solved by Luca Imponenti
Kreyszig 2011 pg.84 problem 5[edit | edit source]
Find a (real) general solution. State which rule you are using. Show each step of your work.
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
We have the following excitation
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
![{\displaystyle y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)]-e^{-x}[Kcos(x)+Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dc8afa7e1aaa426610e9cc0858b5b3e3337a5e2)
![{\displaystyle y_{p}'=e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78815ed74bdefacd675be27b478ef5aceb5a4888)
and
![{\displaystyle y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)]-e^{-x}[Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89b0f54d3a20737a9ff3a05bcb4e57024434ff75)
![{\displaystyle y_{p}''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)+Ksin(x)-Mcos(x)+Kcos(x)+Msin(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f31ce844b5537eb26184c223dc579fef46c5b6c8)
![{\displaystyle y_{p}''=e^{-x}[2Ksin(x)-2Mcos(x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25c2c14176a78437c663b52a21fe4ad61ab7af23)
Plugging these equations back into the differential equation
![{\displaystyle e^{-x}[2Ksin(x)-2Mcos(x)]+4e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]+4e^{-x}[Kcos(x)+Msin(x)]=e^{-x}cos(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8131d28963cb4970b42734eda4f28b1868603f1b)
![{\displaystyle e^{-x}[2Ksin(x)-2Mcos(x)-4Ksin(x)+4Mcos(x)-4Kcos(x)-4Msin(x)+4Kcos(x)+4Msin(x)]=e^{-x}cos(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63fa531d5bca94ce4da328d382ad91b5db945944)
from the above equation it is obvious that 
and 
therefore the particular solution to the differential equation is
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.
Kreyszig 2011 pg.84 problem 6[edit | edit source]
Find a (real) general solution. State which rule you are using. Show each step of your work.
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^{-\alpha x}[Acos(\omega x)+Bsin(\omega x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c459547757ff9092c1f0e8534dbef8f5ce2b0bb8)
![{\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
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 ![{\displaystyle y_{p}(x)=xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5deff0cc65f3ad32c2e73aee462e18d2d483581)
differentiating
![{\displaystyle y_{p}'=\pi xe^{-{\frac {x}{2}}}[-Ksin(\pi x)+Mcos(\pi x)]+(e^{\frac {x}{2}}-{\frac {1}{2}}xe^{-{\frac {x}{2}}})[Kcos(\pi x)+Msin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/094c6bc1c031df22c4d57be314a372ee8f83a129)
![{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}(\pi x[-Ksin(\pi x)+Mcos(\pi x)]+(1-{\frac {1}{2}}x)[Kcos(\pi x)+Msin(\pi x)])\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dcd8d55fd5b31a1192a5656e202ec59d9c953c1)
![{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c75485ee8adfb2f338da772a3e7ec696ed6b3978)
and
![{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)]-{\frac {1}{2}}e^{-{\frac {x}{2}}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-{\frac {1}{2}}xKcos(\pi x)-{\frac {1}{2}}xMsin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a19cb087518b898afe5d091725d5b6de76ca0b72)
![{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[-\pi ^{2}xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^{2}xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{2}}\pi xKsin(\pi x)-{\frac {1}{2}}\pi xMcos(\pi x)-{\frac {1}{2}}Kcos(\pi x)-{\frac {1}{2}}Msin(\pi x)+{\frac {1}{4}}xKcos(\pi x)+{\frac {1}{4}}xMsin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09d33dd5b55d3382057c1fe2d6dc3a0e8e423c6d)
grouping cosine and sine terms we get
![{\displaystyle y_{p}''=e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bea30d590cc0fdf2f67ce0486df14a333fed0277)
and
![{\displaystyle y_{p}'=e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5702430cac89fca3049b9bbdfd3e608bd718e655)
next we substitute the above equations into the ODE
![{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)]+e^{-{\frac {x}{2}}}[(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)]+(\pi ^{2}+{\frac {1}{4}})xe^{-{\frac {x}{2}}}[Kcos(\pi x)+Msin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1498469fe50717c724e7af63aba569bebd84113)
![{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K)sin(\pi x)+(\pi xM+K-{\frac {1}{2}}xK)cos(\pi x)+(-\pi xK+M-{\frac {1}{2}}xM)sin(\pi x)+(\pi ^{2}+{\frac {1}{4}})x(Kcos(\pi x)+Msin(\pi x))]=e^{-{\frac {x}{2}}}sin(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4ae7d7765effd364c468dc365abd736cb301a97)
![{\displaystyle e^{-{\frac {x}{2}}}[(2\pi M-\pi xM+({\frac {1}{4}}-\pi ^{2})xK-K+\pi xM+K-{\frac {1}{2}}xK+(\pi ^{2}+{\frac {1}{4}})xK)cos(\pi x)+(({\frac {1}{4}}-\pi ^{2})xM-M+\pi xK-2\pi K-\pi xK+M-{\frac {1}{2}}xM+(\pi ^{2}+{\frac {1}{4}})xM)sin(\pi x)]=e^{-{\frac {x}{2}}}sin(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3de7db15a930a972135a806746012b6c67ca1fc2)
after cancelling terms; we can equate cosine and sine coefficients to get two equations
so 
and 
and the particular solution to the ODE is
The general solution will be the summation of the homogeneous and particular solutions
![{\displaystyle y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)]-{\frac {1}{2\pi }}xe^{-{\frac {x}{2}}}cos(\pi x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/502abef71fc061be883578be22272f21dc7df2b2)
![{\displaystyle y=e^{-{\frac {x}{2}}}[Acos(\pi x)+Bsin(\pi x)-{\frac {1}{2\pi }}xcos(\pi x)]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0682929f321e66f22831aa0e121a1283484bb0f5)
Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)