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Chapter 3: Problem 13

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

Expert verified
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

01

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 \(y_2(t) = t^{-1}\): - \(y_2'(t) = \frac{d(y_2)}{dt} = -t^{-2}\) - \(y_2''(t) = \frac{d^2(y_2)}{dt^2} = 2t^{-3}\)
02

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.
03

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 \(c_2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥� 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鈥檚 versatility and the variety of scenarios it can represent.

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Most popular questions from this chapter

Use the method of reduction of order to find a second solution of the given differential equation. \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x\)

Show that \(y=\sin t\) is a solution of $$ y^{\prime \prime}+\left(k \sin ^{2} t\right) y^{\prime}+(1-k \cos t \sin t) y=0 $$ for any value of the constant \(k .\) If \(00\) and \(k \sin ^{2} t \geq 0\). Thus observe that even though the coefficients of this variable coefficient differential equation are nonnegative (and the coefficient of \(y^{\prime}\) is zero only at the points \(t=0, \pi, 2 \pi, \ldots\), it has a solution that does not approach zero as \(t \rightarrow \infty .\) Compare this situation with the result of Problem \(38 .\) Thus we observe a not unusual situation in the theory of differential equations: equations that are apparently very similar can have quite different properties.

In the absence of damping the motion of a spring-mass system satisfies the initial value problem $$ m u^{\prime \prime}+k u=0, \quad u(0)=a, \quad u^{\prime}(0)=b $$ (a) Show that the kinetic energy initially imparted to the mass is \(m b^{2} / 2\) and that the potential energy initially stored in the spring is \(k a^{2} / 2,\) so that initially the total energy in the system is \(\left(k a^{2}+m b^{2}\right) / 2\). (b) Solve the given initial value problem. (c) Using the solution in part (b), determine the total energy in the system at any time \(t .\) Your result should confirm the principle of conservation of energy for this system.

The position of a certain undamped spring-mass system satisfies the initial value problem $$ u^{\prime \prime}+2 u=0, \quad u(0)=0, \quad u^{\prime}(0)=2 $$ (a) Find the solution of this initial value problem. (b) Plot \(u\) versus \(t\) and \(u^{\prime}\) versus \(t\) on the same axes. (c) Plot \(u^{\prime}\) versus \(u ;\) that is, plot \(u(t)\) and \(u^{\prime}(t)\) parametrically with \(t\) as the parameter. This plot is known as a phase plot and the \(u u^{\prime}\) -plane is called the phase plane. Observe that a closed curve in the phase plane corresponds to a periodic solution \(u(t) .\) What is the direction of motion on the phase plot as \(t\) increases?

Deal with the initial value problem $$ u^{\prime \prime}+0.125 u^{\prime}+u=F(t), \quad u(0)=2, \quad u^{\prime}(0)=0 $$ (a) Plot the given forcing function \(F(t)\) versus \(t\) and also plot the solution \(u(t)\) versus \(t\) on the same set of axes. Use a \(t\) interval that is long enough so the initial transients are substantially eliminated. Observe the relation between the amplitude and phase of the forcing term and the amplitude and phase of the response. Note that \(\omega_{0}=\sqrt{k / m}=1\). (b) Draw the phase plot of the solution, that is, plot \(u^{\prime}\) versus \(u .\) \(F(t)=3 \cos 3 t\)
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