary constants:** In this solution, \(c_1\) and \(c_2\) are arbitrary constants. These can be modified to fit specific scenarios or initial conditions.- **Structure:** The first part, \(c_1(t+2)\), represents a linear relationship, while the second part, \(c_2\frac{1}{(t+2)^2}\), represents a decay-type relationship.The exceptional point \(t eq -2\) indicates a singularity point where the function is not defined. The general solution thus provides flexibility to adjust by changing these constants and can describe a wide range of behaviors depending on their values.

Differential Equations

Differential equations involve derivatives of a function. They're used to model real-world phenomena where change over time is crucial, like motion, growth, or decay.**Euler's Differential Equations** specifically deal with cases that have polynomial coefficients linked to the variable. They're useful for equations where coefficients are powers of the independent variable, such as time.The differential equation in the exercise is:\[ (t-t_0)^2 y'' + \alpha(t-t_0)y' + \beta y = 0 \]This type of equation often requires techniques beyond simple algebra for a solution, given the powers present. Here:- **Second-Order:** The highest derivative is the second derivative, \(y''\).- **Coefficients:** Polynomial terms \((t-t_0)^2\) make it Euler-specific, indicating the solution's nature might involve terms with powers of \(t\). Using Euler's method allows transforming the problem into a form where standard solution techniques can apply, essentially making complex phenomena approachable.

Initial Conditions

Initial conditions help in determining the specific solution from the general one by setting exact values for the solution and its derivatives at a given point. While the exercise doesn't explicitly present initial conditions, understanding their role is crucial.- **Purpose:** They eliminate the arbitrary constants in the general solution, giving a unique solution that aligns with physical or theoretical constraints.- **Example of Application:** Suppose you knew the value of \(y(t)\) at two points or the values of \(y(t)\) and its derivative \(y'(t)\) at one point. Substituting these into the general solution would help in finding specific values for \(c_1\) and \(c_2\).Ultimately, initial conditions enable solutions to not just exist in theory but to precisely match real-world data or specific cases, making them essential for problem-solving exercises.