## Elementary Differential Equations and Boundary Value Problems 9th Edition

$$y^{\prime \prime}+2 y^{\prime}+1.25 y=0$$ The general solution of that equation is given by $$y(t) =e^{-t}( c_{1} \cos\frac{t}{2} +c_{2} \sin\frac{t}{2} )$$ where $c_{1}$and $c_{2}$ are arbitrary constants.
$$y^{\prime \prime}+2 y^{\prime}+1.25 y=0 \quad\quad\quad \quad (1)$$ We assume that $y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$r^{2}+2 r+1.25=100 r^{2}+200 r+125=0,$$ so its roots are $$\:r_{1,\:2}=\frac{-200\pm \sqrt{200^2-4\cdot \:100\cdot \:125}}{2\cdot \:100}$$ Thus the possible values of $r$ are $$r_{1}=-1+i\frac{1}{2} ,\quad r_{2}=-1-i\frac{1}{2} .$$ Therefore two solutions of Eq. (1) are $$y_{1}(t) = e^{(-1+i\frac{1}{2})t} =e^{-t}(\cos \frac{t}{2} + i \sin \frac{t}{2})$$ and $$y_{1}(t) = e^{(-1-i\frac{1}{2})t} =e^{-t}(\cos \frac{t}{2} - i \sin \frac{t}{2})$$ Thus the general solution of the differential equation is $$y(t) =e^{-t}( c_{1} \cos\frac{t}{2} +c_{2} \sin\frac{t}{2} )$$ where $c_{1}$and $c_{2}$ are arbitrary constants.