Answer to Question #182086 in Differential Equations for Gordian

Question

Formulate a differential equation whose general solution is z(t)=A sin t-B cos t.

Answer 1

$Answer: z''+z=0$

$Check: \\ z'' + z = z''(t) + z(t) = (A \sin(t) - B\cos(t))'' + A\sin(t) - B\cos(t) = \\ = (A\cos(t) + B\sin(t))' + A\sin(t) - B\cos(t) = \\ = -A\sin(t) + B\cos(t) + A\sin(t) - B\cos(t) = 0$

If you decide strictly (optional):

$\begin{vmatrix} z(t) & z(t)' & z(t)'' \\ sin(t) & cos(t) & -sin(t) \\ -cos(t) & sin(t) & cos(t) \end{vmatrix} = 0 \\ \Rightarrow z cos^2(t) + z'sin(t)cos(t) + z''sin^2(t) + z''cos^2(t) + zsin^2(t) - z'sin(t)cos(t) = 0\\ \Rightarrow z (cos^2(t) + sin^2(t)) + z''(sin^2(t) + cos^2(t)) = 0 \;\; \;\; (sin^2(t) + cos^2(t) = 1)\\ \Rightarrow z + z'' = 0$