Author Topic: TUT0602 quiz 4 solution  (Read 610 times)

christine

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TUT0602 quiz 4 solution
« on: October 18, 2019, 02:00:02 PM »
Question: Find the general solution to the given equation: $4y''+9y=0$

Answer:

Assume $y=e^{rt}$ is a solution of this equation, then we can write $4r^2+9=0$
or $r^2=\frac{-9}{4}$
so we get $r=\pm\frac{3}{2}i$

Recall that if the roots are complex, we have them in the form of $\lambda\pm i\mu, \mu\neq 0$
in this case $\lambda=0$ and $\mu=\frac{3}{2}$

Then the general equation is given by the following:
$y=c_1e^{\lambda}cos(\mu t)+c_2e^{\lambda}sin(\mu t)$
$y=c_1e^0cos(\frac{3}{2}t)+ c_2e^0sin(\frac{3}{2}t)$

Therefore, the general solution of the given equation is the following:
$y=c_1cos(\frac{3}{2}t)+c_2sin(\frac{3}{2}t)$