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Chapter 4: Problem 38

Consider the equation \(y^{\mathrm{iv}}-y=0\) (a) Use Abel's formula [Problem \(20(\mathrm{d}) \text { of Section } 4.1]\) to find the Wronskian of a fundamental set of solutions of the given equation. (b) Determine the Wronskian of the solutions \(e^{t}, e^{-t}, \cos t,\) and \(\sin t .\) (c) Determine the Wronskian of the solutions \(\cosh t, \sinh t, \cos t,\) and \(\sin t .\)

Short Answer

Expert verified
In this exercise, we found a fundamental set of solutions for the given fourth-order linear homogeneous differential equation, used Abel's formula to calculate the Wronskian, and then found the Wronskian for two given sets of functions. We determined that the Wronskian of both \(e^{t}, e^{-t}, \cos{t}, \sin{t}\) and \(\cosh{t}, \sinh{t}, \cos{t}, \sin{t}\) is \(-4\).

Step by step solution

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01

Finding a Fundamental Set of Solutions

First, let's identify the characteristic equation for our given differential equation: $$r^4 - 1 = 0$$ To find the roots, we can rewrite it as $$(r^2 + 1)(r^2 - 1) = 0$$ Now we have two quadratic equations, $$r^2 - 1 = 0$$ and $$r^2 + 1 = 0$$ Solving each equation, we get the roots: \(r_1 = 1, r_2 = -1, r_3 = i, r_4 = -i.\) Using these roots, we can construct a fundamental set of solutions for the given differential equation. The solutions are: $$y_1(t) = e^t, y_2(t) = e^{-t}, y_3(t) = \cos t, y_4(t) = \sin t$$
02

Using Abel's formula

Recall that Abel's formula gives the Wronskian W of a linear homogeneous differential equation. We can calculate the Wronskian of the fundamental set of solutions we found without calculating the derivatives. According to Abel's formula, $$W(y_1, y_2, y_3, y_4)(t) = ce^{\int p(t) dt}$$ In our case, since the given equation is \(y^{\mathrm{iv}}-y=0\), there are no coefficients of lower-order derivatives, so \(p(t) = 0\). Therefore, $$W(y_1, y_2, y_3, y_4)(t) = ce^{\int 0 dt} = c$$ for any non-zero constant \(c\).
03

Determining the Wronskian (Part b)

Now we will find the Wronskian of the given solutions $$e^{t}, e^{-t}, \cos t, \text{ and } \sin t$$ using the definition of the Wronskian. The Wronskian is a determinant of their derivatives, so we have: \[W(y_1, y_2, y_3, y_4)(t) = \begin{vmatrix} e^t & e^{-t} & \cos{t} & \sin{t} \\ e^t & -e^{-t} & -\sin{t} & \cos{t} \\ e^t & e^{-t} & -\cos{t} & -\sin{t} \\ e^t & -e^{-t} & \sin{t} & -\cos{t} \end{vmatrix}\] Calculating this determinant, we get: \[W(y_1, y_2, y_3, y_4)(t) = -4\] So, the Wronskian of these solutions is \(-4\).
04

Determining the Wronskian (Part c)

Finally, we will find the Wronskian of the given solutions $$\cosh t, \sinh t, \cos t, \text{ and } \sin t$$ using the definition of the Wronskian. The Wronskian is a determinant of their derivatives, so we have: \[W(y_1, y_2, y_3, y_4)(t) = \begin{vmatrix} \cosh{t} & \sinh{t} & \cos{t} & \sin{t} \\ \sinh{t} & \cosh{t} & -\sin{t} & \cos{t} \\ \cosh{t} & \sinh{t} & -\cos{t} & -\sin{t} \\ \sinh{t} & \cosh{t} & \sin{t} & -\cos{t} \end{vmatrix}\] Calculating this determinant, we get: \[W(y_1, y_2, y_3, y_4)(t) = -4\] So, the Wronskian of these solutions is also \(-4\).

Key Concepts

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

Wronskian
The Wronskian is a determinant used to verify if a set of functions are linearly independent. Linear independence is a crucial property for solutions in differential equations because it ensures that each solution adds new information and none is redundant. The Wronskian, denoted by \( W(y_1, y_2, ..., y_n) \), is a determinant of a matrix where each row consists of successive derivatives of the functions.
  • If the Wronskian is not zero at some point in the interval of interest, the functions are linearly independent.
  • If it is zero everywhere, the independence cannot be confirmed without further investigation.
The computation involves calculating a rather complex determinant, where each element is the derivative of a function in the previous row. Therefore, the size of the Wronskian matrix grows with the number of functions considered.
Fundamental Set of Solutions
A fundamental set of solutions comprises solutions to a differential equation that, together, can express the general solution for the equation. For a linear homogeneous differential equation of order \( n \), this set contains \( n \) linearly independent solutions.When a fundamental set is determined, any solution of the differential equation can be expressed as a linear combination:\[ y(t) = c_1y_1(t) + c_2y_2(t) + ... + c_ny_n(t) \]where \( c_1, c_2, ..., c_n \) are constants. The importance of the fundamental set arises from its ability to cover all possible behaviors of the system described by the differential equation.In our exercise, the solutions \( e^t, e^{-t}, \cos t, \sin t \) form a fundamental set for the differential equation \(y^{\text{iv}} - y = 0\). This means any solution to this differential equation can be synthesized from these four functions.
Abel's Formula
Abel's Formula provides a way to determine the Wronskian of a set of solutions to a second-order linear differential equation without needing to compute derivatives.This formula is especially useful because it relates directly to the solutions of the differential equation and their parameterization over an interval, plus it accounts for initial conditions without diving into the often tedious derivative calculation.The formula states:\[ W(y_1, y_2, \, ..., \, y_n)(t) = c \cdot e^{\int p(t) \,dt} \]Here, \( p(t) \) is the coefficient of the highest-order derivative less one in the differential equation. For our equation, since \( p(t) = 0 \), the Wronskian simplifies to a constant \( c \), implying that it remains unchanged over time. Knowing Abel's Formula allows us to skip the determinant computation when solving differential equations with constant coefficients.
Characteristic Equation
The characteristic equation is a critical step in solving linear homogeneous differential equations. It transforms the differential equation into an algebraic form, making it easier to solve. The characteristic equation is derived by assuming a solution of the form \( y = e^{rt} \), where \( r \) represents the roots to be determined.For example, given the differential equation \( y^{\text{iv}} - y = 0 \), we construct the characteristic equation:\[ r^4 - 1 = 0 \]This equation is then factored into:\[ (r^2 + 1)(r^2 - 1) = 0 \]leading to the roots: \( r = 1, -1, i, -i \). These roots are vital as they allow us to express the general solution of the differential equation. Each root provides insight, such as real roots leading to exponential solutions and complex roots leading to sinusoidal solutions. Solving the characteristic equation is foundational, as it informs the fundamental set of solutions necessary to address any problems dictated by the differential equation.

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Most popular questions from this chapter

Find the solution of the given initial value problem and plot its graph. How does the solution behave as \(t \rightarrow \infty ?\) $$ \begin{array}{l}{2 y^{\mathrm{iv}}-y^{\prime \prime \prime}+4 y^{\prime}+4 y=0 ; \quad y(0)=-2, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=-2} \\\ {y^{\prime \prime \prime}(0)=0}\end{array} $$

The purpose of this problem is to show that if \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right) \neq 0\) for some \(t_{0}\) in an interval \(I,\) then \(y_{1}, \ldots, y_{n}\) are linearly independent on \(I,\) and if they are linearly independent and solutions of $$ L(y]=y^{(n)}+p_{1}(t) y^{(n-1)}+\cdots+p_{n}(t) y=0 $$ on \(I,\) then \(W\left(y_{1}, \ldots, y_{n}\right)\) is nowhere zero in \(I .\) (a) Suppose that \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right) \neq 0,\) and suppose that $$ c_{1} y_{1}(t)+\cdots+c_{n} y_{n}(t)=0 $$ for all \(t\) in \(I\). By writing the equations corresponding to the first \(n-1\) derivatives of Fa. (ii) at \(t_{0}\), show that \(c_{1}=\cdots=c_{n}=0 .\) Therefore, \(y_{1}, \ldots, y_{n}\) are linearly independent. (b) Suppose that \(y_{1}, \ldots, y_{n}\) are linearly independent solutions of Eq. (i). If \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right)=0\) for some \(t_{0},\) show that there is a nonzero solution of Eq. (i) satisfying the initial conditions $$ y\left(t_{0}\right)=y^{\prime}\left(t_{0}\right)=\cdots=y^{(n-1)}\left(t_{0}\right)=0 $$ since \(y=0\) is a solution of this initial value problem, the uniqueness part of Theorem 4. 1. I yields a contradiction. Thus \(W\) is never zero.

Consider the nonhomogeneous \(n\) th order linear differential equation $$ a_{0} y^{(n)}+a_{1} y^{(n-1)}+\cdots+a_{n} y=g(t) $$ where \(a_{0}, \ldots, a_{n}\) are constants. Verify that if \(g(t)\) is of the form $$ e^{\alpha t}\left(b_{0} t^{m}+\cdots+b_{m}\right) $$ then the substitution \(y=e^{\alpha t} u(t)\) reduces the preceding equation to the form $$ k_{0} u^{(n)}+k_{1} u^{(n-1)}+\cdots+k_{n} u=b_{0} t^{m}+\cdots+b_{m} $$ where \(k_{0}, \ldots, k_{n}\) are constants. Determine \(k_{0}\) and \(k_{n}\) in terms of the \(a^{\prime}\) 's and \(\alpha .\) Thus the problem of determining a particular solution of the original equation is reduced to the simpler problem of determining a particular solution of an equation with constant coefficients and a polynomial for the nonhomogeneous term.

verify that the given functions are solutions of the differential equation, and determine their Wronskian. $$ y^{\mathrm{iv}}+y^{\prime \prime}=0 ; \quad 1, \quad t, \quad \cos t, \quad \sin t $$

Find the solution of the given initial value problem. Then plot a graph of the solution. \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3 t+4, \quad y(0)=y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=y^{\prime \prime \prime}(0)=1\)
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