Question

Show that Y=(sinRt-sinKt)/(K^2-R^2) is a particular integral of the equation y"+yK^2=sinRt and investigate the limiting case when R->K.

Answer 1

$$Y=\frac{\sin Rt-\sin Kt}{K^2-R^2}$$gives us:$$Y'=\frac{R\cos Rt-K\cos Kt}{K^2-R^2}\\Y''=-\frac{R^2\sin Rt-K^2\sin Kt}{K^2-R^2}$$so:$$Y''+K^2Y=-\frac{R^2\sin Rt-K^2\sin Kt}{K^2-R^2}+\frac{K^2\sin Rt-K^2\sin Kt}{K^2-R^2}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{K^2-R^2}{K^2-R^2}\sin Rt$$