Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz4 => Topic started by: Xuefen luo on October 18, 2019, 01:59:50 PM

$y''+2y'+y=2e^{t}$
This is nonhomogeneous differential equation, so to find the complimentary solution,
we need to consider $y''+2y'+y=0$.
We assume that $y=e^{rt}$ is a solution of this equation. Then the characteristic equation is:
$r^2+2r+1=0$
$(r+1)^2=0$
$r=1,1$
Then, the complimentary solution is given by
$ y_c(t)=c_1e^{t}+c_2te^{t}$, where $c_1, c_2$ are constants.
To find the particular solution, we assume that $y_p(t)=Ae^{t}$.
However, it fails because $e^{t}$ is a solution of the homogeneous equation.
Also if we assume $y_p(t)=Ate^{t}$, again it fails as $te^{t}$ is also a solution of the homogeneous equation.
Then, we assume $y_p(t)=At^2e^{t}$ is the particular solution,
then it satisfies the equation $y''+2y'+y=2e^{t}$.
Since $y_p=At^2e^{t}$,
$y'=2Ate^{t}At^2e^{t}$
$y''=2Ae^{t}2Ate^{t}2Ate^{t}+At^2e^{t}=2Ae^{t}4Ate^{t}+At^2e^{t}$
Using these values in equation $y''+2y'+y=2e^{t}$, we have:
$2Ae^{t}4Ate^{t}+At^2e^{t}+4Ate^{t}2At^2e^{t}+At^2e^{t}=2e^{t}$
i.e. $2Ae^{t}=2e^{t}$
i.e. $A=1$
Then the particular solution is
$y_p(t)=t^2e^{t}$
Hence the general solution of the equation is
$y=y_c(t)+y_p(t)$
i.e. $y=c_1e^{t}+c_2te^{t}+t^2e^{t}$