Respuesta :
The question is:
Consider the differential equation y'' - 2y' + 37y = 0;
[tex]e^x cos(6x), e^x sin(6x), (-\infty, \infty).[/tex]
Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.
The functions satisfy the differential equation and are linearly independent since
a) [tex]W(e^x cos 6x, e^x sin 6x) \neq 0[/tex]
Answer:
To verify if the given functions form a fundamental set of solutions to the differential equation, we find the Wronskian of the two functions.
The Wronskian of functions [tex]y_1 $and$ y_2[/tex] is given as[tex]W(y_1, y_2) = \left|\begin{array}{cc}y_1&y_2\\y_1'&y_2'\end{array}\right|\\\\y_1= e^x cos 6x \\y_2 = e^x sin 6x \\y_1' = -6e^x sin 6x \\y_2' = 6e^x cos 6x \\W\left(e^x cos 6x, e^x sin 6x \right) = \left|\begin{array}{cc}e^x cos 6x &e^x sin 6x \\ \\ -6e^x sin 6x&6e^x cos 6x \end{array}\right|\\ \\= 6e^{2x} cos^2 6x + e^{2x} sin^2 6x \\ \\ = 6e^{2x}\left( cos^2 6x + sin^2 6x\right)\\ \\$but $cos^2 6x + sin^2 6x = 1\\ \\W\left(y_1, y_2 \right) = 6e^{2x} \neq 0[/tex]
Because the Wronskian is not zero, we say the solutions are linearly independent
