Question

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

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

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: $$y_1(t) = e^t, y_2(t) = e^{-t}, y_3(t) = \cos t, y_4(t) = \sin t$$

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 \(c\).

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 \(-4\).

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 \(-4\).