Author Topic: QUIZ3 TUT 0502  (Read 335 times)

Xinyu Jing

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QUIZ3 TUT 0502
« on: October 11, 2019, 08:06:25 PM »
Question:Find the general solution of the given differential equation.
𝑦″−2𝑦′−2𝑦=0


Solution:
𝑦=$𝑒^{𝑟𝑡}$
and it follows that r must be a root of characteristic equation
$𝑟^{2}$−2𝑟−2=0

𝑟=$\frac{−𝑏±\sqrt{𝑏^{2}−4𝑎𝑐}}{2𝑎}$


$𝑟_{1}$=1+\sqrt{3} $𝑟_{2}$=1−\sqrt{3}


Therefore, the general solution of the given differential equation is:
𝑦=$C_{1}𝑒^{(1+\sqrt{3})𝑡}+C_{2}𝑒^{(1−\sqrt{3})𝑡}$