Author Topic: TUT0202 Quiz 3  (Read 670 times)

Jiwen Bi

  • Jr. Member
  • **
  • Posts: 7
  • Karma: 3
    • View Profile
TUT0202 Quiz 3
« on: October 11, 2019, 02:00:10 PM »
$2y''-3y'+y=0\\
The\,given\,differemtial\,eqution\,is\,\\
2y''-3y'+y=0\\
we\,assume\,that\,y=e^{rt}\,is\,a\,solution\,of\\2y''-3y'+y=0\\
Now \,y=e^{rt}\\
then\,y'=re^{rt}\\
y''=r^{2}e^{rt}\\
using\,these\,values\,in\,eqution\,2y''-3y'+y=0\\
2r^{2}e^{rt}-3re^{rt}+e^{rt}=0\\
e^{rt}(2r^2-3r+1)=0\,since\,e^{rt}\,\neq 0\\
2r^2-3r+1=0\,,2r(r-1)-1(r-1)=0\\
(r-1)(2r-1)=0\\
Then\,r=(1,\frac{1}{2})\\
then\,e^{t}and\,e^{\frac{1}{2}t}\,are\,two\,solution\,of\,eqution\,and\,hence\,the\,general\,solution\,is\,given\,by\\
y=c_{1}e^{t}+c_{2}e^{\frac{t}{2}}$