Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz3 => Topic started 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.

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 =$c_1$ $e^{(1+ \sqrt{3})t}$ + $c_2$ $e^{(1 \sqrt{3})t}$