User:Egm4313.s12.team17.ying/Test Link

Part 2-1 Solution
Using Eq.3.1.8, the function $$r_n(x)$$ can be developed for each value of n (4,7,11). We will first solve for $$y_n(x)$$ when n = 4.


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$$  \displaystyle r_n(x) = log(1+x) = \sum_{n=1}^n \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.1) Where n = 4:
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$$  \displaystyle r_4(x) = \sum_{n=1}^4 \frac{x^n}{n}(-1)^{n-1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$     (Eq.3.2.2) Thus, the particular solution to the Eq. #.#.# in the Part 2 given is then:
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$$  \displaystyle y_{n,p} = \sum_{i=0}^4 c_i x_i = c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.3)
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Deriving Eq.3.2.3 twice will yield the following:


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$$  \displaystyle y'_{n,p} = 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1 $$     (Eq.3.2.4)
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$$  \displaystyle y''_{n,p} = 12c_4x^2 + 6c_3x + 2c_2 $$     (Eq.3.2.5)
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To solve for the particular solution y_{n,p), Eq.3.2.4 and Eq.3.2.5 will be substituted into the following particular equation.


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$$  \displaystyle y''_{n,p} - 3y'_{n,p} + 2y_{n,p} = 12c_4x^2 + 6c_3x + 2c_2 - 3(4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) + 2(c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$     (Eq.3.2.6)
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Combining like terms, Eq.3.2.6 simplifies to:
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$$  \displaystyle (2c_4x^4) + (2c_3x^3 - 12c_4x^3) + (-9c_3x^2 + 2c_2x^2 + 12c_4x^2) + (2c_1x - 6c_2x + 6c_3x) + (2c_2 + c_0 - 3c_1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$     (Eq.3.2.7)
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The coefficients of each corresponding term can be substituted into an upper triangular matrix that can be easily solved by using matrices in matlab.

The following is the code which generated the values $$c_0,c_1,c_2,c_3,c_4$$


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$$  \displaystyle c_0 = -4.0625 $$
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$$  \displaystyle c_1 = -4.1250 $$
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$$  \displaystyle c_2 = -2.1250 $$
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$$  \displaystyle c_3 = -0.5833 $$
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$$  \displaystyle c_4 = -0.1250 $$
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Finally the particular solution has been reached. The solution $$y_n(x)$$ is then:


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$$  \displaystyle y_{n} = y_{n,h}(x) - 0.1250x^4 - 0.5833x^3 - 2.1250x^2 - 4.1250x - 4.0625 $$     (Eq.3.2.8)
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Where $$y_{n,h}(x)$$ is solved by setting Eq.#.#.# to 0 and finding the roots


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$$  \displaystyle y'' - 3y' + 2y = 0 = \lambda^2 - 3\lambda + 2 = 0 $$     (Eq.3.2.9)
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$$  \displaystyle y'' - 3y' + 2y = 0 = \lambda^2 - 3\lambda + 2 = 0 $$     (Eq.3.2.10)
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$$  \displaystyle (\lambda - 2)(\lambda - 2) = 0 $$     (Eq.3.2.11)
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$$  \displaystyle \lambda = 2,1 $$     (Eq.3.2.12)
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Therefore the homogeneous solution is then:


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$$  \displaystyle y_{n,h}(x) = c_6e^{2x} + c_5e^x $$     (Eq.3.2.13)
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Which then generates the final solution $$y_n(x)$$


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$$  \displaystyle y_{n} = c_6e^{2x} + c_5e^x - 0.1250x^4 - 0.5833x^3 - 2.1250x^2 - 4.1250x - 4.0625 $$     (Eq.3.2.14)
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Thus to solve for the final coefficients I used a matlab code to generated matrices to solve for the final two unknowns

The values then come out to be:
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$$  \displaystyle c_6 = -5.5868 $$
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$$  \displaystyle c_5 = 8.9001 $$
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$$  \displaystyle y_{n}(x) = -5.5868e^{2x} + 8.9001e^x - 0.1250x^4 - 0.5833x^3 - 2.1250x^2 - 4.1250x - 4.0625 $$     (Eq.3.2.15)
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Part 2-2 Solution

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$$  \displaystyle r_n(x) = log(1+x) = \sum_{n=1}^n \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.16) Where n = 7:
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$$  \displaystyle r_4(x) = \sum_{n=1}^4 \frac{x^n}{n}(-1)^{n-1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7} $$     (Eq.3.2.17)
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$$  \displaystyle y_{n,p} = \sum_{i=0}^7 c_i x_i = c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.18)
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Deriving Eq.3.2.18 twice will yield the following:


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$$  \displaystyle y'_{n,p} = 7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1 $$     (Eq.3.2.19)
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$$  \displaystyle y''_{n,p} = 42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 $$     (Eq.3.2.20)
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To solve for the particular solution y_{n,p), Eq.3.2.19 and Eq.3.2.20 will be substituted into the following particular equation.


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$$  \displaystyle y''_{n,p} - 3y'_{n,p} + 2y_{n,p} = $$
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$$  \displaystyle 42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 - 3(7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) + 2(c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$
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$$  \displaystyle - 3(7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) + 2(c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$
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$$  \displaystyle + 2(c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$ Combining like terms, Eq.3.2.6 simplifies to:
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$$  \displaystyle 2c_7x^7 + (2c_6x^6 -141c_7x^6) + (2c_5x^5 - 18c_6x^5 + 42c_7x^5) + (2c_4x^4- 15c_5x^4+30c_6x^4 ) + (2c_3x^3 - 12c_4x^3 + 20c_5x^3) + (2c_2x^2 - 9c_3x^2 + 12c_4x^2) $$
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$$  \displaystyle + (2c_1x - 6c_2x + 6c_3x) + (2c_0 - 3c_1 + 2c_2) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7} $$     (Eq.3.2.22) The coefficients of each corresponding term can be substituted into an upper triangular matrix that can be easily solved by using matrices in matlab.
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The following is the code which generated the values $$c_0,c_1,c_2,c_3,c_4,c_5,c_6,c_7$$


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$$  \displaystyle c_0 = 617.6875 $$
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$$  \displaystyle c_1 = 615.3750 $$
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$$  \displaystyle c_2 = 305.3750 $$
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$$  \displaystyle c_3 = 100.4167 $$
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$$  \displaystyle c_4 = 24.3750 $$
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$$  \displaystyle c_5 = 4.6000 $$
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$$  \displaystyle c_6 = 0.6667 $$
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$$  \displaystyle c_7 = 0.0714 $$
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Finally the particular solution has been reached. The solution $$y_n(x)$$ is then:
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$$  \displaystyle y_{n} = y_{n,h}(x) + 0.0714x^7 + 0.6667x^6 + 4.6000x^5 + 24.3750x^4 + 100.4167x^3 + 305.3750x^2 + 615.3750x + 617.6875 $$     (Eq.3.2.23)
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The homogeneous solution is the same:
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$$  \displaystyle y_{n,h}(x) = c_9e^{2x} + c_8e^x $$     (Eq.3.2.24)
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Which then makes the solution:


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$$  \displaystyle y_{n} = c_9e^{2x} + c_8e^x + 0.0714x^7 + 0.6667x^6 + 4.6000x^5 + 24.3750x^4 + 100.4167x^3 + 305.3750x^2 + 615.3750x + 617.6875 $$     (Eq.3.2.25)
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Using the same matlab code but adjusting certain formulas, $$c_9,c_8$$ can be found:

The following values for $$c_9,c_8$$ are obtained:
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$$  \displaystyle c_9 = -3.3774 $$
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$$  \displaystyle c_8 = -615.0789 $$
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The final solution is then:
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$$  \displaystyle y_{n}(x) = -3.3774e^{2x} -615.0789e^x + 0.0714x^7 + 0.6667x^6 + 4.6000x^5 + 24.3750x^4 + 100.4167x^3 + 305.3750x^2 + 615.3750x + 617.6875 $$     (Eq.3.2.26)
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Part 2-3 solution

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$$  \displaystyle r_n(x) = log(1+x) = \sum_{n=1}^n \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.27) Where n = 7:
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$$  \displaystyle r_4(x) = \sum_{n=1}^4 \frac{x^n}{n}(-1)^{n-1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7} - \frac{x^8}{8} + \frac{x^9}{9} - \frac{x^10}{10} + \frac{x^11}{11} $$     (Eq.3.2.28)
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$$  \displaystyle y_{n,p} = \sum_{i=0}^11 c_i x_i = c_{11}x^{11}+c_{10}x^{10}+c_9x^9+c_8x^8 + c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.29)
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Deriving Eq.3.2.28 twice will yield the following:


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$$  \displaystyle y'_{n,p} = 11c_{11}x^{10} + 10c_{10}x^9 + 9c_9x^8 + 8c_8x^7 + 7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1 $$     (Eq.3.2.30)
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$$  \displaystyle y''_{n,p} = 110c_{11}x^9 + 90c_{10}x^8 + 72c_9x^7 + 56c_8x^6 + 42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 $$     (Eq.3.2.31)
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To solve for the particular solution $$y_{n,p}$$, Eq.3.2.29 and Eq.3.2.30 will be substituted into the following particular equation.


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$$  \displaystyle y''_{n,p} - 3y'_{n,p} + 2y_{n,p} = $$
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$$  \displaystyle 110c_{11}x^9 + 90c_{10}x^8 + 72c_9x^7 + 56c_8x^6+42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 $$
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$$  \displaystyle - 3(11c_{11}x^{10} + 10c_{10}x^9 + 9c_9x^8 + 8c_8x^7 + 7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) $$
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$$  \displaystyle + 2(c_{11}x^{11}+c_{10}x^{10}+c_9x^9+c_8x^8 + c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$     (Eq.3.2.32)
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Combining like terms, Eq.3.2.6 simplifies to:
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$$  \displaystyle 2c_{11}x^{11} + (2c_{10}x^10-33c_{11}x^{10}) + (2c_9x^9 - 30c_10x^9 + 110c_11x^9) + (2c_8x^8 - 27c_9x^8 + 90c_10x^8) $$
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$$  \displaystyle + (2c_7x^7 - 24c_8x^7 + 72c_9x^7) + (2c_6x^6 -21c_7x^6 + 56c_8x^6) + (2c_5x^5 - 18c_6x^5 + 42c_7x^5) + (2c_4x^4- 15c_5x^4+30c_6x^4) $$
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$$  \displaystyle + (2c_3x^3 - 12c_4x^3 + 20c_5x^3) + (2c_2x^2 - 9c_3x^2 + 12c_4x^2) + (2c_1x - 6c_2x + 6c_3x) + (2c_0 - 3c_1 + 2c_2) $$     (Eq.3.2.33) The coefficients of each corresponding term can be substituted into an upper triangular matrix that can be easily solved by using matrices in matlab.
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The following is the code which generated the values $$c_0,c_1,c_2,c_3,c_4,c_5,c_6,c_7,c_8,c_9,c_{10},c_{11}$$


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$$  \displaystyle c_0 = -1.4204 \times 10^7 $$
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$$  \displaystyle c_1 = -1.4170 \times 10^7 $$
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$$  \displaystyle c_2 = -0.7052 \times 10^7 $$
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$$  \displaystyle c_3 = -0.2328 \times 10^7 $$
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$$  \displaystyle c_4 = -0.0571 \times 10^7 $$
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$$  \displaystyle c_5 = -0.0110 \times 10^7 $$
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$$  \displaystyle c_6 = -0.0017 \times 10^7 $$
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$$  \displaystyle c_7 = -0.0002 \times 10^7 $$
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$$  \displaystyle c_8 = -0.0000 \times 10^7 $$
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$$  \displaystyle c_9 = 0.0000 \times 10^7 $$
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$$  \displaystyle c_{10} = 0.0000 \times 10^7 $$
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$$  \displaystyle c_{11} = 0.0000 \times 10^7 $$
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Finally the particular solution has been reached. The solution $$y_n(x)$$ is then:
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$$  \displaystyle y_{n} = y_{n,h}(x) -0.0002 \times 10^7x^7 -0.0017 \times 10^7x^6 -0.0110 \times 10^7x^5 -0.0571 \times 10^7x^4 -0.2328 \times 10^7x^3 -0.7052 \times 10^7x^2 -1.4170 \times 10^7x -1.4204 \times 10^7 $$     (Eq.3.2.34)
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The homogeneous solution is the same:
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$$  \displaystyle y_{n,h}(x) = c_9e^{2x} + c_8e^x $$     (Eq.3.2.35)
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Which then makes the solution simplifies Eq. 3.2.34


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$$  \displaystyle c_{13}e^{2x} + c_{12}e^x -0.0002 \times 10^7x^7 -0.0017 \times 10^7x^6 -0.0110 \times 10^7x^5 -0.0571 \times 10^7x^4 -0.2328 \times 10^7x^3 -0.7052 \times 10^7x^2 -1.4170 \times 10^7x -1.4204 \times 10^7 $$     (Eq.3.2.36)
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Using the same matlab code but adjusting certain formulas, $$c_{13},c_{12}$$ can be found:

The following values for $$c_{13},c_{12}$$ are obtained:
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$$  \displaystyle c_{13} = -0.0034 \times 10^7 $$
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$$  \displaystyle c_{12} = 1.4237 \times 10^7 $$
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The final solution is then:
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$$  \displaystyle y_{n}(x) = -0.0034 \times 10^7e^{2x} + 1.4237 \times 10^7e^x -0.0002 \times 10^7x^7 -0.0017 \times 10^7x^6 -0.0110 \times 10^7x^5 -0.0571 \times 10^7x^4 -0.2328 \times 10^7x^3 -0.7052 \times 10^7x^2 $$ $$   \displaystyle -1.4170 \times 10^7x -1.4204 \times 10^7 $$     (Eq.3.2.37)
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Below are the three corresponding graphs to the n=4,7,11