solved by Luca Imponenti
Find a (real) general solution. State which rule you are using. Show each step of your work.
![{\displaystyle y''+4y'+4y=e^{-x}cos(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aebe9d4d1cbd9c69e2f31ee8603e74fc96ea029d)
To find the homogeneous solution,
, we must find the roots of the equation
We know the homogeneous solution for the case of a double root to be
![{\displaystyle y_{h}=c_{1}e^{-2x}+c_{2}xe^{-2x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53e70f6138fd8f5f0ba59b1716688fcdf64a308c)
We have the following excitation
![{\displaystyle r(x)=e^{-x}cos(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25b3073f436f1d5d7593b66392ccf3c55c180c81)
From table 2.1, K 2011, pg. 82, we have
![{\displaystyle y_{p}(x)=e^{-x}[Kcos(x)+Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2950ed00892386d583a528f359c482e1736bc975)
Since this does not correspond to our homogeneous solution we can use the Basic Rule (a), K 2011, pg. 81 to solve for the particular solution
where
and
Plugging these equations back into the differential equation
from the above equation it is obvious that
and
therefore the particular solution to the differential equation is
![{\displaystyle y_{p}(x)={\frac {1}{2}}e^{-x}sin(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/018f73294cf5482f614b503079f88e3a538682ad)
The general solution will be the summation of the homogeneous and particular solutions
The coefficients
and
can be readily solved for given either initial conditions or boundary value conditions.
Find a (real) general solution. State which rule you are using. Show each step of your work.
![{\displaystyle y''+y'+(\pi ^{2}+{\frac {1}{4}})y=e^{-{\frac {x}{2}}}sin(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8018e8641d3cd09c5ac8e4a4c88eff888b731946)
To find the homogeneous solution,
, we must find the roots of the equation
with
We know the homogeneous solution for the case of a double root to be
![{\displaystyle y_{h}=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f5e8dd31a33fa3927a13b4134f19ac28ecabe58)
We have the following excitation
![{\displaystyle r(x)=e^{-{\frac {x}{2}}}sin(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/037e72e1d9016404f6d7655c605062f832e86dd7)
From table 2.1, K 2011, pg. 82, we have
![{\displaystyle y_{p}(x)=e^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5abfa767eda5faf832c68f258fa7b90642c7bbf)
Since this corresponds to our homogeneous solution we must use the Modification Rule (b), K 2011, pg. 81 to solve for the particular solution
so
differentiating
and
grouping cosine and sine terms we get
and
next we substitute the above equations into the ODE
after cancelling terms; we can equate cosine and sine coefficients to get two equations
so
and
and the particular solution to the ODE is
![{\displaystyle y_{p}(x)=-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28ca5c890ede1194333f43596b9e348a3b682b6b)
The general solution will be the summation of the homogeneous and particular solutions
Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)