Author Topic: quiz4 tut0601  (Read 595 times)

Mingzhu Yu

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quiz4 tut0601
« on: October 19, 2019, 02:08:46 AM »
Give the general solution for
$$4y''+9y'= 0$$
We assume that $y=e^{r t}$ is a solution of equation,Now $y=e^{r t}$
Then $y^{\prime}=r e^{r t}$
And $y^{\prime \prime}=r^{2} e^{n}$
$$4 r^{2} +9=0$$
$$4r^{2}=9$$
$$r^{2}=\frac{-9}{4}$$
$$r=\pm \frac{3i}{2}$$
$$\therefore \lambda=0, \mu=\frac{3}{2}$$
Hence,The general solution is
$$y=c_{1}\cos{\frac{3}{2}t} +c_{2}\sin{\frac{3}{2}t}$$