Problem R2.1
Given the two roots and the initial conditions:
![{\displaystyle \lambda _{1}=-2,\lambda _{2}=5\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c7c4c03bb87a3a40b4c751f4c217be3e606ff06)
![{\displaystyle y(0)=1,y'(0)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d358729028ad1e6ffc817e7b90208281056b9220)
Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation
.
Consider no excitation:
![{\displaystyle r(x)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc9d112d4e8c94ff6ad494c7b6604a47366ca9ee)
Plot the solution
![{\displaystyle (\lambda +2)(\lambda -5)=\lambda ^{2}+2\lambda -5\lambda -10=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df79607e5edd66251e0c2c823732f9f4bd85968f)
![{\displaystyle \lambda ^{2}-3\lambda -10=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/979be5ae597112e5d458a01661d52f1ba0169dc3)
![{\displaystyle y''-3y'-10=r(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c9da6c526d99596c83f64055c85deff5a58d237)
![{\displaystyle y_{h}(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdfcb52e4aba8d8f0c8f8069fc3ac477b940f653)
![{\displaystyle y(x)=c_{1}e^{-2x}+c_{2}e^{5x}+y_{p}(x)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11d845723716527f219ce50d4fa8cefb39bf0b85)
Since there is no excitation,
![{\displaystyle y_{p}(x)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49cb59bbf42e39ad1dc18dc62e70ea8f2f3bd8aa)
![{\displaystyle y(x)=c_{1}e^{-2x}+c_{2}e^{5x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6998327eb030be07cf770bfaa560cd50909e12dc)
![{\displaystyle y(0)=1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/995f2574abb9838c6479c36e2a828b0bb990c6f0)
![{\displaystyle 1=c_{1}+c_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2f94febf7094e1cb8dd2f36e1cf792a8a1b616f)
![{\displaystyle y'(0)=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e2c7cc83a57d75b1c0eec87fb9833db40b60c62)
![{\displaystyle 0=-2c_{1}+5c_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdb137a2430b1c7f318ca88d64e483a640d29430)
Solving these two equations for
and
yields:
![{\displaystyle y(x)=(5/4)e^{-2x}-(1/4)e^{5x}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e467865d38a37b82022a5549f77f9d01f6f8e5cf)
Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values in (3a) p.3-7 as the 2 roots of the corresponding characteristic equation.
![{\displaystyle 2(\lambda +2)(\lambda -5)=2\lambda ^{2}+4\lambda -10\lambda -20=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01bcbd467a290b7819775defe235371c9edb5e6d)
![{\displaystyle 2\lambda ^{2}-6\lambda -20=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9341f81efb281c248217d5999c211d6021c08e12)
![{\displaystyle 3(\lambda +2)(\lambda -5)=3\lambda ^{2}+6\lambda -15\lambda -30=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb4f6ca2b2853dfca5ccc750092d9a4f59497480)
![{\displaystyle 3\lambda ^{2}-9\lambda -30=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d92cf941a7890023ced2175e3192c5fbdf46b2e3)
![{\displaystyle 4(\lambda +2)(\lambda -5)=4\lambda ^{2}+8\lambda -20\lambda -40=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ff0d65933f0ac95857e9d22162204065445fabf)
![{\displaystyle 4\lambda ^{2}-12\lambda -40=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/408949b9b40cbddedcbd97cee19fdb3b67844834)
--Egm4313.s12.team11.gooding 02:01, 7 February 2012 (UTC)