Author Topic: TUT0602 QUIZ3  (Read 18 times)

Kun Zheng

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TUT0602 QUIZ3
« on: October 12, 2019, 02:01:37 AM »
Hi everyone, my question is to get the solution of y"-4y"=0, use the initial points of y'(-2)=1 and y(-2)=-1

First of all, we assume that y=e^(rt), and r must be a root of the characteristic equation.
Hence, we rewrite it as:
$r^2 -4r=0$
$r(r-4)=0$
$r_1=0, r_2=4$

Then we have the general structure as:
$y=C_1e^{r_1t}+C_2e^{r_1t}$
$y=C_1+C_2e^{4t}$

Derivative $y=C_1+C_2e^{4t}$
$y'=4C_2e^{4t}$

Use the initial values to plug in y'(-2)=1, y(-2)=-1
Got $C_2=e^8/4$
Then $y=C_1+e^{4t+8}/4$
Got $C_1=3/4$

Therefore, the initial equation is $y=3/4+e^{4t+8}/4$
Note: $y\rightarrow \infty, t\rightarrow \infty$

Correct me if I made a wrong solution or wrong question!
Have a good weekend!

Vickyyy

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Re: TUT0602 QUIZ3
« Reply #1 on: October 13, 2019, 12:57:42 AM »
hi, is the question y''-4y'=0 instead of y''-4y''=0?