Question

Verify that \(y_{1}(t)=t^{2}\) and \(y_{2}(t)=t^{-1}\) are two solutions of the differential equation \(t^{2} y^{\prime \prime}-\) \(2 y=0\) for \(t>0 .\) Then show that \(c_{1} t^{2}+c_{2} t^{-1}\) is also a solution of this equation for any \(c_{1}\) and \(c_{2} .\)

Short answer

If yes, what is the general solution? Answer: Yes, \(y_1(t) = t^2\) and \(y_2(t) = t^{-1}\) are solutions of the given differential equation. The general solution is \(y(t) = c_1t^2 + c_2t^{-1}\), where \(c_1\) and \(c_2\) are constants.

Step-by-step solution

Find second derivatives of \(y_1(t)\) and \(y_2(t)\)

To check if the functions \(y_1(t)\) and \(y_2(t)\) are solutions of the given differential equation, we need to find their second derivatives. Let's do it separately: First, for \(y_1(t) = t^2\): - \(y_1'(t) = \frac{d(y_1)}{dt} = 2t\) - \(y_1''(t) = \frac{d^2(y_1)}{dt^2} = 2\) Now, for \(y_2(t) = t^{-1}\): - \(y_2'(t) = \frac{d(y_2)}{dt} = -t^{-2}\) - \(y_2''(t) = \frac{d^2(y_2)}{dt^2} = 2t^{-3}\)

Verify the solutions

To verify if \(y_1(t)\) and \(y_2(t)\) are solutions of the given differential equation, we will plug their corresponding second derivatives and functions into the equation \(t^2 y'' - 2y = 0\). For \(y_1(t)\): - \(t^2(2) - 2(t^2) = 0\), which holds true. So, \(y_1(t) = t^2\) is a solution. For \(y_2(t)\): - \(t^2(2t^{-3}) - 2(t^{-1}) = 2t - 2t = 0\), which also holds true. So, \(y_2(t) = t^{-1}\) is a solution too.

Show the general solution with constants

Now that we have verified \(y_1(t)\) and \(y_2(t)\) as solutions, we will check whether a linear combination of these functions, represented as \(y(t) = c_1t^2 + c_2t^{-1}\), is also a solution. First, find the first and second derivatives of \(y(t)\): - \(y'(t) = \frac{d(y)}{dt} = 2c_1t - c_2t^{-2}\) - \(y''(t) = \frac{d^2(y)}{dt^2} = 2c_1 + 2c_2t^{-3}\) Now, plug the second derivative and \(y(t)\) into the given equation: - \(t^2(2c_1 + 2c_2t^{-3}) - 2(c_1t^2 + c_2t^{-1}) = 2c_1t^2 + 2c_2 - 2c_1t^2 - 2c_2 = 0\) Since the equation holds true, the linear combination \(y(t) = c_1t^2 + c_2t^{-1}\) is also a solution of the given equation for any constants \(c_1\) and \(c_2\).

Key concepts

Second Derivative

When working with differential equations, the concept of a second derivative plays a crucial role in understanding the behavior of solutions. In mathematical terms, the second derivative is the derivative of the derivative, and it represents the rate at which the first derivative is changing.

For example, in motion physics, if the position of an object is given by a function, the first derivative represents velocity, and the second derivative represents acceleration. In our textbook exercise, by calculating the second derivatives of the given solutions, we establish the functions’ concavity and assess whether they satisfy the specified differential equation. This determination is a pivotal step in verifying the behavior of the solutions in relation to the acceleration of the system described by the differential equation.

Linear Combination of Solutions

A powerful tool in solving differential equations is the principle of superposition, which states that the linear combination of solutions is itself a solution, as long as the differential equation is linear and homogeneous. This means that if we have two solutions, say \(y_{1}(t)\) and \(y_{2}(t)\), to a given differential equation, we can form a new solution by taking \(c_{1}y_{1}(t) + c_{2}y_{2}(t)\), where \(c_{1}\) and \(c_{2}\) are arbitrary constants.

In the exercise, after verifying that \(y_{1}(t)=t^{2}\) and \(y_{2}(t)=t^{-1}\) are indeed solutions, we demonstrate that any linear combination of these functions also solves the original equation. This property is essential when we need to find a general solution that includes all possible specific solutions to the differential equation.

General Solution of Differential Equations

The general solution to a differential equation represents an entire family of possible solutions, typically involving arbitrary constants. This differs from a particular solution, which corresponds to one specific instance obtained by giving particular values to the arbitrary constants. The general solution captures all possible behaviors allowed by the differential equation under various initial conditions.

The last step of the textbook exercise shows how to express the general solution using the verified individual solutions and their linear combination. By including the arbitrary constants \(c_{1}\) and \(c_{2}\), we accommodate for any initial conditions that might apply to the problem at hand. Thus, the general solution \(y(t) = c_{1}t^{2} + c_{2}t^{-1}\) embodies all the particular solutions of the differential equation \(t^{2} y'' - 2y = 0\) for \(t > 0\), highlighting the equation’s versatility and the variety of scenarios it can represent.