Report 4
Problem Statement
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Obtain equations (2), (3), (n-2), (n-1), (n), and set up the matrix A as in (1) p.7-21
for the general case, with the matrix coefficients for rows 1, 2, 3, (n-2), (n-1), n, filled in,
as obtained from equations (1), (2), (3), (n-2), (n-1), (n).
As shown in p.7-21, the first equation is:
2 C 2 + a c 1 + b c 0 = d 0 {\displaystyle 2C_{2}+ac_{1}+bc_{0}=d_{0}\!}
According to p.7-20, the general form of the series is:
∑ j = 0 n − 2 [ c j + 2 ( j + 2 ) ( j + 1 ) + a c j + 1 + b c j ] x j + a c n n x n − 1 + b [ c n − 1 x n − 1 + c n x n ] = ∑ j = 0 n d j x j {\displaystyle \sum _{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}+bc_{j}]x^{j}+ac_{n}nx^{n-1}+b[c_{n-1}x^{n-1}+c_{n}x^{n}]=\sum _{j=0}^{n}d_{j}x^{j}\!}
From (2) p.7-20, we can obtain n+1 equations for n+1 unknown coefficients c 0 , . . . , c n {\displaystyle {c_{0},...,c_{n}}\!}
After referring to p.7-22, it can be determined that the matrix to be set up is of the following form:
A = [ X X X 0 0 0 0 X X 0 0 0 0 0 X 0 0 0 0 0 0 X X X 0 0 0 0 X X 0 0 0 0 0 X ] {\displaystyle A={\begin{bmatrix}X&&X&&X&&0&&0&0\\0&&X&&X&&0&&0&0\\0&&0&&X&&0&&0&0\\0&&0&&0&&X&&X&X\\0&&0&&0&&0&&X&X\\0&&0&&0&&0&&0&X\end{bmatrix}}\!}
where the rows signify the coefficients c 0 , c 1 , c 2 , c n − 2 , c n − 1 , c n {\displaystyle c_{0},c_{1},c_{2},c_{n-2},c_{n-1},c_{n}\!} d 0 , d 1 , d 2 , d n − 2 , d n − 1 , d n {\displaystyle d_{0},d_{1},d_{2},d_{n-2},d_{n-1},d_{n}\!}
Solution
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Building the coefficient matrix as shown in p.7-22 of the class notes, we can begin to solve for the coefficients
as follows:
Equation associated with d 0 {\displaystyle d_{0}\!}
j=0: d 0 = 2 C 2 + a c 1 + b c 0 {\displaystyle d_{0}=2C_{2}+ac_{1}+bc_{0}\!}
Equation associated with d 1 {\displaystyle d_{1}\!}
j=1: d 1 = 6 c 3 + 2 a c 2 + b c 1 {\displaystyle d_{1}=6c_{3}+2ac_{2}+bc_{1}\!}
Equation associated with d 2 {\displaystyle d_{2}\!}
j=2: d 2 = 12 c 4 + 3 a c 3 + b c 2 {\displaystyle d_{2}=12c_{4}+3ac_{3}+bc_{2}\!}
Equation associated with d n − 2 {\displaystyle d_{n-2}\!}
j=n-2: d n − 2 = [ c n ( n ) ( n − 1 ) + a c n − 1 ( n − 1 ) + b c n − 2 ] {\displaystyle d_{n-2}=[c_{n}(n)(n-1)+ac_{n-1}(n-1)+bc_{n-2}]\!}
Equation associated with d n − 1 {\displaystyle d_{n-1}\!}
j=n-1: d n − 1 = a c n n + b c n − 1 {\displaystyle d_{n-1}=ac_{n}n+bc_{n-1}\!}
Equation associated with d n {\displaystyle d_{n}\!}
j=n: d n = b c n {\displaystyle d_{n}=bc_{n}\!}
Using all of the above equations, (1), (2), (3), (n-2), (n-1), (n), we can then determine the A matrix to be:
A = [ b a 2 0 0 0 0 b 2 a 0 0 0 0 0 b 0 0 0 0 0 0 b a ( n − 1 ) n ( n − 1 ) 0 0 0 0 b a n 0 0 0 0 0 b ] {\displaystyle A={\begin{bmatrix}b&&a&&2&&0&&0&0\\0&&b&&2a&&0&&0&0\\0&&0&&b&&0&&0&0\\0&&0&&0&&b&&a(n-1)&n(n-1)\\0&&0&&0&&0&&b&an\\0&&0&&0&&0&&0&b\end{bmatrix}}\!}
Solved by Gonzalo Perez
Solved by Jonathan Sheider
Problem Statement
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Given: y ″ − 3 y ′ + 2 y = s i n ( x ) {\displaystyle y^{''}-3y^{'}+2y=sin(x)\!} y ( 0 ) = 1 , y ′ ( 0 ) = 0 {\displaystyle y(0)=1,y'(0)=0\!}
Solution
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The Taylor series expansion for sin(x) that is plotted with n = 13 is as follows:y = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + x 9 9 ! − x 11 11 ! + x 13 13 ! − x 15 15 ! + x 17 17 ! − x 19 19 ! + x 21 21 ! − x 23 23 ! + x 25 25 ! {\displaystyle y=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-{\frac {x^{11}}{11!}}+{\frac {x^{13}}{13!}}-{\frac {x^{15}}{15!}}+{\frac {x^{17}}{17!}}-{\frac {x^{19}}{19!}}+{\frac {x^{21}}{21!}}-{\frac {x^{23}}{23!}}+{\frac {x^{25}}{25!}}\!}
Solved by Jonathan Sheider
Problem Statement
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Given: y ″ − 3 y ′ + 2 y = r ( x ) {\displaystyle y^{''}-3y^{'}+2y=r(x)\!} y ( 0 ) = 1 , y ′ ( 0 ) = 0 {\displaystyle y(0)=1,y'(0)=0\!} r ( x ) {\displaystyle r(x)\!} s i n ( x ) {\displaystyle sin(x)\!} r ( x ) = ∑ k = 0 n ( − 1 ) k t 2 k + 1 ( 2 k + 1 ) ! {\displaystyle r(x)=\sum _{k=0}^{n}{\frac {(-1)^{k}t^{2k+1}}{(2k+1)!}}\!} y n ( x ) {\displaystyle y_{n}(x)\!} n = 3 , 5 , 9 {\displaystyle n=3,5,9\!} [ 0 , 4 π ] {\displaystyle [0,4\pi ]\!}
Solution
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First finding the homogenous solution to the ODE:λ 2 − 3 λ + 2 = 0 {\displaystyle \lambda ^{2}-3\lambda +2=0\!} ( λ − 2 ) ( λ − 1 ) = 0 {\displaystyle (\lambda -2)(\lambda -1)=0\!} λ = 1 , 2 {\displaystyle \lambda =1,2\!} y h = c 1 e x + c 2 e 2 x {\displaystyle y_{h}=c_{1}e^{x}+c_{2}e^{2x}\!} For n = 3: r x = x − x 3 3 ! + x 5 5 ! − x 7 7 ! {\displaystyle r_{x}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}\!} y p = K 7 x 7 + K 6 x 6 + K 5 x 5 + K 4 x 4 + K 3 x 3 + K 2 x 2 + K 1 x + K 0 {\displaystyle y_{p}=K_{7}x^{7}+K_{6}x^{6}+K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}\!} y p ′ = 7 K 7 x 6 + 6 K 6 x 5 + 5 K 5 x 4 + 4 K 4 x 3 + 3 K 3 x 2 + 2 K 2 x + 1 K 1 {\displaystyle y_{p}^{'}=7K_{7}x^{6}+6K_{6}x^{5}+5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+1K_{1}\!} y p ″ = 42 K 7 x 5 + 30 K 6 x 4 + 20 K 5 x 3 + 12 K 4 x 2 + 6 K 3 x + 2 K 2 {\displaystyle y_{p}^{''}=42K_{7}x^{5}+30K_{6}x^{4}+20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!} 42 K 7 x 5 + 30 K 6 x 4 + 20 K 5 x 3 + 12 K 4 x 2 + 6 K 3 x + 2 K 2 {\displaystyle 42K_{7}x^{5}+30K_{6}x^{4}+20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!} − 3 ( 7 K 7 x 6 + 6 K 6 x 5 + 5 K 5 x 4 + 4 K 4 x 3 + 3 K 3 x 2 + 2 K 2 x + 1 K 1 ) {\displaystyle -3(7K_{7}x^{6}+6K_{6}x^{5}+5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+1K_{1})\!} + 2 ( K 7 x 7 + K 6 x 6 + K 5 x 5 + K 4 x 4 + K 3 x 3 + K 2 x 2 + K 1 x + K 0 ) = x − x 3 3 ! + x 5 5 ! − x 7 7 ! {\displaystyle +2(K_{7}x^{7}+K_{6}x^{6}+K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0})=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}\!} 42 K 7 x 5 + 30 K 6 x 4 + 20 K 5 x 3 + 12 K 4 x 2 + 6 K 3 x + 2 K 2 {\displaystyle 42K_{7}x^{5}+30K_{6}x^{4}+20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!} − 21 K 7 x 6 − 18 K 6 x 5 − 15 K 5 x 4 − 12 K 4 x 3 − 9 K 3 x 2 − 6 K 2 x − 3 K 1 {\displaystyle -21K_{7}x^{6}-18K_{6}x^{5}-15K_{5}x^{4}-12K_{4}x^{3}-9K_{3}x^{2}-6K_{2}x-3K_{1}\!} + 2 K 7 x 7 + 2 K 6 x 6 + 2 K 5 x 5 + 2 K 4 x 4 + 2 K 3 x 3 + 2 K 2 x 2 + 2 K 1 x + 2 K 0 = x − x 3 3 ! + x 5 5 ! − x 7 7 ! {\displaystyle +2K_{7}x^{7}+2K_{6}x^{6}+2K_{5}x^{5}+2K_{4}x^{4}+2K_{3}x^{3}+2K_{2}x^{2}+2K_{1}x+2K_{0}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}\!} [ 2 − 3 2 0 0 0 0 0 0 2 − 6 6 0 0 0 0 0 0 2 − 9 12 0 0 0 0 0 0 2 − 12 20 0 0 0 0 0 0 2 − 15 30 0 0 0 0 0 0 2 − 18 42 0 0 0 0 0 0 2 − 21 0 0 0 0 0 0 0 2 ] [ K 0 K 1 K 2 K 3 K 4 K 5 K 6 K 7 ] = [ 0 1 0 − 1 3 ! 0 1 5 ! 0 − 1 7 ! ] {\displaystyle {\begin{bmatrix}2&-3&2&0&0&0&0&0\\0&2&-6&6&0&0&0&0\\0&0&2&-9&12&0&0&0\\0&0&0&2&-12&20&0&0\\0&0&0&0&2&-15&30&0\\0&0&0&0&0&2&-18&42\\0&0&0&0&0&0&2&-21\\0&0&0&0&0&0&0&2\end{bmatrix}}{\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\\K_{4}\\K_{5}\\K_{6}\\K_{7}\end{bmatrix}}={\begin{bmatrix}0\\1\\0\\-{\frac {1}{3!}}\\0\\{\frac {1}{5!}}\\0\\-{\frac {1}{7!}}\end{bmatrix}}\!} y p = − 1 10080 x 7 + − 1 960 x 6 + − 1 320 x 5 + − 1 128 x 4 + − 19 192 x 3 + − 51 128 x 2 + − 51 128 x + − 51 256 {\displaystyle y_{p}={\frac {-1}{10080}}x^{7}+{\frac {-1}{960}}x^{6}+{\frac {-1}{320}}x^{5}+{\frac {-1}{128}}x^{4}+{\frac {-19}{192}}x^{3}+{\frac {-51}{128}}x^{2}+{\frac {-51}{128}}x+{\frac {-51}{256}}\!} y n = c 1 e x + c 2 e 2 x + − 1 10080 x 7 + − 1 960 x 6 + − 1 320 x 5 + − 1 128 x 4 + − 19 192 x 3 + − 51 128 x 2 + − 51 128 x + − 51 256 {\displaystyle y_{n}=c_{1}e^{x}+c_{2}e^{2x}+{\frac {-1}{10080}}x^{7}+{\frac {-1}{960}}x^{6}+{\frac {-1}{320}}x^{5}+{\frac {-1}{128}}x^{4}+{\frac {-19}{192}}x^{3}+{\frac {-51}{128}}x^{2}+{\frac {-51}{128}}x+{\frac {-51}{256}}\!} y n ( 0 ) = c 1 + c 2 + − 51 256 = 1 {\displaystyle y_{n}(0)=c_{1}+c_{2}+{\frac {-51}{256}}=1\!} y n ′ ( 0 ) = c 1 + 2 c 2 + − 51 128 = 0 {\displaystyle y_{n}^{'}(0)=c_{1}+2c_{2}+{\frac {-51}{128}}=0\!} c 1 = 2 , c 2 = − 205 256 {\displaystyle c_{1}=2,c_{2}={\frac {-205}{256}}\!} y n = 2 e x + − 205 256 e 2 x + − 1 10080 x 7 + − 1 960 x 6 + − 1 320 x 5 + − 1 128 x 4 + − 19 192 x 3 + − 51 128 x 2 + − 51 128 x + − 51 256 {\displaystyle y_{n}=2e^{x}+{\frac {-205}{256}}e^{2x}+{\frac {-1}{10080}}x^{7}+{\frac {-1}{960}}x^{6}+{\frac {-1}{320}}x^{5}+{\frac {-1}{128}}x^{4}+{\frac {-19}{192}}x^{3}+{\frac {-51}{128}}x^{2}+{\frac {-51}{128}}x+{\frac {-51}{256}}\!}
For n = 5: r x = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + x 9 9 ! − x 11 11 ! {\displaystyle r_{x}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-{\frac {x^{11}}{11!}}\!} y p = K 11 x 11 + K 10 x 10 + K 9 x 9 + K 8 x 8 + K 7 x 7 + K 6 x 6 + K 5 x 5 + K 4 x 4 + K 3 x 3 + K 2 x 2 + K 1 x + K 0 {\displaystyle y_{p}=K_{11}x^{11}+K_{10}x^{10}+K_{9}x^{9}+K_{8}x^{8}+K_{7}x^{7}+K_{6}x^{6}+K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}\!} y p ′ = 11 K 11 x 10 + 10 K 10 x 9 + 9 K 9 x 8 + 8 K 8 x 7 + 7 K 7 x 6 + 6 K 6 x 5 + 5 K 5 x 4 + 4 K 4 x 3 + 3 K 3 x 2 + 2 K 2 x + 1 K 1 {\displaystyle y_{p}^{'}=11K_{11}x^{10}+10K_{10}x^{9}+9K_{9}x^{8}+8K_{8}x^{7}+7K_{7}x^{6}+6K_{6}x^{5}+5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+1K_{1}\!} y p ″ = 110 K 11 x 9 + 90 K 10 x 8 + 72 K 9 x 7 + 56 K 8 x 6 + 42 K 7 x 5 + 30 K 6 x 4 + 20 K 5 x 3 + 12 K 4 x 2 + 6 K 3 x + 2 K 2 {\displaystyle y_{p}^{''}=110K_{11}x^{9}+90K_{10}x^{8}+72K_{9}x^{7}+56K_{8}x^{6}+42K_{7}x^{5}+30K_{6}x^{4}+20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!} y p = − 1 79833600 x 11 + − 1 4838400 x 10 + − 1 967680 x 9 + − 1 215040 x 8 + − 7 59419 x 7 + − 17 15360 x 6 {\displaystyle y_{p}={\frac {-1}{79833600}}x^{11}+{\frac {-1}{4838400}}x^{10}+{\frac {-1}{967680}}x^{9}+{\frac {-1}{215040}}x^{8}+{\frac {-7}{59419}}x^{7}+{\frac {-17}{15360}}x^{6}\!} + − 17 5120 x 5 + − 17 2048 x 4 + − 307 3072 x 3 + − 819 2048 x 2 + − 819 2048 x + − 819 4096 {\displaystyle +{\frac {-17}{5120}}x^{5}+{\frac {-17}{2048}}x^{4}+{\frac {-307}{3072}}x^{3}+{\frac {-819}{2048}}x^{2}+{\frac {-819}{2048}}x+{\frac {-819}{4096}}\!} y n = c 1 e x + c 2 e 2 x + − 1 79833600 x 11 + − 1 4838400 x 10 + − 1 967680 x 9 + − 1 215040 x 8 + − 7 59419 x 7 + − 17 15360 x 6 {\displaystyle y_{n}=c_{1}e^{x}+c_{2}e^{2x}+{\frac {-1}{79833600}}x^{11}+{\frac {-1}{4838400}}x^{10}+{\frac {-1}{967680}}x^{9}+{\frac {-1}{215040}}x^{8}+{\frac {-7}{59419}}x^{7}+{\frac {-17}{15360}}x^{6}\!} + − 17 5120 x 5 + − 17 2048 x 4 + − 307 3072 x 3 + − 819 2048 x 2 + − 819 2048 x + − 819 4096 {\displaystyle +{\frac {-17}{5120}}x^{5}+{\frac {-17}{2048}}x^{4}+{\frac {-307}{3072}}x^{3}+{\frac {-819}{2048}}x^{2}+{\frac {-819}{2048}}x+{\frac {-819}{4096}}\!} y n ( 0 ) = c 1 + c 2 + − 819 4096 = 1 {\displaystyle y_{n}(0)=c_{1}+c_{2}+{\frac {-819}{4096}}=1\!} y n ′ ( 0 ) = c 1 + 2 c 2 + − 819 2048 = 0 {\displaystyle y_{n}^{'}(0)=c_{1}+2c_{2}+{\frac {-819}{2048}}=0\!} c 1 = 2 , c 2 = − 3277 4096 {\displaystyle c_{1}=2,c_{2}={\frac {-3277}{4096}}\!} y n = 2 e x + − 3277 4096 e 2 x + − 1 79833600 x 11 + − 1 4838400 x 10 + − 1 967680 x 9 + − 1 215040 x 8 + − 7 59419 x 7 + − 17 15360 x 6 {\displaystyle y_{n}=2e^{x}+{\frac {-3277}{4096}}e^{2x}+{\frac {-1}{79833600}}x^{11}+{\frac {-1}{4838400}}x^{10}+{\frac {-1}{967680}}x^{9}+{\frac {-1}{215040}}x^{8}+{\frac {-7}{59419}}x^{7}+{\frac {-17}{15360}}x^{6}\!} + − 17 5120 x 5 + − 17 2048 x 4 + − 307 3072 x 3 + − 819 2048 x 2 + − 819 2048 x + − 819 4096 {\displaystyle +{\frac {-17}{5120}}x^{5}+{\frac {-17}{2048}}x^{4}+{\frac {-307}{3072}}x^{3}+{\frac {-819}{2048}}x^{2}+{\frac {-819}{2048}}x+{\frac {-819}{4096}}\!}
Plotting this equation in Matlab yields:
For n = 9: r x = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + x 9 9 ! − x 11 11 ! + x 13 13 ! − x 15 15 ! + x 17 17 ! − x 19 19 ! {\displaystyle r_{x}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-{\frac {x^{11}}{11!}}+{\frac {x^{13}}{13!}}-{\frac {x^{15}}{15!}}+{\frac {x^{17}}{17!}}-{\frac {x^{19}}{19!}}\!} y p = K 19 x 19 + K 18 x 18 + K 17 x 17 + K 16 x 16 + K 15 x 15 + K 14 x 14 + K 13 x 13 + K 12 x 12 + K 11 x 11 + K 10 x 10 {\displaystyle y_{p}=K_{19}x^{19}+K_{18}x^{18}+K_{17}x^{17}+K_{16}x^{16}+K_{15}x^{15}+K_{14}x^{14}+K_{13}x^{13}+K_{12}x^{12}+K_{11}x^{11}+K_{10}x^{10}\!} + K 9 x 9 + K 8 x 8 + K 7 x 7 + K 6 x 6 + K 5 x 5 + K 4 x 4 + K 3 x 3 + K 2 x 2 + K 1 x + K 0 {\displaystyle +K_{9}x^{9}+K_{8}x^{8}+K_{7}x^{7}+K_{6}x^{6}+K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}\!} y p ′ = 19 K 19 x 18 + 18 K 18 x 17 + 17 K 17 x 16 + 16 K 16 x 15 + 15 K 15 x 14 + 14 K 14 x 13 + 13 K 13 x 12 + 12 K 12 x 11 + 11 K 11 x 10 + 10 K 10 x 9 {\displaystyle y_{p}^{'}=19K_{19}x^{18}+18K_{18}x^{17}+17K_{17}x^{16}+16K_{16}x^{15}+15K_{15}x^{14}+14K_{14}x^{13}+13K_{13}x^{12}+12K_{12}x^{11}+11K_{11}x^{10}+10K_{10}x^{9}\!} + 9 K 9 x 8 + 8 K 8 x 7 + 7 K 7 x 6 + 6 K 6 x 5 + 5 K 5 x 4 + 4 K 4 x 3 + 3 K 3 x 2 + 2 K 2 x + 1 K 1 {\displaystyle +9K_{9}x^{8}+8K_{8}x^{7}+7K_{7}x^{6}+6K_{6}x^{5}+5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+1K_{1}\!} y p ″ = 342 K 19 x 17 + 306 K 18 x 16 + 272 K 17 x 15 + 240 K 16 x 14 + 210 K 15 x 13 + 182 K 14 x 12 + 156 K 13 x 11 + 132 K 12 x 10 + 110 K 11 x 9 + 90 K 10 x 8 {\displaystyle y_{p}^{''}=342K_{19}x^{17}+306K_{18}x^{16}+272K_{17}x^{15}+240K_{16}x^{14}+210K_{15}x^{13}+182K_{14}x^{12}+156K_{13}x^{11}+132K_{12}x^{10}+110K_{11}x^{9}+90K_{10}x^{8}\!} + 72 K 9 x 7 + 56 K 8 x 6 + 42 K 7 x 5 + 30 K 6 x 4 + 20 K 5 x 3 + 12 K 4 x 2 + 6 K 3 x + 2 K 2 {\displaystyle +72K_{9}x^{7}+56K_{8}x^{6}+42K_{7}x^{5}+30K_{6}x^{4}+20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!} y p = − 1 2.433 E 17 x 19 + − 1 8.536 E 15 x 18 + − 1 9.485 E 14 x 17 + − 1 1.112 E 14 x 16 + − 1 2.202 E 12 x 15 + − 1 1.094 E 11 x 14 {\displaystyle y_{p}={\frac {-1}{2.433E17}}x^{19}+{\frac {-1}{8.536E15}}x^{18}+{\frac {-1}{9.485E14}}x^{17}+{\frac {-1}{1.112E14}}x^{16}+{\frac {-1}{2.202E12}}x^{15}+{\frac {-1}{1.094E11}}x^{14}\!} + − 1 1.563 E 10 x 13 + − 1 2.404 E 9 x 12 + − 1 6.657 E 7 x 11 + − 1 4.537 E 6 x 10 + − 1 9.074 E 5 x 9 + − 4 8.066 E 5 x 8 + − 38 3.192 E 5 x 7 {\displaystyle +{\frac {-1}{1.563E10}}x^{13}+{\frac {-1}{2.404E9}}x^{12}+{\frac {-1}{6.657E7}}x^{11}+{\frac {-1}{4.537E6}}x^{10}+{\frac {-1}{9.074E5}}x^{9}+{\frac {-4}{8.066E5}}x^{8}+{\frac {-38}{3.192E5}}x^{7}\!} + − 73 657010 x 6 + − 218 65401 x 5 + − 218 65401 x 4 + − 3441 49157 x 3 + − 1198 4061 x 2 + − 1431 8177 x + 145 4462 {\displaystyle +{\frac {-73}{657010}}x^{6}+{\frac {-218}{65401}}x^{5}+{\frac {-218}{65401}}x^{4}+{\frac {-3441}{49157}}x^{3}+{\frac {-1198}{4061}}x^{2}+{\frac {-1431}{8177}}x+{\frac {145}{4462}}\!} y n = c 1 e x + c 2 e 2 x + − 1 2.433 E 17 x 19 + − 1 8.536 E 15 x 18 + − 1 9.485 E 14 x 17 + − 1 1.112 E 14 x 16 + − 1 2.202 E 12 x 15 + − 1 1.094 E 11 x 14 {\displaystyle y_{n}=c_{1}e^{x}+c_{2}e^{2x}+{\frac {-1}{2.433E17}}x^{19}+{\frac {-1}{8.536E15}}x^{18}+{\frac {-1}{9.485E14}}x^{17}+{\frac {-1}{1.112E14}}x^{16}+{\frac {-1}{2.202E12}}x^{15}+{\frac {-1}{1.094E11}}x^{14}\!} + − 1 1.563 E 10 x 13 + − 1 2.404 E 9 x 12 + − 1 6.657 E 7 x 11 + − 1 4.537 E 6 x 10 + − 1 9.074 E 5 x 9 + − 4 8.066 E 5 x 8 + − 38 3.192 E 5 x 7 {\displaystyle +{\frac {-1}{1.563E10}}x^{13}+{\frac {-1}{2.404E9}}x^{12}+{\frac {-1}{6.657E7}}x^{11}+{\frac {-1}{4.537E6}}x^{10}+{\frac {-1}{9.074E5}}x^{9}+{\frac {-4}{8.066E5}}x^{8}+{\frac {-38}{3.192E5}}x^{7}\!} + − 73 657010 x 6 + − 218 65401 x 5 + − 218 65401 x 4 + − 3441 49157 x 3 + − 1198 4061 x 2 + − 1431 8177 x + 145 4462 {\displaystyle +{\frac {-73}{657010}}x^{6}+{\frac {-218}{65401}}x^{5}+{\frac {-218}{65401}}x^{4}+{\frac {-3441}{49157}}x^{3}+{\frac {-1198}{4061}}x^{2}+{\frac {-1431}{8177}}x+{\frac {145}{4462}}\!} y n ( 0 ) = c 1 + c 2 + 145 4462 = 1 {\displaystyle y_{n}(0)=c_{1}+c_{2}+{\frac {145}{4462}}=1\!} y n ′ ( 0 ) = c 1 + 2 c 2 + − 1431 8177 = 0 {\displaystyle y_{n}^{'}(0)=c_{1}+2c_{2}+{\frac {-1431}{8177}}=0\!} c 1 = 1.76 , c 2 = − 0.7925 {\displaystyle c_{1}=1.76,c_{2}=-0.7925\!} y n = 1.76 e x + − 0.7925 e 2 x + − 1 2.433 E 17 x 19 + − 1 8.536 E 15 x 18 + − 1 9.485 E 14 x 17 + − 1 1.112 E 14 x 16 + − 1 2.202 E 12 x 15 + − 1 1.094 E 11 x 14 {\displaystyle y_{n}=1.76e^{x}+-0.7925e^{2x}+{\frac {-1}{2.433E17}}x^{19}+{\frac {-1}{8.536E15}}x^{18}+{\frac {-1}{9.485E14}}x^{17}+{\frac {-1}{1.112E14}}x^{16}+{\frac {-1}{2.202E12}}x^{15}+{\frac {-1}{1.094E11}}x^{14}\!} + − 1 1.563 E 10 x 13 + − 1 2.404 E 9 x 12 + − 1 6.657 E 7 x 11 + − 1 4.537 E 6 x 10 + − 1 9.074 E 5 x 9 + − 4 8.066 E 5 x 8 + − 38 3.192 E 5 x 7 {\displaystyle +{\frac {-1}{1.563E10}}x^{13}+{\frac {-1}{2.404E9}}x^{12}+{\frac {-1}{6.657E7}}x^{11}+{\frac {-1}{4.537E6}}x^{10}+{\frac {-1}{9.074E5}}x^{9}+{\frac {-4}{8.066E5}}x^{8}+{\frac {-38}{3.192E5}}x^{7}\!} + − 73 65701 x 6 + − 218 65401 x 5 + − 218 65401 x 4 + − 3441 49157 x 3 + − 1198 4061 x 2 + − 1431 8177 x + 145 4462 {\displaystyle +{\frac {-73}{65701}}x^{6}+{\frac {-218}{65401}}x^{5}+{\frac {-218}{65401}}x^{4}+{\frac {-3441}{49157}}x^{3}+{\frac {-1198}{4061}}x^{2}+{\frac {-1431}{8177}}x+{\frac {145}{4462}}\!}
Note: On the interval from 0 {\displaystyle 0\!} 4 π {\displaystyle 4\pi \!} 10 10 {\displaystyle 10^{10}\!}
--Egm4313.s12.team11.sheider (talk ) 01:09, 14 March 2012 (UTC)
Solved by Daniel Suh
Problem Statement
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Using the particular solution from Table 2.1 on Kreyszig, find the overall solution y ( x ) {\displaystyle y(x)\!} y n ( x ) {\displaystyle y_{n}(x)\!}
y ″ − 3 y ′ + 2 y = r ( x ) {\displaystyle y''-3y'+2y=r(x)\!} r ( x ) = s i n ( x ) {\displaystyle r(x)=sin(x)\!} y ( 0 ) = 1 {\displaystyle y(0)=1\!} y ′ ( 0 ) = 0 {\displaystyle y'(0)=0\!}
Homogenous Solution
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To find y h ( x ) {\displaystyle y_{h}(x)\!} y ″ − 3 y ′ + 2 y = r ( x ) {\displaystyle y''-3y'+2y=r(x)\!} λ 2 − 3 λ + 2 = 0 {\displaystyle \lambda ^{2}-3\lambda +2=0\!} ( λ − 2 ) ( λ − 1 ) = 0 {\displaystyle (\lambda -2)(\lambda -1)=0\!} λ 1 = 2 , λ 2 = 1 {\displaystyle \lambda _{1}=2,\lambda _{2}=1\!} y h ( x ) = c 1 e 2 x + c 2 e x {\displaystyle y_{h}(x)=c_{1}e^{2x}+c_{2}e^{x}\!}
Particular Solution
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To find y p ( x ) {\displaystyle y_{p}(x)\!} r ( x ) = s i n ( x ) {\displaystyle r(x)=sin(x)\!} y p = K c o s ( ω x ) + M s i n ( ω x ) {\displaystyle y_{p}=Kcos(\omega x)+Msin(\omega x)\!} y p ′ = − K s i n ( ω x ) + M c o s ( ω x ) {\displaystyle y'_{p}=-Ksin(\omega x)+Mcos(\omega x)\!} y p ″ = − K c o s ( ω x ) − M s i n ( ω x ) {\displaystyle y''_{p}=-Kcos(\omega x)-Msin(\omega x)\!}
Plugging it back into the equation,( − K c o s ( x ) − M s i n ( x ) ) − 3 ( − K s i n ( x ) + M c o s ( x ) ) + 2 ( K c o s ( x ) + M s i n ( x ) ) = s i n ( x ) {\displaystyle (-Kcos(x)-Msin(x))-3(-Ksin(x)+Mcos(x))+2(Kcos(x)+Msin(x))=sin(x)\!} ( 3 K + M ) s i n ( x ) + ( K − 3 M ) c o s ( x ) = s i n ( x ) {\displaystyle (3K+M)sin(x)+(K-3M)cos(x)=sin(x)\!}
Solving for coefficients,( 3 K + M ) = 1 {\displaystyle (3K+M)=1\!} ( K − 3 M ) = 0 {\displaystyle (K-3M)=0\!} K = 3 10 {\displaystyle K={\frac {3}{10}}\!} M = 1 10 {\displaystyle M={\frac {1}{10}}\!} y p ( x ) = 3 10 c o s ( x ) + 1 10 s i n ( x ) {\displaystyle y_{p}(x)={\frac {3}{10}}cos(x)+{\frac {1}{10}}sin(x)\!}
Overall Solution
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y ( x ) = y h ( x ) + y p ( x ) {\displaystyle y(x)=y_{h}(x)+y_{p}(x)\!} y ( x ) = c 1 e 2 x + c 2 e x + 3 10 c o s ( x ) + 1 10 s i n ( x ) {\displaystyle y(x)=c_{1}e^{2x}+c_{2}e^{x}+{\frac {3}{10}}cos(x)+{\frac {1}{10}}sin(x)\!}
Solving with initial conditions,y ( x ) = c 1 e 2 x + c 2 e x + 3 10 c o s ( x ) + 1 10 s i n ( x ) {\displaystyle y(x)=c_{1}e^{2x}+c_{2}e^{x}+{\frac {3}{10}}cos(x)+{\frac {1}{10}}sin(x)\!} y ′ ( x ) = 2 c 1 e 2 x + c 2 e x − 3 10 s i n ( x ) + 1 10 c o s ( x ) {\displaystyle y'(x)=2c_{1}e^{2x}+c_{2}e^{x}-{\frac {3}{10}}sin(x)+{\frac {1}{10}}cos(x)\!}
1 = c 1 + c 2 + 3 10 {\displaystyle 1=c_{1}+c_{2}+{\frac {3}{10}}\!} 0 = 2 c 1 + c 2 + 1 10 {\displaystyle 0=2c_{1}+c_{2}+{\frac {1}{10}}\!} c 1 = − 4 5 {\displaystyle c_{1}={\frac {-4}{5}}\!} c 2 = 3 2 {\displaystyle c_{2}={\frac {3}{2}}\!}
y ( x ) = − 4 5 e 2 x + 3 2 e x + 3 10 c o s ( x ) + 1 10 s i n ( x ) {\displaystyle y(x)={\frac {-4}{5}}e^{2x}+{\frac {3}{2}}e^{x}+{\frac {3}{10}}cos(x)+{\frac {1}{10}}sin(x)\!} {\displaystyle \!}
Plotting in matlab yields:
Created by [Daniel Suh ] 20:43, 13 March 2012 (UTC)
Solved by Francisco Arrieta
Problem Statement
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Develop log ( x + 1 ) {\displaystyle \log(x+1)\!} x = 0 {\displaystyle x=0\!} f ( x ) = ∑ n = 0 ∞ f ( n ) ( x ^ ) n ! ( x − x ^ ) n {\displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}({\hat {x}})}{n!}}(x-{\hat {x}})^{n}\!} Solution
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For n=4:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}\!} For n=7:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}\!} For n=11:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) + x 11 log ( 10 11 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}+{\frac {x^{11}}{\log(10^{11})}}\!} For n=16:
log ( x + 1 ) = x log ( 10 ) − x 2 log ( 10 2 ) + x 3 log ( 10 3 ) − x 4 log ( 10 4 ) + x 5 log ( 10 5 ) − x 6 log ( 10 6 ) + x 7 log ( 10 7 ) − x 8 log ( 10 8 ) + x 9 log ( 10 9 ) − x 10 log ( 10 10 ) + x 11 log ( 10 11 ) {\displaystyle \log(x+1)={\frac {x}{\log(10)}}-{\frac {x^{2}}{\log(10^{2})}}+{\frac {x^{3}}{\log(10^{3})}}-{\frac {x^{4}}{\log(10^{4})}}+{\frac {x^{5}}{\log(10^{5})}}-{\frac {x^{6}}{\log(10^{6})}}+{\frac {x^{7}}{\log(10^{7})}}-{\frac {x^{8}}{\log(10^{8})}}+{\frac {x^{9}}{\log(10^{9})}}-{\frac {x^{10}}{\log(10^{10})}}+{\frac {x^{11}}{\log(10^{11})}}\!} − x 12 log ( 10 12 ) + x 13 log ( 10 13 ) − x 14 log ( 10 14 ) + x 15 log ( 10 15 ) − x 16 log ( 10 16 ) {\displaystyle -{\frac {x^{12}}{\log(10^{12})}}+{\frac {x^{13}}{\log(10^{13})}}-{\frac {x^{14}}{\log(10^{14})}}+{\frac {x^{15}}{\log(10^{15})}}-{\frac {x^{16}}{\log(10^{16})}}\!}
Using MATLAB
--Egm4313.s12.team11.arrieta (talk ) 00:44, 14 March 2012 (UTC)
Solved by: --Egm4313.s12.team11.vargas.aa (talk ) 01:28, 14 March 2012 (UTC)
Problem Statement
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Given:
y ″ − 3 y ′ + 2 y = r ( x ) {\displaystyle y''-3y'+2y=r(x)\!}
where r ( x ) = log ( 1 + x ) {\displaystyle r(x)=\log(1+x)\!}
With initial conditions: y ( − 3 4 ) = 1 , y ′ ( − 3 4 ) = 0 {\displaystyle y(-{\frac {3}{4}})=1,y'(-{\frac {3}{4}})=0\!} y n ( x ) {\displaystyle y_{n}(x)\!} n = 4 , 7 , 11 {\displaystyle n=4,7,11\!} [ − 3 4 , 3 ] {\displaystyle [-{\frac {3}{4}},3]\!}
Solution
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First we find the homogeneous solution to the ODE:λ 2 − 3 λ + 2 = 0 {\displaystyle \lambda ^{2}-3\lambda +2=0\!} ( λ − 2 ) ( λ − 1 ) = 0 {\displaystyle (\lambda -2)(\lambda -1)=0\!} λ = 1 , 2 {\displaystyle \lambda =1,2\!} y h = C 1 e ( 2 x ) + c 2 e x {\displaystyle y_{h}=C_{1}e^{(}2x)+c_{2}e^{x}\!}
Now to find the particulate solutionFor n=4
r ( x ) = ∑ 0 n − ( − 1 ) n x n n ln ( 10 ) {\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}
r ( x ) = x ln ( 10 ) − x 2 2 ln ( 10 ) + x 3 3 ln ( 10 ) − x 4 4 ln ( 10 ) {\displaystyle r(x)={\frac {x}{\ln(10)}}-{\frac {x^{2}}{2\ln(10)}}+{\frac {x^{3}}{3\ln(10)}}-{\frac {x^{4}}{4\ln(10)}}\!}
We can then use a matrix to organize the known coefficients:
[ 2 − 3 2 0 0 0 2 − 6 6 0 0 0 2 − 9 12 0 0 0 2 − 12 0 0 0 0 2 ] [ K 0 K 1 K 2 K 3 K 4 ] = [ 0 1 l n ( 10 ) 1 2 l n ( 10 ) 1 3 l n ( 10 ) 1 4 l n ( 10 ) ] {\displaystyle {\begin{bmatrix}2&-3&2&0&0\\0&2&-6&6&0\\0&0&2&-9&12\\0&0&0&2&-12\\0&0&0&0&2\\\end{bmatrix}}{\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\\K_{4}\end{bmatrix}}={\begin{bmatrix}0\\{\frac {1}{ln(10)}}\\{\frac {1}{2ln(10)}}\\{\frac {1}{3ln(10)}}\\{\frac {1}{4ln(10)}}\end{bmatrix}}\!} y p 4 = 4.0444 + 3.7458 x + 1.5743 x 2 + 0.3981 x 3 + 0.0543 x 4 {\displaystyle y_{p4}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}\!} y n = 4.0444 + 3.7458 x + 1.5743 x 2 + 0.3981 x 3 + 0.0543 x 4 + C 1 e 2 x + C 2 e x {\displaystyle y_{n}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}+C_{1}e^{2x}+C_{2}e^{x}\!} y n ′ = 3.7458 + 3.1486 x + 1.1943 x 2 + 0.2172 x 3 + 2 C 1 e 2 x + C 2 e x {\displaystyle y'_{n}=3.7458+3.1486x+1.1943x^{2}+0.2172x^{3}+2C_{1}e^{2x}+C_{2}e^{x}\!} y ( − 0.75 ) = 0.9698261719 + 0.231301601 C ! + 0.4723665527 C 2 = 1 {\displaystyle y(-0.75)=0.9698261719+0.231301601C_{!}+0.4723665527C_{2}=1\!} y ′ ( − 0.75 ) = 1.9645125 + 0.4462603203 C 1 + 0.4723665527 C 2 {\displaystyle y'(-0.75)=1.9645125+0.4462603203C_{1}+0.4723665527C_{2}\!} C 1 = − 4.46 {\displaystyle C_{1}=-4.46\!} C 2 = 0.055 {\displaystyle C_{2}=0.055\!} y n = 4.0444 + 3.7458 x + 1.5743 x 2 + 0.3981 x 3 + 0.0543 x 4 − 4.46 e 2 x + 0.055 e x {\displaystyle y_{n}=4.0444+3.7458x+1.5743x^{2}+0.3981x^{3}+0.0543x^{4}-4.46e^{2x}+0.055e^{x}\!}
For n=7
r ( x ) = ∑ 0 n − ( − 1 ) n x n n ln ( 10 ) {\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}
r ( x ) = x ln ( 10 ) − x 2 2 ln ( 10 ) + x 3 3 ln ( 10 ) − x 4 4 ln ( 10 ) + x 5 5 ln ( 10 ) − x 6 6 ln ( 10 ) + x 7 7 ln ( 10 ) {\displaystyle r(x)={\frac {x}{\ln(10)}}-{\frac {x^{2}}{2\ln(10)}}+{\frac {x^{3}}{3\ln(10)}}-{\frac {x^{4}}{4\ln(10)}}+{\frac {x^{5}}{5\ln(10)}}-{\frac {x^{6}}{6\ln(10)}}+{\frac {x^{7}}{7\ln(10)}}\!}
We can then use a matrix to organize the known coefficients:
[ 2 − 3 2 0 0 0 0 0 0 2 − 6 6 0 0 0 0 0 0 2 − 9 12 0 0 0 0 0 0 2 − 12 20 0 0 0 0 0 0 2 − 15 30 0 0 0 0 0 0 2 − 18 42 0 0 0 0 0 0 2 − 21 0 0 0 0 0 0 0 2 ] [ K 0 K 1 K 2 K 3 K 4 K 5 K 6 K 7 ] [ 0 1 l n ( 10 ) 1 2 l n ( 10 ) 1 3 l n ( 10 ) 1 4 l n ( 10 ) 1 5 l n ( 10 ) 1 6 l n ( 10 ) 1 7 l n ( 10 ) ] {\displaystyle {\begin{bmatrix}2&-3&2&0&0&0&0&0\\0&2&-6&6&0&0&0&0\\0&0&2&-9&12&0&0&0\\0&0&0&2&-12&20&0&0\\0&0&0&0&2&-15&30&0\\0&0&0&0&0&2&-18&42\\0&0&0&0&0&0&2&-21\\0&0&0&0&0&0&0&2\\\end{bmatrix}}{\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\\K_{4}\\K_{5}\\K_{6}\\K_{7}\end{bmatrix}}{\begin{bmatrix}0\\{\frac {1}{ln(10)}}\\{\frac {1}{2ln(10)}}\\{\frac {1}{3ln(10)}}\\{\frac {1}{4ln(10)}}\\{\frac {1}{5ln(10)}}\\{\frac {1}{6ln(10)}}\\{\frac {1}{7ln(10)}}\end{bmatrix}}\!} y p 7 = 377.4833 + 375.3933 x + 185.6066 x 2 + 60.5479 x 3 + 14.4946 x 4 + 2.6492 x 5 + 0.3619 x 6 + 0.0310 x 7 {\displaystyle y_{p7}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}\!} y n = 377.4833 + 375.3933 x + 185.6066 x 2 + 60.5479 x 3 + 14.4946 x 4 + 2.6492 x 5 + 0.3619 x 6 + 0.0310 x 7 + C 1 e 2 x + c 2 e x {\displaystyle y_{n}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}+C_{1}e^{2x}+c_{2}e^{x}\!} y n ′ = 375.3933 + 371.213 x + 181.644 x 2 + 57.9784 x 3 + 13.46 x 4 + 2.1714 x 5 + 0.214 x 6 + 2 C 1 e 2 x + C 2 e x {\displaystyle y'_{n}=375.3933+371.213x+181.644x^{2}+57.9784x^{3}+13.46x^{4}+2.1714x^{5}+0.214x^{6}+2C_{1}e^{2x}+C_{2}e^{x}\!} y ( − 0.75 ) = 178.816 + 0.2231301601 C ! + 0.4723665527 C 2 = 1 {\displaystyle y(-0.75)=178.816+0.2231301601C_{!}+0.4723665527C_{2}=1\!} y ′ ( − 0.75 ) = 178.413 + 0.4462603203 C 1 + 0.4723665527 C 2 {\displaystyle y'(-0.75)=178.413+0.4462603203C_{1}+0.4723665527C_{2}\!} C 1 = − 2.6757 {\displaystyle C_{1}=-2.6757\!} C 2 = − 375.173 {\displaystyle C_{2}=-375.173\!} y n = 377.4833 + 375.3933 x + 185.6066 x 2 + 60.5479 x 3 + 14.4946 x 4 + 2.6492 x 5 + 0.3619 x 6 + 0.0310 x 7 + − 2.6757 e 2 x − 375.173 e x {\displaystyle y_{n}=377.4833+375.3933x+185.6066x^{2}+60.5479x^{3}+14.4946x^{4}+2.6492x^{5}+0.3619x^{6}+0.0310x^{7}+-2.6757e^{2x}-375.173e^{x}\!} For n=11
r ( x ) = ∑ 0 n − ( − 1 ) n x n n ln ( 10 ) {\displaystyle r(x)=\sum _{0}^{n}-{\frac {(-1)^{n}x^{n}}{n\ln(10)}}\!}
r ( x ) = x ln ( 10 ) − x 2 2 ln ( 10 ) + x 3 3 ln ( 10 ) − x 4 4 ln ( 10 ) + x 5 5 ln ( 10 ) − x 6 6 ln ( 10 ) + x 7 7 ln ( 10 ) {\displaystyle r(x)={\frac {x}{\ln(10)}}-{\frac {x^{2}}{2\ln(10)}}+{\frac {x^{3}}{3\ln(10)}}-{\frac {x^{4}}{4\ln(10)}}+{\frac {x^{5}}{5\ln(10)}}-{\frac {x^{6}}{6\ln(10)}}+{\frac {x^{7}}{7\ln(10)}}\!}
We can then use a matrix to organize the known coefficients: [ K 0 K 1 K 2 K 3 K 4 K 5 K 6 K 7 K 8 K 9 K 10 K 11 ] = [ 0 1 l n ( 10 ) 1 2 l n ( 10 ) 1 3 l n ( 10 ) 1 4 l n ( 10 ) 1 5 l n ( 10 ) 1 6 l n ( 10 ) 1 7 l n ( 10 ) 1 8 l n ( 10 ) 1 9 l n ( 10 ) 1 10 l n ( 10 ) 1 11 l n ( 10 ) ] {\displaystyle {\begin{bmatrix}K_{0}\\K_{1}\\K_{2}\\K_{3}\\K_{4}\\K_{5}\\K_{6}\\K_{7}\\K_{8}\\K_{9}\\K_{10}\\K_{11}\end{bmatrix}}={\begin{bmatrix}0\\{\frac {1}{ln(10)}}\\{\frac {1}{2ln(10)}}\\{\frac {1}{3ln(10)}}\\{\frac {1}{4ln(10)}}\\{\frac {1}{5ln(10)}}\\{\frac {1}{6ln(10)}}\\{\frac {1}{7ln(10)}}\\{\frac {1}{8ln(10)}}\\{\frac {1}{9ln(10)}}\\{\frac {1}{10ln(10)}}\\{\frac {1}{11ln(10)}}\end{bmatrix}}\!} y p 11 = 1753158.594 + 1752673.419 x + 875851.535 x 2 + 291627.134 x 3 + 72745.1129 x 4 + 14484.362 x 5 + 2392.510 x 6 + 335.632 x 7 + 40.417 x 8 {\displaystyle y_{p11}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!} + 4.1499 x 9 + 0.3474 x ( 10 ) + 0.0197 x ( 11 ) {\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)\!} y n = 1753158.594 + 1752673.419 x + 875851.535 x 2 + 291627.134 x 3 + 72745.1129 x 4 + 14484.362 x 5 + 2392.510 x 6 + 335.632 x 7 + 40.417 x 8 {\displaystyle y_{n}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!} + 4.1499 x 9 + 0.3474 x ( 10 ) + 0.0197 x ( 11 ) + C 1 e 2 x + C 2 e ( x ) {\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)+C_{1}e^{2x}+C_{2}e^{(}x)\!} y n ′ = 0.2167 x 1 0 + 3.474 x 9 + 37.3491 x 8 + 323.336 x 7 + 2349.42 x 6 + 14355.1 x 5 + 72421.8 x 4 + 290980. x 3 + 874881. x 2 {\displaystyle y'_{n}=0.2167x^{1}0+3.474x^{9}+37.3491x^{8}+323.336x^{7}+2349.42x^{6}+14355.1x^{5}+72421.8x^{4}+290980.x^{3}+874881.x^{2}\!} + 1.7517 x 10 6 x + 1.75267 x 10 6 + 2 C 1 e 2 x + C 2 e x {\displaystyle +1.7517x10^{6}x+1.75267x10^{6}+2C_{1}e^{2x}+C_{2}e^{x}\!} y ( − 0.75 ) = 828254 + 0.2231301601 C ! + 0.4723665527 C 2 = 1 {\displaystyle y(-0.75)=828254+0.2231301601C_{!}+0.4723665527C_{2}=1\!} y ′ ( − 0.75 ) = 828145 + 0.4462603203 C 1 + 0.4723665527 C 2 = 0 {\displaystyle y'(-0.75)=828145+0.4462603203C_{1}+0.4723665527C_{2}=0\!} C 1 = − 484.022 {\displaystyle C_{1}=-484.022\!} C 2 = − 1753750 {\displaystyle C_{2}=-1753750\!} y n = 1753158.594 + 1752673.419 x + 875851.535 x 2 + 291627.134 x 3 + 72745.1129 x 4 + 14484.362 x 5 + 2392.510 x 6 + 335.632 x 7 + 40.417 x 8 {\displaystyle y_{n}=1753158.594+1752673.419x+875851.535x^{2}+291627.134x^{3}+72745.1129x^{4}+14484.362x^{5}+2392.510x^{6}+335.632x^{7}+40.417x^{8}\!} + 4.1499 x 9 + 0.3474 x ( 10 ) + 0.0197 x ( 11 ) − 484.022 e 2 x − 1753750 e x {\displaystyle +4.1499x^{9}+0.3474x^{(}10)+0.0197x^{(}11)-484.022e^{2x}-1753750e^{x}\!}
Plot for part 2 and part 3. Please note that because of the broad scale, it is almost impossible to distinguish between the graphs for n=4,7,11 nad the ODE45 operator. Only when zoomed in is there any noticeable difference
Problem Statement
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Find n sufficiently high so that y n ( x 1 ) , y n ′ ( x 1 ) {\displaystyle y_{n}(x_{1}),y'_{n}(x_{1})} 10 − 5 {\displaystyle 10^{-5}} x 1 = 0.9 {\displaystyle x_{1}=0.9}
Solution
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Using a program in MATLAB that iteratively added terms onto the taylor series of l o g ( 1 + x ) {\displaystyle log(1+x)} 10 − 5 {\displaystyle 10^{-5}}
It was found after trial and error that n = 39 {\displaystyle n=39} 10 − 5 {\displaystyle 10^{-5}} 9.7422e-005
Similarly, for y n ′ ( x 1 ) {\displaystyle y'_{n}(x_{1})}
It was found after trial and error that n = 74 {\displaystyle n=74} 10 − 5 {\displaystyle 10^{-5}} 9.3967e-005
Problem Statement
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Develop l o g ( 1 + x ) {\displaystyle log(1+x)} x ^ = 1 {\displaystyle {\hat {x}}=1} n = 4 , 7 , 11 {\displaystyle n=4,7,11}
Solution
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A MATLAB program was created, which calculated the Taylor series of each n value, along with the exact function, then plotted these together to show the comparison of all the series.n = 7 {\displaystyle n=7} x ^ = 1 {\displaystyle {\hat {x}}=1} x − 1 2 ln ( 10 ) − ( x − 1 ) 2 8 ln ( 10 ) + ( x − 1 ) 3 24 ln ( 10 ) − ( x − 1 ) 4 64 ln ( 10 ) + ( x − 1 ) 5 160 ln ( 10 ) − ( x − 1 ) 6 384 ln ( 10 ) + ln ( 2 ) ln ( 10 ) {\displaystyle {\frac {x-1}{2\,\ln \!\left(10\right)}}-{\frac {{\left(x-1\right)}^{2}}{8\,\ln \!\left(10\right)}}+{\frac {{\left(x-1\right)}^{3}}{24\,\ln \!\left(10\right)}}-{\frac {{\left(x-1\right)}^{4}}{64\,\ln \!\left(10\right)}}+{\frac {{\left(x-1\right)}^{5}}{160\,\ln \!\left(10\right)}}-{\frac {{\left(x-1\right)}^{6}}{384\,\ln \!\left(10\right)}}+{\frac {\ln \!\left(2\right)}{\ln \!\left(10\right)}}}
egm4313.s12.team11.gooding (talk ) 03:48, 14 March 2012 (UTC)
Solved by Luca Imponenti
Find y n ( x ) {\displaystyle y_{n}(x)\!} n = 4 , 7 , 11 {\displaystyle n=4,7,11\!}
y n ″ − 3 y n ′ + 2 y n = r n ( x ) {\displaystyle y_{n}''-3y_{n}'+2y_{n}=r_{n}(x)\!} for x {\displaystyle x\!} [ 0.9 , 3 ] {\displaystyle [0.9,3]\!}
Plot y n ( x ) {\displaystyle y_{n}(x)\!} n = 4 , 7 , 11 {\displaystyle n=4,7,11\!} x {\displaystyle x\!} [ 0.9 , 3 ] {\displaystyle [0.9,3]\!}
Homogeneous Solution
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The homogeneous case is shown below:
y h ″ − 3 y h ′ + 2 y h = 0 {\displaystyle y''_{h}-3y'_{h}+2y_{h}=0\!} This equation has the following roots:
λ 1 = 1 , λ 2 = 2 {\displaystyle \lambda _{1}=1,\lambda _{2}=2\!}
Which gives yields the homogeneous solution
y h = c 1 e x + c 2 e 2 x {\displaystyle y_{h}=c_{1}e^{x}+c_{2}e^{2x}\!} General Solution, n=4
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Using the taylor series approximation from earlier with n = 4 {\displaystyle n=4\!}
r 4 ( x ) = l o g ( 2 ) + ( x − 1 ) 2 l n ( 10 ) − ( x − 1 ) 2 8 l n ( 10 ) + ( x − 1 ) 3 24 l n ( 10 ) − ( x − 1 ) 4 64 l n ( 10 ) {\displaystyle r_{4}(x)=log(2)+{\frac {(x-1)}{2ln(10)}}-{\frac {(x-1)^{2}}{8ln(10)}}+{\frac {(x-1)^{3}}{24ln(10)}}-{\frac {(x-1)^{4}}{64ln(10)}}\!}
We know the particular solution, y p 4 ( x ) {\displaystyle y_{p4}(x)\!}
y p 4 ( x ) = a 4 ( x − 1 ) 4 + a 3 ( x − 1 ) 3 + a 2 ( x − 1 ) 2 + a 1 ( x − 1 ) + a 0 {\displaystyle y_{p4}(x)=a_{4}(x-1)^{4}+a_{3}(x-1)^{3}+a_{2}(x-1)^{2}+a_{1}(x-1)+a_{0}\!}
taking the derivatives of this solution
y p 4 ′ ( x ) = d d x y p 4 ( x ) = 4 a 4 ( x − 1 ) 3 + 3 a 3 ( x − 1 ) 2 + 2 a 2 ( x − 1 ) + a 1 {\displaystyle y'_{p4}(x)={\frac {d}{dx}}y_{p4}(x)=4a_{4}(x-1)^{3}+3a_{3}(x-1)^{2}+2a_{2}(x-1)+a_{1}\!}
and
y p 4 ″ ( x ) = d d x y p 4 ′ ( x ) = 12 a 4 ( x − 1 ) 3 + 6 a 3 ( x − 1 ) 2 + 2 a 2 {\displaystyle y''_{p4}(x)={\frac {d}{dx}}y'_{p4}(x)=12a_{4}(x-1)^{3}+6a_{3}(x-1)^{2}+2a_{2}\!}
Plugging the above equations into the original ODE yields the following matrix equation:
[ 2 0 0 0 0 − 12 2 0 0 0 12 − 9 2 0 0 0 6 − 6 2 0 0 0 2 − 3 2 ] ∗ [ a 4 a 3 a 2 a 1 a 0 ] = [ − 1 64 l n ( 10 ) 1 24 l n ( 10 ) − 1 8 l n ( 10 ) 1 2 l n ( 10 ) l o g ( 2 ) ] {\displaystyle {\begin{bmatrix}2&0&0&0&0\\-12&2&0&0&0\\12&-9&2&0&0\\0&6&-6&2&0\\0&0&2&-3&2\end{bmatrix}}*{\begin{bmatrix}a_{4}\\a_{3}\\a_{2}\\a_{1}\\a_{0}\end{bmatrix}}={\begin{bmatrix}-{\frac {1}{64ln(10)}}\\{\frac {1}{24ln(10)}}\\-{\frac {1}{8ln(10)}}\\{\frac {1}{2ln(10)}}\\log(2)\end{bmatrix}}\!} The unknown vector a {\displaystyle a\!}
a 4 = − .0034 , a 3 = − .0113 , a 2 = − .0577 , a 1 = − .1624 , a 0 = .1624 {\displaystyle a_{4}=-.0034,a_{3}=-.0113,a_{2}=-.0577,a_{1}=-.1624,a_{0}=.1624\!}
So the particular solution y p 4 {\displaystyle y_{p4}\!}
y p 4 = 0.1624 − 0.1624 ∗ ( x − 1 ) − .0577 ∗ ( x − 1 ) 2 − .0113 ∗ ( x − 1 ) 3 − .0034 ∗ ( x − 1 ) 4 {\displaystyle y_{p4}=0.1624-0.1624*(x-1)-.0577*(x-1)^{2}-.0113*(x-1)^{3}-.0034*(x-1)^{4}\!}
We can now find the general solution for n=4, y 4 ( x ) {\displaystyle y_{4}(x)\!}
y 4 ( x ) = y h ( x ) + y p 4 ( x ) {\displaystyle y_{4}(x)=y_{h}(x)+y_{p4}(x)\!}
y 4 ( x ) = c 1 e x + c 2 e 2 x + 0.1624 − 0.1624 ∗ ( x − 1 ) {\displaystyle y_{4}(x)=c_{1}e^{x}+c_{2}e^{2x}+0.1624-0.1624*(x-1)\!} − .0577 ∗ ( x − 1 ) 2 − .0113 ∗ ( x − 1 ) 3 − .0034 ∗ ( x − 1 ) 4 {\displaystyle -.0577*(x-1)^{2}-.0113*(x-1)^{3}-.0034*(x-1)^{4}\!}
Solving using the initial conditions yields;
y 4 ( x ) = .0595 e x − .0076 e 2 x + 0.1624 − 0.1624 ∗ ( x − 1 ) {\displaystyle y_{4}(x)=.0595e^{x}-.0076e^{2x}+0.1624-0.1624*(x-1)\!} − .0577 ∗ ( x − 1 ) 2 − .0113 ∗ ( x − 1 ) 3 − .0034 ∗ ( x − 1 ) 4 {\displaystyle -.0577*(x-1)^{2}-.0113*(x-1)^{3}-.0034*(x-1)^{4}\!}
General Solution, n=7
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Using the taylor series approximation from earlier with n = 7 {\displaystyle n=7\!}
r 7 ( x ) = l o g ( 2 ) + ( x − 1 ) 2 l n ( 10 ) − ( x − 1 ) 2 8 l n ( 10 ) + ( x − 1 ) 3 24 l n ( 10 ) − ( x − 1 ) 4 64 l n ( 10 ) + ( x − 1 ) 5 160 l n ( 10 ) − ( x − 1 ) 6 384 l n ( 10 ) + ( x − 1 ) 7 896 l n ( 10 ) {\displaystyle r_{7}(x)=log(2)+{\frac {(x-1)}{2ln(10)}}-{\frac {(x-1)^{2}}{8ln(10)}}+{\frac {(x-1)^{3}}{24ln(10)}}-{\frac {(x-1)^{4}}{64ln(10)}}+{\frac {(x-1)^{5}}{160ln(10)}}-{\frac {(x-1)^{6}}{384ln(10)}}+{\frac {(x-1)^{7}}{896ln(10)}}\!}
In a similar fashion we construct a matrix equation for n=7:
[ 2 0 0 0 0 0 0 0 − 21 2 0 0 0 0 0 0 42 − 18 2 0 0 0 0 0 0 30 − 15 2 0 0 0 0 0 0 20 − 12 2 0 0 0 0 0 0 12 − 9 2 0 0 0 0 0 0 6 − 6 2 0 0 0 0 0 0 2 − 3 2 ] ∗ [ a 7 a 6 a 5 a 4 a 3 a 2 a 1 a 0 ] = [ 1 896 l n ( 10 ) − 1 384 l n ( 10 ) 1 160 l n ( 10 ) − 1 64 l n ( 10 ) 1 24 l n ( 10 ) − 1 8 l n ( 10 ) 1 2 l n ( 10 ) l o g ( 2 ) ] {\displaystyle {\begin{bmatrix}2&0&0&0&0&0&0&0\\-21&2&0&0&0&0&0&0\\42&-18&2&0&0&0&0&0\\0&30&-15&2&0&0&0&0\\0&0&20&-12&2&0&0&0\\0&0&0&12&-9&2&0&0\\0&0&0&0&6&-6&2&0\\0&0&0&0&0&2&-3&2\end{bmatrix}}*{\begin{bmatrix}a_{7}\\a_{6}\\a_{5}\\a_{4}\\a_{3}\\a_{2}\\a_{1}\\a_{0}\end{bmatrix}}={\begin{bmatrix}{\frac {1}{896ln(10)}}\\-{\frac {1}{384ln(10)}}\\{\frac {1}{160ln(10)}}\\-{\frac {1}{64ln(10)}}\\{\frac {1}{24ln(10)}}\\-{\frac {1}{8ln(10)}}\\{\frac {1}{2ln(10)}}\\log(2)\end{bmatrix}}\!} Solving:
a 7 = .0002 , a 6 = .0020 , a 5 = .0141 , a 4 = .0725 , a 3 = .3034 , a 2 = .9029 , a 1 = 1.9072 , a 0 = 2.1084 {\displaystyle a_{7}=.0002,a_{6}=.0020,a_{5}=.0141,a_{4}=.0725,a_{3}=.3034,a_{2}=.9029,a_{1}=1.9072,a_{0}=2.1084\!}
So the particular solution y p 7 {\displaystyle y_{p7}\!}
y p 7 = 2.1084 + 1.9072 ∗ ( x − 1 ) + .9029 ∗ ( x − 1 ) 2 + .3034 ∗ ( x − 1 ) 3 + .0725 ∗ ( x − 1 ) 4 + .0141 ∗ ( x − 1 ) 5 + .0020 ∗ ( x − 1 ) 6 + .0002 ∗ ( x − 1 ) 7 {\displaystyle y_{p7}=2.1084+1.9072*(x-1)+.9029*(x-1)^{2}+.3034*(x-1)^{3}+.0725*(x-1)^{4}+.0141*(x-1)^{5}+.0020*(x-1)^{6}+.0002*(x-1)^{7}\!}
We can now find the general solution for n=7, y 7 ( x ) {\displaystyle y_{7}(x)\!}
y 7 ( x ) = y h ( x ) + y p 7 ( x ) {\displaystyle y_{7}(x)=y_{h}(x)+y_{p7}(x)\!}
y 7 ( x ) = c 1 e x + c 2 e 2 x + 2.1084 + 1.9072 ∗ ( x − 1 ) + .9029 ∗ ( x − 1 ) 2 + {\displaystyle y_{7}(x)=c_{1}e^{x}+c_{2}e^{2x}+2.1084+1.9072*(x-1)+.9029*(x-1)^{2}+\!}
.3034 ∗ ( x − 1 ) 3 + .0725 ∗ ( x − 1 ) 4 + .0141 ∗ ( x − 1 ) 5 + .0020 ∗ ( x − 1 ) 6 + .0002 ∗ ( x − 1 ) 7 {\displaystyle .3034*(x-1)^{3}+.0725*(x-1)^{4}+.0141*(x-1)^{5}+.0020*(x-1)^{6}+.0002*(x-1)^{7}\!}
Solving using our initial conditions yields
y 7 ( x ) = − .7271 e x + .0233 e 2 x + 2.1084 + 1.9072 ∗ ( x − 1 ) + {\displaystyle y_{7}(x)=-.7271e^{x}+.0233e^{2x}+2.1084+1.9072*(x-1)+\!} .9029 ∗ ( x − 1 ) 2 + .3034 ∗ ( x − 1 ) 3 + .0725 ∗ ( x − 1 ) 4 + .0141 ∗ ( x − 1 ) 5 + {\displaystyle .9029*(x-1)^{2}+.3034*(x-1)^{3}+.0725*(x-1)^{4}+.0141*(x-1)^{5}+\!} .0020 ∗ ( x − 1 ) 6 + .0002 ∗ ( x − 1 ) 7 {\displaystyle .0020*(x-1)^{6}+.0002*(x-1)^{7}\!}
General Solution, n=11
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Using the taylor series approximation from earlier with n = 11 {\displaystyle n=11\!}
r 11 ( x ) = l o g ( 2 ) + ( x − 1 ) 2 l n ( 10 ) − ( x − 1 ) 2 8 l n ( 10 ) + ( x − 1 ) 3 24 l n ( 10 ) − ( x − 1 ) 4 64 l n ( 10 ) + ( x − 1 ) 5 160 l n ( 10 ) − ( x − 1 ) 6 384 l n ( 10 ) + {\displaystyle r_{11}(x)=log(2)+{\frac {(x-1)}{2ln(10)}}-{\frac {(x-1)^{2}}{8ln(10)}}+{\frac {(x-1)^{3}}{24ln(10)}}-{\frac {(x-1)^{4}}{64ln(10)}}+{\frac {(x-1)^{5}}{160ln(10)}}-{\frac {(x-1)^{6}}{384ln(10)}}+\!}
( x − 1 ) 7 896 l n ( 10 ) − ( x − 1 ) 8 2048 l n ( 10 ) + ( x − 1 ) 9 4608 l n ( 10 ) − ( x − 1 ) 10 10240 l n ( 10 ) + ( x − 1 ) 11 22528 l n ( 10 ) {\displaystyle {\frac {(x-1)^{7}}{896ln(10)}}-{\frac {(x-1)^{8}}{2048ln(10)}}+{\frac {(x-1)^{9}}{4608ln(10)}}-{\frac {(x-1)^{10}}{10240ln(10)}}+{\frac {(x-1)^{11}}{22528ln(10)}}\!}
Finally, we write out the matrix equation for n=11:
A ∗ [ a 11 a 10 a 9 a 8 a 7 a 6 a 5 a 4 a 3 a 2 a 1 a 0 ] = [ 1 22528 l n ( 10 ) − 1 10240 l n ( 10 ) 1 4608 l n ( 10 ) − 1 2048 l n ( 10 ) 1 896 l n ( 10 ) − 1 384 l n ( 10 ) 1 160 l n ( 10 ) − 1 64 l n ( 10 ) 1 24 l n ( 10 ) − 1 8 l n ( 10 ) 1 2 l n ( 10 ) l o g ( 2 ) ] {\displaystyle A*{\begin{bmatrix}a_{11}\\a_{10}\\a_{9}\\a_{8}\\a_{7}\\a_{6}\\a_{5}\\a_{4}\\a_{3}\\a_{2}\\a_{1}\\a_{0}\end{bmatrix}}={\begin{bmatrix}{\frac {1}{22528ln(10)}}\\-{\frac {1}{10240ln(10)}}\\{\frac {1}{4608ln(10)}}\\-{\frac {1}{2048ln(10)}}\\{\frac {1}{896ln(10)}}\\-{\frac {1}{384ln(10)}}\\{\frac {1}{160ln(10)}}\\-{\frac {1}{64ln(10)}}\\{\frac {1}{24ln(10)}}\\-{\frac {1}{8ln(10)}}\\{\frac {1}{2ln(10)}}\\log(2)\end{bmatrix}}\!}
Solving the system in matlab:
a 11 = 0 , a 10 = .0002 , a 9 = .0019 , a 8 = .0181 , a 7 = .15 , a 6 = 1.0675 , a 5 = 6.4597 , {\displaystyle a_{11}=0,a_{10}=.0002,a_{9}=.0019,a_{8}=.0181,a_{7}=.15,a_{6}=1.0675,a_{5}=6.4597,\!}
a 4 = 32.4318 , a 3 = 130.0033 , a 2 = 390.3968 , a 1 = 781.289 , a 0 = 781.6873 {\displaystyle a_{4}=32.4318,a_{3}=130.0033,a_{2}=390.3968,a_{1}=781.289,a_{0}=781.6873\!}
So the particular solution y p 11 {\displaystyle y_{p11}\!}
y p 11 = 781.6873 + 781.289 ∗ ( x − 1 ) + 390.3968 ∗ ( x − 1 ) 2 + 130.0033 ∗ ( x − 1 ) 3 + 32.4318 ∗ ( x − 1 ) 4 + {\displaystyle y_{p11}=781.6873+781.289*(x-1)+390.3968*(x-1)^{2}+130.0033*(x-1)^{3}+32.4318*(x-1)^{4}+\!}
6.4597 ∗ ( x − 1 ) 5 + 1.0675 ∗ ( x − 1 ) 6 + .15 ∗ ( x − 1 ) 7 + .0181 ∗ ( x − 1 ) 8 + .0019 ∗ ( x − 1 ) 9 + .0002 ∗ ( x − 1 ) 10 {\displaystyle 6.4597*(x-1)^{5}+1.0675*(x-1)^{6}+.15*(x-1)^{7}+.0181*(x-1)^{8}+.0019*(x-1)^{9}+.0002*(x-1)^{10}\!}
We can now find the general solution for n=11, y 1 1 ( x ) {\displaystyle y_{1}1(x)\!}
y 11 ( x ) = y h ( x ) + y p 11 ( x ) {\displaystyle y_{11}(x)=y_{h}(x)+y_{p11}(x)\!}
y 11 ( x ) = c 1 e x + c 2 e 2 x + 781.6873 + 781.289 ∗ ( x − 1 ) {\displaystyle y_{11}(x)=c_{1}e^{x}+c_{2}e^{2x}+781.6873+781.289*(x-1)\!}
+ 390.3968 ∗ ( x − 1 ) 2 + 130.0033 ∗ ( x − 1 ) 3 + 32.4318 ∗ ( x − 1 ) 4 {\displaystyle +390.3968*(x-1)^{2}+130.0033*(x-1)^{3}+32.4318*(x-1)^{4}\!}
6.4597 ∗ ( x − 1 ) 5 + 1.0675 ∗ ( x − 1 ) 6 + .15 ∗ ( x − 1 ) 7 + .0181 ∗ ( x − 1 ) 8 + .0019 ∗ ( x − 1 ) 9 + .0002 ∗ ( x − 1 ) 10 {\displaystyle 6.4597*(x-1)^{5}+1.0675*(x-1)^{6}+.15*(x-1)^{7}+.0181*(x-1)^{8}+.0019*(x-1)^{9}+.0002*(x-1)^{10}\!}
Solving using our initial conditions yields
y 11 ( x ) = − 287.5907 e x + .05 e 2 x + 781.6873 + 781.289 ∗ ( x − 1 ) {\displaystyle y_{11}(x)=-287.5907e^{x}+.05e^{2x}+781.6873+781.289*(x-1)\!} + 390.3968 ∗ ( x − 1 ) 2 + 130.0033 ∗ ( x − 1 ) 3 + 32.4318 ∗ ( x − 1 ) 4 {\displaystyle +390.3968*(x-1)^{2}+130.0033*(x-1)^{3}+32.4318*(x-1)^{4}\!} 6.4597 ∗ ( x − 1 ) 5 + 1.0675 ∗ ( x − 1 ) 6 + {\displaystyle 6.4597*(x-1)^{5}+1.0675*(x-1)^{6}+\!} .15 ∗ ( x − 1 ) 7 + .0181 ∗ ( x − 1 ) 8 + .0019 ∗ ( x − 1 ) 9 + .0002 ∗ ( x − 1 ) 10 {\displaystyle .15*(x-1)^{7}+.0181*(x-1)^{8}+.0019*(x-1)^{9}+.0002*(x-1)^{10}\!}
y 4 {\displaystyle y_{4}\!}
y 7 {\displaystyle y_{7}\!}
y 11 {\displaystyle y_{11}\!}
Solved by Luca Imponenti
Use the matlab command ode45 to integrate numerically y ″ − 3 y ′ + 2 y = r ( x ) {\displaystyle y''-3y'+2y=r(x)\!} r ( x ) = l o g ( 1 + x ) {\displaystyle r(x)=log(1+x)\!}
and the initial conditions from Part 3 to obtain the numerical solution for y(x).
Plot y(x) in the same figure as above.
Matlab Solution
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The numerical solution calculated using the matlab ode45 command is shown below:
ans =
0.2788
0.2854
0.2923
0.2997
0.3074
0.3229
0.3401
0.3592
0.3804
0.4040
0.4302
0.4595
0.4921
0.5285
0.5691
0.6145
0.6651
0.7218
0.7850
0.8557
0.9346
1.0228
1.1213
1.2313
1.3542
1.4914
1.6445
1.8155
2.0063
2.2193
2.4569
2.7219
3.0175
3.3471
3.7146
4.1243
4.5809
5.0898
5.6568
6.2885
6.9921
7.3442
7.7142
8.1032
8.5119
Plotting the aboved vector of y-values,along with the results from earlier yields the following graph:
where the answer calculated in matlab is shown in yellow.
Egm4313.s12.team11.imponenti (talk ) 08:04, 14 March 2012 (UTC)
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