MAT244--2019F > Quiz-3

QUIZ3 TUT 0502

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**Xinyu Jing**:

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=$C_{1}𝑒^{(1+\sqrt{3})𝑡}+C_{2}𝑒^{(1−\sqrt{3})𝑡}$

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