Author Topic: TUT0301  (Read 290 times)

Xuefen luo

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« on: October 18, 2019, 01:59:50 PM »

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:


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 $y_p=At^2e^{-t}$,

Using these values in equation $y''+2y'+y=2e^{-t}$, we have:

i.e. $2Ae^{-t}=2e^{-t}$
i.e. $A=1$

Then the particular solution is

Hence the general solution of the equation is
i.e. $y=c_1e^{-t}+c_2te^{-t}+t^2e^{-t}$