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

$$y^{\prime \prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1$$ The general solution of the given initial value problem is $$y(t) =\frac{1}{2} \sin 2t$$ For increasing $t$, $y$ will be steadily oscillating
$$y^{\prime \prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1 \quad (i)$$ We assume that $y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$r^{2}+4=0,$$ so its roots are $$\quad \:r_{1,\:2}=\frac{\pm \sqrt{-4\cdot \:1\cdot \:4}}{2\cdot \:1}$$ Thus the possible values of $r$ are $$r_{1}=2i ,\quad r_{2}=-2i .$$ Therefore two solutions of Eq. (1) are $$y_{1}(t) = e^{2i}=(\cos 2t + i \sin 2t)$$ and $$y_{2}(t) = e^{-2i}=(\cos 2t - i \sin 2t)$$ Thus the general solution of the differential equation is $$y(t) =( c_{1} \cos 2t+c_{2} \sin 2t) \quad (ii)$$ where $c_{1}$ and $c_{2}$ are arbitrary constants. To apply the first initial condition, we set $t = 0$ in Eq. (ii); this gives $y(0) = c_{1} = 0.$ For the second initial condition we must differentiate Eq. (ii) as follows $$y^{\prime}(t) =( -2c_{1} \sin 2t+2c_{2} \cos 2t)$$ and then set $t = 0$. In this way we find that $$y^{\prime}(0) =2c_{2} =1$$ from which $c_{2} = \frac{1}{2}$. Using these values of $c_{1}=0$ and $c_{2}=\frac{1}{2}$ in Eq. (ii), we obtain $$y(t) =\frac{1}{2} \sin 2t$$ as the solution of the initial value problem (i). For increasing $t$, $y$ will be steadily oscillating