Author Topic: TUT0302 Quiz3  (Read 238 times)

Aiting Zhang

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TUT0302 Quiz3
« on: October 11, 2019, 02:00:00 PM »
$$\mbox{Find the solution of the given initial value problem }$$
$${2y}''+{y}'-4y=0$$
$$y\left(0 \right)=1 \mbox{ and } {y}'\left(0 \right)=0$$
$$2r^2+r-4=0$$
$$r=\frac{-1\pm\sqrt{1-4\times2\times(-4)}}{4}=\frac{-1\pm\sqrt{33}}{4}$$
$$\mbox{Hence, }r_{1}=\frac{-1+\sqrt{33}}{4} \mbox{ and }  r_{2}=\frac{-1-\sqrt{33}}{4}$$
$$\mbox{Then the general solution of the given differential equation is }$$
$$y=c_{1}e^{\frac{-1+\sqrt{33}}{4} t}+c_{2}e^{\frac{-1-\sqrt{33}}{4} t}$$
$$\because y\left(0 \right)=1$$
$$\therefore c_{1}+c_{2}=1$$
$${y}'=\frac{-1+\sqrt{33}}{4}c_{1}e^{\frac{-1+\sqrt{33}}{4} t}+\frac{-1-\sqrt{33}}{4}c_{2}e^{\frac{-1-\sqrt{33}}{4} t}$$
$$\because {y}'\left(0 \right)=0$$
$$\therefore \frac{-1+\sqrt{33}}{4}c_{1}+\frac{-1-\sqrt{33}}{4}c_{2}=0$$
$$\mbox{Since }c_{1}+c_{2}=1, \mbox{so }c_{1}=1-c_{2}$$
$$\therefore \frac{-1+\sqrt{33}}{4}(1-c_{2})+\frac{-1-\sqrt{33}}{4}c_{2}=0$$
$$ \frac{-1+\sqrt{33}}{4}-c_{2}\frac{2\sqrt{33}}{4}=0$$
$$\therefore c_{2}=\frac{33-\sqrt{33}}{66}$$
$$c_{1}=1-\frac{33-\sqrt{33}}{66}=\frac{33+\sqrt{33}}{66}$$
$$\mbox{Therefore, the solution of the given initial value problem is }$$
$$y=\frac{33+\sqrt{33}}{66}e^{\frac{-1+\sqrt{33}}{4} t}+\frac{33-\sqrt{33}}{66}e^{\frac{-1-\sqrt{33}}{4} t}$$