Solved by: Andrea Vargas
Given
![{\displaystyle y''-3y'+2y=4x^{2}-6x^{5}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfca61778477658797760ef24d6f5533b482134f)
1. Obtain the coefficients of ![{\displaystyle x,x^{2},x^{3},x^{5}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a080834d0b4fd0bb5353b540eaff989fbba1834)
2.Verify all equations by long-hand expansion. Use the series before adjusting the indexes.
3. Put the system of equations in an upper triangular matrix.
4. Solve for
by using back substitution.
5. Using the initial conditions
find
and plot it
1. To obtain the equations for the coefficients we use the following equation from (1) p7-11:
![{\displaystyle \sum _{j=0}^{3}[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_{j}]x^{j}-3c_{5}(5)x^{4}+2[c_{4}x^{4}+c^{5}x^{5}]=4x^{2}-6x^{5}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66548e831aed1df2c0af03f920d318c9ff821faf)
Finding the coefficients of
where
:
![{\displaystyle [c_{3}(3)(2)-3(c_{2})(2)+2c_{1}]x^{1}=[6c_{3}-6c_{2}+2c_{1}]x\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/588b04bbf17c37133fad9347ca49ae12d1e4f135)
Finding the coefficients of
where
:
![{\displaystyle [c_{4}(4)(3)-3(c_{3})(3)+2c_{2}]x^{2}=[12c_{4}-9c_{3}+2c_{2}]x^{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9977462a0e7926e2648ec2ec9a0880243279768)
Finding the coefficients of
where
:
![{\displaystyle [c_{5}(5)(4)-3(c_{4})(4)+2c_{3}]x^{3}=[20c_{4}-12c_{3}+2c_{2}]x^{3}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a721f2df736c522657a58a8a965ad35f0ead0d79)
Finding the coefficients of
where
:
![{\displaystyle c_{5}x^{5}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/747fa4147f651c79aa42284771f0c332acbbb14c)
2. To verify all the above equations by long hand expansion we use the following equation from (4) p7-12:
![{\displaystyle \sum _{j=2}^{5}c_{j}\times j\times (j-1)\times x^{j-2}-3\sum _{j=1}^{5}c_{j}\times j\times x^{j-1}+2\sum _{j=0}^{5}c_{j}\times x^{j}=4x^{2}-6x^{5}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4276e220b91c35b21b09daf0ccb4a23dadb3fce6)
Finding the coefficients when
:
![{\displaystyle 2c_{0}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38abeb24736582dade7af4f65fd9ac043233c316)
Finding the coefficients when
:
![{\displaystyle -3c_{1}+2c_{1}x_{1}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a614c2df4d2563a8b6ef1b42757d20414e308b76)
Finding the coefficients when
:
![{\displaystyle 2c_{2}-6c_{2}x+2c_{2}x^{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0dc29c01f71df4628a3675401817f68a6e609a09)
Finding the coefficients when
:
![{\displaystyle 6c_{3}x-9c_{3}x^{2}+2c_{3}x^{3}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a43e661b6fa2f02a702c54e5d4aebdee13674fb)
Finding the coefficients when
:
![{\displaystyle 20c_{5}x^{3}-15c_{5}x^{4}+2c_{5}x^{5}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/daa676f3952bc932a5661119fea5f8267516382f)
By collecting these terms we can compare them to the equations of part 1.
Coefficients of
:
![{\displaystyle [c_{3}(3)(2)-3(c_{2})(2)+2c_{1}]x^{1}=[6c_{3}-6c_{2}+2c_{1}]x\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/588b04bbf17c37133fad9347ca49ae12d1e4f135)
Coefficients of
:
![{\displaystyle [c_{4}(4)(3)-3(c_{3})(3)+2c_{2}]x^{2}=[12c_{4}-9c_{3}+2c_{2}]x^{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9977462a0e7926e2648ec2ec9a0880243279768)
Coefficients of
:
![{\displaystyle [c_{5}(5)(4)-3(c_{4})(4)+2c_{3}]x^{3}=[20c_{4}-12c_{3}+2c_{2}]x^{3}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a721f2df736c522657a58a8a965ad35f0ead0d79)
Coefficients of
:
![{\displaystyle c_{5}x^{5}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/747fa4147f651c79aa42284771f0c332acbbb14c)
We can see that we obtain the same system of equations to solve for the coefficients with both methods.
3.Constructing the coefficients matrix:
![{\displaystyle {\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/972619c05bb26c3118bb81e27210ded53d92546b)
Then, the system becomes:
![{\displaystyle {\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\\c_{4}\\c_{5}\\\end{bmatrix}}={\begin{bmatrix}0\\0\\4\\0\\0\\-6\\\end{bmatrix}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a40ad2b27b1102804a590b3e1d4c7f3c18901b98)
4. Solving for the coefficients:
![{\displaystyle 2c_{5}=-6\rightarrow c_{5}=-3\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2aeacfb4e47cf7bb0c275b8aaf81225f67190736)
![{\displaystyle 2c_{4}-(15)(-3)=0\rightarrow c_{4}={\frac {-45}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b33c74606e2ee2726ac609a40a25e1170b79659)
![{\displaystyle 2c_{3}-12(-{\frac {45}{2}})+20(-3)=0\rightarrow c_{3}=-105\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3385bf0524438e714c9822a4e199e4d7f5b0e199)
![{\displaystyle 2c_{2}-9(-105)+12(-{\frac {45}{2}})=4\rightarrow c_{2}=-{\frac {671}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35675b573deeeaffc350b2c21fb1a27df8ebc58c)
![{\displaystyle 2c_{1}-6(-{\frac {671}{2}})+6(-105)=0\rightarrow c_{1}=-{\frac {1383}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/273ac45109293f4609e51676429735a5821540d1)
![{\displaystyle 2c_{0}-3(-{\frac {1383}{2}})+2(-{\frac {671}{2}})=0\rightarrow c_{0}=-{\frac {2807}{4}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aefbea6124fe68faf9e180521c8481bbbf9c14f1)
This yields the particular solution:
![{\displaystyle y{}''_{h}-y{}'_{h}+2y_{h}=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac0d501d1de4af3c2b39599a1e4beca0303f4aab)
![{\displaystyle \lambda ^{2}-3\lambda +2=0\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/acda7e2f9dc09ae2212b455504353bd189576ce3)
Then, we can find the characteristic equation:
![{\displaystyle (\lambda -2)(\lambda -1)\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a8cf2e4cc2245277a698bb29ed34abb62bd549a)
![{\displaystyle \lambda =2,1\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8830e3b79291b7ff3617964e7c21171a715ba43a)
Then the solution for the homogeneous equation becomes:
Using the given initial conditions
we find the overall solution:
![{\displaystyle y(x)=y_{h}+y_{p}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35020114217a1d6a96d9a9d047a36f4f6aabfb3b)
![{\displaystyle y(x)=C_{1}e^{2x}+C_{2}e^{x}-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bb7eef5b85480b25fa15608dffd814ab7588557)
![{\displaystyle y'(x)=2C_{1}e^{2x}+C_{2}e^{x}-15x^{4}-90x^{3}-315x^{2}-671x-{\frac {1383}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75790c9b93516eb95488e304336bdfa6c8e47a9f)
Using the initial conditions to solve for
and ![{\displaystyle C_{2}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65a839a192c05765b3233f5fbb8a32a0afa9d0bc)
![{\displaystyle 1=C_{1}+C_{2}-{\frac {2807}{4}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e2084f05add5a03ccbb7fc0cb57b9d5e32c45c8)
![{\displaystyle 0=C_{1}+C_{2}-{\frac {1383}{2}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a600bf327324e609062f15d701b209cf18e75e1a)
![{\displaystyle C_{1}=-{\frac {45}{4}}\;\;\;C_{2}=714\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1604ce7c677380438e6d63de74c6d5dcdd3154b)
The general solution becomes
![{\displaystyle y(x)=-{\frac {45}{4}}e^{2x}+714e^{x}-3x^{5}-{\frac {45}{2}}x^{4}-105x^{3}-{\frac {671}{2}}x^{2}-{\frac {1383}{2}}x-{\frac {2807}{4}}\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec81565a0f7ac7e14925f267edcbe4e4e509136d)
Below is a plot of the solution:
![](//upload.wikimedia.org/wikiversity/en/b/bb/R3_5.jpg)
--Egm4313.s12.team11.vargas.aa 05:56, 21 February 2012 (UTC)