# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: Ranran Wang on October 11, 2019, 03:33:38 PM

Title: TUT 0501 Quiz 3
Post by: Ranran Wang on October 11, 2019, 03:33:38 PM
Here is my question and solution for the question I met in Quiz 3.
Title: Re: TUT 0501 Quiz 3
Post by: Nan Yang on October 11, 2019, 04:11:12 PM
we got the same question and I type the solution of this question.
Question: find the general solution of the given differential equation
y'' $-$ 2y' $-$ 2y = 0

Solution:
Let y'' $-$ 2y' $-$ 2y = 0 be equation (1)
we assume that $y= e^{rt}$ is solution of (1)
Then we have:
$y= e^{rt}$
$y'= re^{rt}$
$y''= r^2e^{rt}$
we substitute them into equation (1),
we have $r^2e^{rt}$ $-$ 2$re^{rt}$ $-$ 2$e^{rt}$ = 0 ,
$e^{rt}$$(r^2 - 2r -2) = 0$
since $e^{rt}$ is not zero
so $(r^2 - 2r -2) = 0$
we can get r = $\frac{2 \pm \sqrt{4 +8}}{2}$
simplify it we get  r = $\frac{2 \pm 2\sqrt{3}}{2}$
Then r = $1+ \sqrt{3}$ r = $1- \sqrt{3}$
Then two roots of equation (1) is  $e^{(1+ \sqrt{3})t}$ and $e^{(1- \sqrt{3})t}$
Therefore, the general solution is  y =$c_1$ $e^{(1+ \sqrt{3})t}$ + $c_2$ $e^{(1- \sqrt{3})t}$