Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz3 => Topic started by: Kun Zheng 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:
$r^2 4r=0$
$r(r4)=0$
$r_1=0, r_2=4$
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!

hi, is the question y''4y'=0 instead of y''4y''=0?

hi, is the question y''4y'=0 instead of y''4y''=0?
Thank you! My mistake.