Answer
$$ y^{\prime \prime}-2 y^{\prime}+5 y=0, \quad y(\pi / 2)=0, \quad y^{\prime}(\pi / 2)=2 $$ The general solution of the given initial value problem is $$ y(t) =e^{t }( -e^{-\pi / 2 } \sin 2t)= -e^{t-\pi / 2 } \sin 2t $$ For increasing $t$, $ y(t) $ has growing oscillation.
Work Step by Step
$$ y^{\prime \prime}-2 y^{\prime}+5 y=0, \quad y(\pi / 2)=0, \quad y^{\prime}(\pi / 2)=2 \quad (i) $$ We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$ r^{2}-2r+5=0,$$ so its roots are $$ \:r_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1} $$ Thus the possible values of $r$ are $$r_{1}=1+2i ,\quad r_{2}=1-2i .$$ Therefore two solutions of Eq. (1) are $$ y_{1}(t) = e^{(1+2i )t}=e^{t }(\cos 2t + i \sin 2t) $$ and $$ y_{2}(t) = e^{(1-2i )t}=e^{t }(\cos 2t - i \sin 2t) $$ Thus the general solution of the differential equation is $$ y(t) =e^{t }( c_{1} \cos 2t+c_{2} \sin 2t) \quad (ii) $$ where $ c_{1} $and $c_{2}$ are arbitrary constants. To apply the first initial condition, we set $t = \pi / 2$ in Eq. (ii); this gives
$ y(\pi / 2) =-e^{\pi / 2 } c_{1} = 0, $ this implies that $ c_{1} = 0 .$ For the second initial condition we must differentiate Eq. (ii) as follows $$ y^{\prime}(t) =e^{t }[( c_{1}+2c_{2}) \sin 2t+(c_{2} -2c_{1}) \cos 2t ] $$ and then set $t =\pi / 2 $. In this way we find that $$ y^{\prime}(\pi / 2) =-e^{ \pi / 2 }(c_{2}-2c_{1}) =2, c_{1} = 0 $$ and we get $ c_{2} = -e^{-\pi / 2 }$. Using these values of $c_{2}=-e^{-\pi / 2 }$ and $c_{1}=0$ in Eq. (ii), we obtain $$ y(t) =e^{t }( -e^{-\pi / 2 } \sin 2t)= -e^{t-\pi / 2 } \sin 2t $$ as the solution of the initial value problem (i). Since $ \sin 2t $ is bounded and $ -e^{t-\pi / 2 }$ goes to $−∞$ as $ t → ∞$ , $ y(t) $ goes to $−∞ $ as $ t → ∞ $. From the graph we can tell that the function $ y(t) $ has growing oscillation.
