# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-4 => Topic started by: Xuefen luo on October 18, 2019, 01:59:50 PM

Title: TUT0301
Post by: Xuefen luo on October 18, 2019, 01:59:50 PM
$y''+2y'+y=2e^{-t}$

This is non-homogeneous 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. $y=c_1e^{-t}+c_2te^{-t}+t^2e^{-t}$