text:
User:Egm4313.s12.team17/Report 4

=Problem R4.1 - General Formulation for Polynomial Excitation=

Given
Given the equation:
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$$  \displaystyle y''+ay'+by= \sum_{j=0}^{n}k_jx^j $$     (Eq.1.1)
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The equation for the coefficients of x^0 \,:
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$$  \displaystyle 2c_2+ac_1+bc_0=k_0 $$     (Eq.1.2)
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Problem Statement
Obtain equations for x^1 \,, x^2 \,, x^3 \,, x^{n-1} \,, x^{n-2} \,, and x^n \, and insert those equations as well as the given equation for x^0 \, into a tri-diagonal matrix.

Solution
The order of the given equation is defined as the highest power of $$ x $$ in Eq. 1.1 below:
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$$  \displaystyle y''+ay'+by=\sum_{j=0}^{n}k_jx^j $$     (Eq.1.3) Therefore, the order of Eq. 1.3 is $$n$$ and the coefficients of the solution may be found using the below sum rule:
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$$  \displaystyle y(x)=\sum_{j=0}^s c_j x^j $$     (Eq.1.4) Where $$ s $$ in Eq. 1.4 is equal to the order of the Eq. 1.3 which was defined above to be $$n$$:
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$$  \displaystyle y(x)=\sum_{j=0}^n c_j x^j $$     (Eq.1.5) Eq. 1.5 may be substituted into Eq. 1.3 for y \, and below are the corresponding substitutions for y^\prime \, and :
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$$  \displaystyle y'(x)=\sum_{j=0}^n j c_j x^{(j-1)}=\sum_{j=1}^n j c_j x^{(j-1)} $$     (Eq.1.6)
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$$  \displaystyle y''(x)=\sum_{j=0}^n j(j-1) c_j x^{(j-2)}=\sum_{j=2}^n j(j-1) c_j x^{(j-2)} $$     (Eq.1.7) Substituting Eq. 1.5, Eq. 1.6, and Eq. 1.7 into Eq. 1.3 yields the below equation:
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$$  \displaystyle ( \sum_{j=2}^n j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=1}^n j c_j x^{(j-1)} )+b( \sum_{j=0}^n c_j x^j )=\sum_{j=0}^n k_jx^j $$     (Eq.1.8) To obtain equation for the coefficients of x^1 \,. The $$ j \,$$ values of Eq. 1.8 are adjusted for each summation such that the exponent of x \, will be equal to 1:
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$$  \displaystyle ( \sum_{j=3}^3 j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=2}^2 j c_j x^{(j-1)} )+b( \sum_{j=1}^1 c_j x^j )=\sum_{j=1}^1k_jx^j $$     (Eq.1.9) Which is further simplified below:
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$$  \displaystyle (3(3-1) c_3 x^{(3-2)} ) +a( 2 c_2 x^{(2-1)} )+b(c_1 x^1 )=k_1x^1 $$     (Eq.1.10)
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$$  \displaystyle (3)2c_3x^1+a(2c_2x^1)+bc_1x^1=k_1x $$     (Eq.1.11)
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$$  \displaystyle 6c_3x^1+2ac_2x^1+bc_1x^1=k_1x $$     (Eq.1.12)
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$$  \displaystyle 2(3c_3+ac_2+\frac{1}{2}bc_1)x^1=k_1x^1 $$     (Eq.1.13) And therefore the coefficients of x^1 \, are:
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$$  \displaystyle 2(3c_3+ac_2+\frac{1}{2}bc_1)=k_1 $$     (Eq.1.14) To obtain equation for the coefficients of x^2 \,. The $$ j \,$$ values of Eq. 1.8 are adjusted for each summation such that the exponent of x \, will be equal to 2:
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$$  \displaystyle ( \sum_{j=4}^4 j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=3}^3 j c_j x^{(j-1)} )+b( \sum_{j=2}^2 c_j x^j )=\sum_{j=2}^2 k_2x^2 $$     (Eq.1.15) Which is further simplified below:
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$$  \displaystyle ( 4(4-1) c_4 x^{(4-2)} ) +a( 3 c_3 x^{(3-1)} )+b(c_2 x^2 )=k_2x^2 $$     (Eq.1.16)
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$$  \displaystyle ( (4)3 c_4 x^{(4-2)} ) +a( 3 c_3 x^{(2)} )+bc_2 x^2=k_2x^2 $$     (Eq.1.17)
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$$  \displaystyle 12c_4 x^{(2)} +3a c_3 x^{(2)} +bc_2 x^2 =k_2x^2 $$     (Eq.1.18)
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$$  \displaystyle (12c_4 +3a c_3 +bc_2 )x^2 =k_2x^2 $$     (Eq.1.19) And therefore the coefficients of x^2 \, are:
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$$  \displaystyle 12c_4 +3a c_3 +bc_2=k_2 $$     (Eq.1.20) To obtain equation for the coefficients of x^3 \,. The $$ j \,$$ values of Eq. 1.8 are adjusted for each summation such that the exponent of x \, will be equal to 3:
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$$  \displaystyle ( \sum_{j=5}^{5} j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=4}^{4} j c_j x^{(j-1)} )+b( \sum_{j=3}^{3} c_j x^j )=\sum_{j=3}^{3} k_{n-2}x^{n-2} $$     (Eq.1.21) Which is further simplified below:
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$$  \displaystyle ( 5(5-1) c_{5} x^{(5-2)} ) +a( 4 c_{4} x^{(4-1)} )+b( c_{3} x^{3} )= k_{3}x^{3} $$     (Eq.1.22)
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$$  \displaystyle ( 5(4) c_{5} x^{(3)} ) +a( 4 c_{4} x^{3} )+b( c_{3} x^{3} )= k_{3}x^{3} $$     (Eq.1.23)
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$$  \displaystyle (20c_{5} +4ac_{4} +bc_{3} )x^{3}= k_{3}x^{3} $$     (Eq.1.24) And therefore the coefficients of x^3 \, are:
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$$  \displaystyle 20 c_{5} +4ac_{4} +bc_{3} = k_{3} $$     (Eq.1.25) To obtain equation for the coefficients of x^{n-2} \,. The $$ j \,$$ values of Eq. 1.8 are adjusted for each summation such that the exponent of x \, will be equal to n-2 \,:
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$$  \displaystyle ( \sum_{j=n}^{n} j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=n-1}^{n-1} j c_j x^{(j-1)} )+b( \sum_{j=n-2}^{n-2} c_j x^j )=\sum_{j=n-2}^{n-2} k_{n-2}x^{n-2} $$     (Eq.1.26) Which is further simplified below:
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$$  \displaystyle ( n(n-1) c_{n} x^{(n-2)} ) +a( (n-1) c_{n-1} x^{(n-1-1)} )+b( c_{n-2} x^{n-2} )= k_{n-2}x^{n-2} $$     (Eq.1.27)
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$$  \displaystyle ( n(n-1) c_{n} x^{(n-2)} ) +a( (n-1) c_{n-1} x^{(n-2)} )+b( c_{n-2} x^{n-2} )= k_{n-2}x^{n-2} $$     (Eq.1.28)
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$$  \displaystyle ((n-1)( n c_{n} +ac_{n-1} )+bc_{n-2} )x^{n-2}= k_{n-2}x^{n-2} $$     (Eq.1.29) And therefore the coefficients of x^{n-2} \, are:
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$$  \displaystyle ((n-1)( n c_{n} +ac_{n-1} )+bc_{n-2} )= k_{n-2} $$     (Eq.1.30) To obtain equation for the coefficients of x^{n-1} \,. The $$ j \,$$ values of Eq. 1.8 are adjusted for each summation such that the exponent of x \, will be equal to n-1 \,:
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$$  \displaystyle ( \sum_{j=n+1}^{n} j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=n}^{n} j c_j x^{(j-1)} )+b( \sum_{j=n-1}^{n-1} c_j x^j )=\sum_{j=n-1}^{n-1} k_j x^j $$     (Eq.1.31) Since, the bounds of a summation in Eq. 1.31 is exceeded in one case that summation is simplified to zero as shown below:
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$$  \displaystyle a( n c_n x^{(n-1)} )+b(c_{n-1} x^{n-1} )=k_{n-1} x^{n-1} $$     (Eq.1.32) Which is further simplified below:
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$$  \displaystyle (an c_n +bc_{n-1}) x^{n-1} =k_{n-1} x^{n-1} $$     (Eq.1.33) And therefore the coefficients of x^{n-1} \, are:
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$$  \displaystyle (an c_n +bc_{n-1}) =k_{n-1} $$     (Eq.1.34)
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To obtain equation for the coefficients of x^n \,. The $$ j \,$$ values of Eq. 1.8 are adjusted for each summation such that the exponent of x \, will be equal to n \,:
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$$  \displaystyle ( \sum_{j=n+2}^{n} j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=n+1}^{n} j c_j x^{(j-1)} )+b( \sum_{j=n}^{n} c_j x^j )=\sum_{j=n}^{n} k_j x^j $$     (Eq.1.35) Since, the bounds of the summations in Eq. 1.35 are exceeded in two cases those summation are simplified to zero as shown below:
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$$  \displaystyle b(c_n x^n) =k_n x^n $$     (Eq.1.36) Which is further simplified below:
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$$  \displaystyle (bc_n) x^n =k_n x^n $$     (Eq.1.37) And therefore the coefficients of x^n \, are:
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$$  \displaystyle bc_n=k_n $$     (Eq.1.38) The following equations Eq. 1.14, Eq. 1.20, Eq. 1.25, Eq. 1.30, Eq. 1.34, abd Eq. 1.38 as well as the given equation for the coefficients of x^0 \, can have the left hand side of the equations inserted into a tri-diagonal matrix as shown below:
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$$  \displaystyle \begin{bmatrix} b & a & 2 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & b & a & 6 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & b & 3a & 12 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & b & 4a & 20 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & b & a(n-1) & n(n-1)\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & b & an\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & b\\ \end{bmatrix} $$     (Eq.1.39)
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Author & Proofreaders
Author: Egm4313.s12.team17.axelrod.a 08:51, 11 March 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.hintz 21:30, 12 March 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.deaver.md 15:36, 14 March 2012 (UTC))

Proofreader 3: Egm4313.s12.team17.ying 10:10, 12 March 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 15:36, 14 March 2012 (UTC)

=Problem R4.2 - Taylor Series Expansion for the Excitations=

Part 1 Given
Given the equation:
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$$  \displaystyle y''-3y'+2y=r(x) $$     (Eq.2.1.1)
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And that the following relations are true for this problem:
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$$  \displaystyle r(x)=sin(x) $$     (Eq.2.1.2)
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$$  \displaystyle y(0)=1 $$     (Eq.2.1.3)
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$$  \displaystyle y'(0)=0 $$     (Eq.2.1.4)
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Part 1 Problem Statement
Reproduce the graph shown below where, using the Taylor series expansion for using the Taylor series expansion for r_n(x) \, shown below:

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$$  \displaystyle r_n(x):=\sum_{k=0}^ \frac{(-1)^kx^{2k+1}}{(2k+1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^n x^{2n+1}}{(2n+1)!} $$     (Eq.2.1.5)
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Part 1 Solution
Note: The n \, is for the highest power of x \,.
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$$  \displaystyle r_n(x)=\sum_{k=1}^undefined \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!} $$     (Eq.2.1.6)
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$$  \displaystyle r_n(x)=\sum_{k=1}^undefined \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.7)
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$$  \displaystyle r_n(x)=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.8) For n=1 \, :
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$$  \displaystyle r_1(x)=\sum_{k=1}^ \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.9)
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$$  \displaystyle r_1(x)=x $$     (Eq.2.1.10) For n=3 \, :
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$$  \displaystyle r_3(x)=\sum_{k=1}^ \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.11)
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$$  \displaystyle r_3(x)=x-\frac{x^3}{3!} $$     (Eq.2.1.12) For n=5 \, :
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$$  \displaystyle r_5(x)=\sum_{k=1}^ \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.13)
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$$  \displaystyle r_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!} $$     (Eq.2.1.14) For n=7 \, :
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$$  \displaystyle r_7(x)=\sum_{k=1}^ \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.15)
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$$  \displaystyle r_7(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} $$     (Eq.2.1.16) For n=9 \, :
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$$  \displaystyle r_9(x)=\sum_{k=1}^ \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.17)
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$$  \displaystyle r_9(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!} $$     (Eq.2.1.18) For n=11 \, :
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$$  \displaystyle r_{11}(x)=\sum_{k=1}^ \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.19)
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$$  \displaystyle r_{11}(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!} $$     (Eq.2.1.20) For n=13 \, :
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$$  \displaystyle r_{13}(x)=\sum_{k=1}^ \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^{t-1} x^{2t-1}}{(2t-1)!} $$     (Eq.2.1.21)
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$$  \displaystyle r_{13}(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!} $$     (Eq.2.1.22) The following are represented in the below graph using MatLab: Eq. 2.1.10, Eq. 2.1.12, Eq. 2.1.14, Eq. 2.1.16, Eq. 2.1.18, Eq. 2.1.20, Eq. 2.1.22.
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MATLAB code is shown below:

Part 2 Given

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$$  \displaystyle \sum_{k=0}^undefined \frac{(-1)^kx^{2k+1}}{(2k+1)!}=x-\frac{x^3}{3!} +...+ \frac{(-1)^t x^{2t+1}}{(2t+1)!} $$     (Eq.2.2.1)
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$$  \displaystyle y_{p,n}''+ay'_{p,n}+by_{p,n}=r_n(x) $$     (Eq.2.2.2)
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$$  \displaystyle y(0)=1 $$     (Eq.2.2.3)
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$$  \displaystyle y'(0)=0 $$     (Eq.2.2.4)
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Part 2 Problem Statement
Find y_n \, for n=3,5,9 \, and plot the solutions for.

Part 2 Problem Solution
From Part 1 we know that:
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$$  \displaystyle r_{3}(x)=x-\frac{x^3}{3!} $$     (Eq.2.1.12) Setting up the method of undetermined coefficients:
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$$  \displaystyle y''-3y'+2y=x-\frac{x^3}{3!} $$     (Eq.2.2.5) The particular solution will take the general form shown below:
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$$  \displaystyle y_{p}(x)= \sum_{j=0}^n c_{j}x^j $$     (Eq.2.2.6) This was solved in | Problem R3.4
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Solving for the particular solution of Eq. 2.2.5:
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$$  \displaystyle y_p=Ax^3+Bx^2+Cx+D $$     (Eq.2.2.7)
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$$  \displaystyle y_p'=3Ax^2+2Bx+C $$     (Eq.2.2.8)
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$$  \displaystyle y_p''=6Ax+2B $$     (Eq.2.2.9) For the x^3 \, terms:
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$$  \displaystyle 2A=-\frac{1}{6} $$     (Eq.2.2.10)
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$$  \displaystyle A=-\frac{1}{12} $$     (Eq.2.2.11) For the x^2 \, terms:
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$$  \displaystyle -9A+2B=0 $$     (Eq.2.2.12)
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$$  \displaystyle -9(-\frac{1}{12})+2B=0 $$     (Eq.2.2.13)
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$$  \displaystyle \frac{3}{4}+2B=0 $$     (Eq.2.2.14)
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$$  \displaystyle 2B=-\frac{3}{4} $$     (Eq.2.2.15)
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$$  \displaystyle B=-\frac{3}{8} $$     (Eq.2.2.16) For the x^1 \, terms:
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$$  \displaystyle 6A-6B+2C=1 $$     (Eq.2.2.17)
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$$  \displaystyle 6(-\frac{1}{12})-6(-\frac{3}{8})+2C=1 $$     (Eq.2.2.18)
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$$  \displaystyle -\frac{1}{2}+\frac{9}{4}+2C=1 $$     (Eq.2.2.19)
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$$  \displaystyle 2C=-\frac{3}{4} $$     (Eq.2.2.20)
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$$  \displaystyle C=-\frac{3}{8} $$     (Eq.2.2.21) For the x^0 \, terms:
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$$  \displaystyle 2B-3C+2D=0 $$     (Eq.2.2.22)
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$$  \displaystyle 2(-\frac{3}{8})-3(-\frac{3}{8})+2D=0 $$     (Eq.2.2.23)
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$$  \displaystyle -\frac{6}{8}+\frac{9}{8}+2D=0 $$     (Eq.2.2.24)
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$$  \displaystyle 2D=-\frac{3}{8} $$     (Eq.2.2.25)
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$$  \displaystyle D=-\frac{3}{16} $$     (Eq.2.2.26) So the particular solution for n=3 \, is:
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$$  \displaystyle y_{p,3}=-\frac{1}{12}x^3-\frac{3}{8}x^2-\frac{3}{8}x-\frac{3}{16} $$     (Eq.2.2.27) The homogeneous equation for n=3 \, is also the homogeneous equation for n=5 \, and n=9 \,:
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$$  \displaystyle y_{h,3}=C_1e^{\lambda_1}+C_2e^{\lambda_2} $$     (Eq.2.2.28) The and  are found using below:
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$$  \displaystyle y''-3y'+2y=0 $$     (Eq.2.2.29)
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$$  \displaystyle \lambda^2-3\lambda+2=0 $$     (Eq.2.2.30)
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$$  \displaystyle (\lambda-2)(\lambda-1)=0 $$     (Eq.2.2.31)
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$$  \displaystyle \lambda_1=2 $$     (Eq.2.2.32)
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$$  \displaystyle \lambda_2=1 $$     (Eq.2.2.33) So therefore the homogeneous equation is:
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$$  \displaystyle y_h=C_1e^{2x}+C_2e^x $$     (Eq.2.2.34) From part 1 we know that:
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$$  \displaystyle r_{5}(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!} $$     (Eq.2.1.14) Setting up the method of undetermined coefficients:
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$$  \displaystyle y''-3y'+2y=x-\frac{x^3}{3!}+\frac{x^5}{5!} $$     (Eq.2.2.35) Solving for the particular solution of Eq.2.2.35:
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$$  \displaystyle y_p=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F $$     (Eq.2.2.36)
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$$  \displaystyle y_p'=5Ax^4+4Bx^3+3Cx^2+2Dx+E $$     (Eq.2.2.37)
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$$  \displaystyle y_p''=20Ax^3+12Bx^2+6Cx+2D $$     (Eq.2.2.38) Which upon inserting Eq.2.2.36, Eq.2.2.37, and Eq.2.2.38 into Eq.2.2.35 yields the following:
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For the x^5 \, terms:
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$$  \displaystyle 2A=1\frac{1}{120} $$     (Eq.2.2.39)
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$$  \displaystyle A=\frac{1}{240} $$     (Eq.2.2.40) For the x^4 \, terms:
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$$  \displaystyle -15A+2B=0 $$     (Eq.2.2.41)
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$$  \displaystyle -15(\frac{1}{240})+2B=0 $$     (Eq.2.2.42)
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$$  \displaystyle -\frac{1}{16}+2B=0 $$     (Eq.2.2.43)
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$$  \displaystyle 2B=\frac{1}{16} $$     (Eq.2.2.44)
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$$  \displaystyle B=\frac{1}{32} $$     (Eq.2.2.45) For the x^3 \, terms:
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$$  \displaystyle 20A-12B+2C=-\frac{1}{6} $$     (Eq.2.2.46)
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$$  \displaystyle 20(\frac{1}{240})-6(\frac{1}{32})+2C=-\frac{1}{6} $$     (Eq.2.2.47)
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$$  \displaystyle \frac{1}{12}-\frac{3}{8}+2C=-\frac{1}{6} $$     (Eq.2.2.48)
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$$  \displaystyle 2C=\frac{1}{8} $$     (Eq.2.2.49)
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$$  \displaystyle C=\frac{1}{16} $$     (Eq.2.2.50) For the x^2 \, terms:
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$$  \displaystyle 12B-9C+2D=0 $$     (Eq.2.2.51)
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$$  \displaystyle 12(\frac{1}{32})-9(\frac{1}{16})+2D=0 $$     (Eq.2.2.52)
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$$  \displaystyle \frac{3}{8}-\frac{9}{16}+2D=0 $$     (Eq.2.2.53)
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$$  \displaystyle 2D=\frac{3}{16} $$     (Eq.2.2.54)
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$$  \displaystyle D=\frac{3}{32} $$     (Eq.2.2.55) For the x^1 \, terms:
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$$  \displaystyle 6C-6D+2E=1 $$     (Eq.2.2.56)
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$$  \displaystyle 6(\frac{1}{16})-6(\frac{3}{32})+2E=1 $$     (Eq.2.2.57)
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$$  \displaystyle \frac{3}{8}-\frac{9}{16}+2E=1 $$     (Eq.2.2.58)
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$$  \displaystyle -\frac{3}{16}+2E=1 $$     (Eq.2.2.59)
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$$  \displaystyle 2E=1+\frac{3}{16} $$     (Eq.2.2.60)
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$$  \displaystyle 2E=\frac{19}{16} $$     (Eq.2.2.61)
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$$  \displaystyle E=\frac{19}{32} $$     (Eq.2.2.62) For the x^0 \, terms:
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$$  \displaystyle 2D-3E+2F=0 $$     (Eq.2.2.63)
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$$  \displaystyle 2(\frac{3}{32})-3(\frac{19}{32})+2F=0 $$     (Eq.2.2.64)
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$$  \displaystyle \frac{3}{16}-\frac{57}{32}+2F=0 $$     (Eq.2.2.65)
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$$  \displaystyle 2F=\frac{51}{32} $$     (Eq.2.2.66)
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$$  \displaystyle F=\frac{51}{64} $$     (Eq.2.2.67) So the particular solution for n=5 \, is:
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$$  \displaystyle y_{p,5}=\frac{1}{240}x^5+\frac{1}{32}x^4+\frac{1}{16}x^3+\frac{3}{32}x^2+\frac{19}{32}x+\frac{51}{54} $$     (Eq.2.2.68) From Part 1 we know that:
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$$  \displaystyle r_{9}(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!} $$     (Eq.2.1.18) Setting up the method of undetermined coefficients:
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$$  \displaystyle y''-3y'+2y=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!} $$     (Eq.2.2.69) Solving for the particular solution for Eq.2.2.69:
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$$  \displaystyle y_p=Ax^9+Bx^8+Cx^7+Dx^6+Ex^5+Fx^4+Gx^3+Hx^2+Ix+J $$     (Eq.2.2.70)
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$$  \displaystyle y_p'=9Ax^8+8Bx^7+7Cx^6+6Dx^5+5Ex^4+4Fx^3+3Gx^2+2Hx+I $$     (Eq.2.2.71)
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$$  \displaystyle y_p''=72Ax^7+56Bx^6+42Cx^5+30Dx^4+20Ex^3+12Fx^2+6Gx+2H $$     (Eq.2.2.72) Inserting Eq. 2.2.70, Eq. 2.2.71, and Eq. 2.2.72 into Eq. 2.2.69 yields:
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For the x^9 \, terms:
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$$  \displaystyle A=1\frac{1}{2(9!)} $$     (Eq.2.2.73) For the x^8 \, terms:
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$$  \displaystyle 2B-27A=0 $$     (Eq.2.2.74)
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$$  \displaystyle 2B-\frac{27}{2(9!)}=0 $$     (Eq.2.2.75)
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$$  \displaystyle B=\frac{27}{4(9!)} $$     (Eq.2.2.76) For the x^7 \, terms:
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$$  \displaystyle 2C-24B+72A=-\frac{1}{7!} $$     (Eq.2.2.77)
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$$  \displaystyle 2C-24(\frac{27}{4(9!)})+\frac{72}{2(9!)}=-\frac{1}{7!} $$     (Eq.2.2.78)
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$$  \displaystyle 2C-\frac{1}{2880}=-\frac{1}{7!} $$     (Eq.2.2.79)
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$$  \displaystyle C=\frac{1}{1344} $$     (Eq.2.2.80) For the x^6 \, terms:
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$$  \displaystyle 2D+56B-21C=0 $$     (Eq.2.2.81)
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$$  \displaystyle 2D+56(\frac{27}{4(9!)})-\frac{21}{13440}=0 $$     (Eq.2.2.81)
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$$  \displaystyle D=\frac{1}{3840} $$     (Eq.2.2.82) For the x^5 \, terms:
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$$  \displaystyle 2E-18D+42C=\frac{1}{120} $$     (Eq.2.2.83)
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$$  \displaystyle 2E-\frac{18}{3840}+\frac{42}{13440}=\frac{1}{120} $$     (Eq.2.2.84)
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$$  \displaystyle E-\frac{1}{640}=\frac{1}{120} $$     (Eq.2.2.85)
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$$  \displaystyle E=\frac{19}{3840} $$     (Eq.2.2.86) For the x^4 \, terms:
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$$  \displaystyle 2F-15E+30D=0 $$     (Eq.2.2.87)
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$$  \displaystyle 2F-\frac{15(19)}{3840}\frac{30}{3840}=0 $$     (Eq.2.2.88)
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$$  \displaystyle F=\frac{17}{512} $$     (Eq.2.2.89) For the x^3 \, terms:
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$$  \displaystyle 2G-12F+20E=\frac{1}{6} $$     (Eq.2.2.90)
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$$  \displaystyle 2G-\frac{12(17)}{512}+\frac{20(19)}{3840}=-\frac{1}{6} $$     (Eq.2.2.91)
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$$  \displaystyle G=\frac{12}{256} $$     (Eq.2.2.92) For the x^2 \, terms:
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$$  \displaystyle 2H-9G+12F=0 $$     (Eq.2.2.93)
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$$  \displaystyle 2H-\frac{9(17)}{256}+\frac{12(17)}{512}=0 $$     (Eq.2.2.94)
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$$  \displaystyle H=\frac{51}{512} $$     (Eq.2.2.95) For the x^1 \, terms:
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$$  \displaystyle 2I-6H+6G=1 $$     (Eq.2.2.96)
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$$  \displaystyle 2I-\frac{6(51)}{512}+\frac{6(17)}{256}=1 $$     (Eq.2.2.97)
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$$  \displaystyle I=\frac{307}{512} $$     (Eq.2.2.98) For the x^0 \, terms:
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$$  \displaystyle 2J-3I+2H=0 $$     (Eq.2.2.99)
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$$  \displaystyle 2J-\frac{3(307)}{512}+\frac{2(51)}{512}=0 $$     (Eq.2.2.100)
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$$  \displaystyle J=\frac{819}{1024} $$     (Eq.2.2.101) So the particular solution for n=9 \, is:
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$$  \displaystyle y_{9}=\frac{x^9}{2(9!)}+\frac{27x^8}{4(9!)}+\frac{x^7}{13440}+\frac{x^6}{3840}+\frac{19x^5}{3840}+\frac{17x^4}{512}+\frac{17x^3}{256}+\frac{51x^2}{512}+\frac{307x}{512}+\frac{819}{1024} $$     (Eq.2.2.102) The overall solution for n=3 \, is:
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$$  \displaystyle y_{3}=C_1e^{2x}+C_2e^x-\frac{x^3}{12}-\frac{3x^2}{8}-\frac{3x}{8}-\frac{3}{16} $$     (Eq.2.2.103) The overall solution for n=5 \, is:
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$$  \displaystyle y_{5}=C_1e^{2x}+C_2e^x+\frac{x^5}{240}+\frac{x^4}{32}+\frac{x^3}{16}+\frac{3x^2}{32}+\frac{19x}{32}+\frac{51}{64} $$     (Eq.2.2.104) The overall solution for n=9 \, is:
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$$  \displaystyle y_{9}=C_1e^{2x}+C_2e^x+\frac{x^9}{2(9!)}+\frac{27x^8}{4(9!)}+\frac{x^7}{13440}+\frac{x^6}{3840}+\frac{19x^5}{3840}+\frac{17x^4}{512}+\frac{17x^3}{256}+\frac{51x^2}{512}+\frac{307x}{512}+\frac{819}{1024} $$     (Eq.2.2.105)
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Use the initial conditions to solve for the final solutions.

For n=3 \,:
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$$  \displaystyle 1=C_1+C_2-\frac{3}{16} $$     (Eq.2.2.106)
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$$  \displaystyle 0=2C_1+C_2-\frac{3}{8} $$     (Eq.2.2.107)
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$$  \displaystyle C_1=-\frac{13}{16} $$     (Eq.2.2.108)
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$$  \displaystyle C_2=2 $$     (Eq.2.2.109) The final solution for n=3 \, is:
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$$  \displaystyle y_{3}=-\frac{13}{16}e^{2x}+2e^x-\frac{x^3}{12}-\frac{3x^2}{8}-\frac{3x}{8}-\frac{3}{16} $$     (Eq.2.2.110)
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For n=5 \,:
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$$  \displaystyle 1=C_1+C_2+\frac{51}{64} $$     (Eq.2.2.111)
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$$  \displaystyle 0=2C_1+C_2+\frac{19}{32} $$     (Eq.2.2.112)
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$$  \displaystyle C_1=-\frac{51}{64} $$     (Eq.2.2.113)
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$$  \displaystyle C_2=1 $$     (Eq.2.2.114) The final solution for n=5 \, is:
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$$  \displaystyle y_{5}=-\frac{51}{64}e^{2x}+e^x+\frac{x^5}{240}+\frac{x^4}{32}+\frac{x^3}{16}+\frac{3x^2}{32}+\frac{19x}{32}+\frac{51}{64} $$     (Eq.2.2.115)
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For n=9 \,:
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$$  \displaystyle 1=C_1+C_2+\frac{819}{1024} $$     (Eq.2.2.116)
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$$  \displaystyle 0=2C_1+C_2+\frac{307}{512} $$     (Eq.2.2.117)
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$$  \displaystyle C_1=-\frac{819}{1024} $$     (Eq.2.2.118)
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$$  \displaystyle C_2=1 $$     (Eq.2.2.119) The final solution for n=9 \, is:
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$$  \displaystyle y_{5}=-\frac{819}{1024}e^{2x}+e^{x}+\frac{x^9}{2(9!)}+\frac{27x^8}{4(9!)}+\frac{x^7}{13440}+\frac{x^6}{3840}+\frac{19x^5}{3840}+\frac{17x^4}{512}+\frac{17x^3}{256}+\frac{51x^2}{512}+\frac{307x}{512}+\frac{819}{1024} $$     (Eq.2.2.120)
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Part 3 Given
Refer to Part 1 Given and Part 2 Given.

Part 3 Problem Statement
Find the exact overall solution of $$y(x)$$ and plot it. Then compare that graph to the graphs in Part 2.

Part 3 Solution
The general solution for the particular take the form shown below:
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$$  \displaystyle y_p=K\cos(\omega x)+M\sin(\omega x) $$ (Eq.2.3.1) Therefore, the particular solution is:
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$$  \displaystyle y_p=K\cos(x)+M\sin(x) $$     (Eq.2.3.2) Take the first and second derivative of Eq. 2.3.2:
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$$  \displaystyle y_p^\prime=-K\sin(x)+M\cos(x) $$     (Eq.2.3.3)
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$$  \displaystyle y_p^{\prime\prime}=-K\cos(x)-M\sin(x) $$     (Eq.2.3.4) Then substitute Eq. 2.3.2, 2.3.3, and 2.3.4 into the equation below:
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$$  \displaystyle y_p^{\prime\prime}-3y_p^\prime+2y_p=\sin(x) $$     (Eq.2.3.5)
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$$  \displaystyle -K\cos(x)-M\sin(x)+3K\sin(x)-3M\cos(x)+2K\cos(x)+2M\sin(x)=\sin(x) $$     (Eq.2.3.6)
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$$  \displaystyle K\cos(x)+M\sin(x)+3K\sin(x)-3M\cos(x)=\sin(x) $$     (Eq.2.3.7) Solve for M and K by like terms (ie. all the \sin(x) terms from each side of Eq. 2.3.7):
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$$  \displaystyle K-3M=0 $$     (Eq.2.3.8)
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$$  \displaystyle M+3K=1 $$     (Eq.2.3.9)
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$$  \displaystyle K=\frac{3}{10} $$     (Eq.2.3.10)
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$$  \displaystyle M=\frac{1}{10} $$     (Eq.2.3.11) Therefore, the final particular solution is:
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$$  \displaystyle y_p=\frac{3}{10}\cos(x)+\frac{1}{10}\sin(x) $$     (Eq.2.3.12) The homogeneous solution is Eq. 2.2.34.
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Using superimposing the homogeneous and particular solution results in:
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$$  \displaystyle y=\frac{3}{10}\cos(x)+\frac{1}{10}\sin(x)+C_1e^{2x}+C_2e^x $$     (Eq.2.3.13) Now, using the initial conditions from the previous part to find the final solution:
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$$  \displaystyle 1=\frac{3}{10}+C_1+C_2 $$     (Eq.2.3.14)
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$$  \displaystyle 0=\frac{1}{10}+2C_1+C_2 $$     (Eq.2.3.15)
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$$  \displaystyle C_1=-\frac{8}{10} $$     (Eq.2.3.16)
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$$  \displaystyle C_2=\frac{15}{10} $$     (Eq.2.3.17) Therefore:
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$$  \displaystyle y=\frac{3}{10}\cos(x)+\frac{1}{10}\sin(x)-\frac{8}{10}e^{2x}+\frac{15}{10}e^x $$     (Eq.2.3.13)
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The graphs are almost identical to one another.

Author & Proofreaders
Author 1: Egm4313.s12.team17.axelrod.a 20:41, 11 March 2012 (UTC)

Author 2: Egm4313.s12.team17.deaver.md 12:18, 14 March 2012 (UTC))

Proofreader 1: Egm4313.s12.team17.hintz 13:30, 14 March 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.Li 14:12, 14 March 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 12:18, 14 March 2012 (UTC)

=Problem R4.3 - Taylor Series Expansion for the Excitations=

Given

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$$  \displaystyle y'' - 3y' + 2y = r(x) $$     (Eq.3.1)
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$$  \displaystyle r(x) = log(1+x) $$     (Eq.3.2)
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$$  \displaystyle y(-3/4) = 1, y'(-3/4) = 0 $$     (Eq.3.3)
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Part 1 Problem Statement
Develop in Taylor series about x=0 \, to reproduce the figure below.

Part 1 Solution
The general function for a Taylor series function about \hat x \,:
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$$  \displaystyle f(x) = \sum_{n=0}^\infty \frac{f^{n} (\hat x)}{n!} (x - \hat x)^n $$     (Eq.3.1.1)
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Therefore, the Taylor series for the function can be approximated using equation 3.1.1 about :
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$$  \displaystyle r(x) = \log(1+x) = \sum_{n=0}^\infty \frac{f^{0}(0)}{n!}x^n = \frac{f^{0}(0)}{0!}x^0 + \frac{f^{1}(0)}{1!}x^1 + \frac{f^{2}(0)}{2!}x^2 + \frac{f^{3}(0)}{3!}x^3 + \frac{f^{4}(0)}{4!}x^4 + \frac{f^{5}(0)}{5!}x^5 + \ldots $$     (Eq.3.1.2)
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The derivative of is defined as:
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$$  \displaystyle \frac{d}{dx}\log_b(x) = \frac{1}{x\ln(b)} $$     (Eq.3.1.3)
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Therefore the derivative of is:
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$$  \displaystyle \frac{d}{dx}log(1+x) = \frac{1}{(x+1)\ln(10)} $$     (Eq.3.1.4)
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Using Eq. 3.1.4 in Eq. 3.1.2 will then give
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$$  \displaystyle r(x) = \log(1+x) = \sum_{n=0}^\infty \frac{f^{0}(0)}{n!}x^0 = \frac{1}{0!}\frac{1}{\ln(10)}x^0 + \frac{1}{1!}\frac{1}{\ln(10)}x^1 - \frac{1}{2!}\frac{1}{\ln(10)}x^2 + \frac{1}{3!}\frac{2}{\ln(10)}x^3 - \frac{1}{4!}\frac{6}{\ln(10)}x^4 + \frac{1}{5!}\frac{24}{\ln(10)}x^5 + \frac{1}{6!}\frac{120}{\ln(10)}x^6 $$     (Eq.3.1.5)
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Taking out the common log term (which simplifies to 1) and simplifying Eq. 3.1.5 will give:
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$$  \displaystyle r(x) =\log(1+x) = \frac{1}{\ln(10)} \sum_{n=0}^\infty \frac{f^{0}(0)}{n!}x^n = \frac{1}{\ln10} (0 + \frac{1}{1}x^1 - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 - \frac{1}{6}x^6 + \cdots) $$     (Eq.3.1.6)
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The Taylor series equation for the excitation is then:
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$$  \displaystyle r(x) = \log(1+x) = \frac{1}{\ln(10)} \sum_{n=1}^\infty \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.1.7)
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Using the Taylor function in MATLAB, the from above can be reproduced.

Below is the MATLAB code used to generate the above graph:

Part 2 Problem Statement
Let r_n(x) \, be the truncated Taylor series, with n \, terms, which is also the highest degree of the Taylor (power) series, of.

Find y_n(x) \, for that satisfies Eq. 3.1 with the use of Eq. 3.3.

Plot y_n(x) \, for for.

Part 2 Solution
Using Eq.3.1.7, 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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$$  \displaystyle r_n(x) = \log(1+x) = \frac{1}{\ln(10)} \sum_{n=1}^n \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.1) Where n=4\,:
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$$  \displaystyle r_4(x) = \frac{1}{\ln(10)} \sum_{n=1}^4 \frac{x^n}{n}(-1)^{n-1} = \frac{1}{\ln(10)}(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}) $$     (Eq.3.2.2) Thus, the particular solution to the Eq. 3.2.2 in the Part 2 given is then:
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$$  \displaystyle y_{4,p} = \sum_{i=0}^4 c_i x_i = c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.3)
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Deriving Eq. 3.2.3 twice will yield the following:

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

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$$  \displaystyle y''_{4,p} - 3y'_{n,p} + 2y_{4,p} = 12c_4x^2 + 6c_3x + 2c_2 - 3(4c_4x^3 + 3c_3x^2 + 2c_2x + c_1) + 2(c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0) $$     (Eq.3.2.6)
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Combining like terms, Eq. 3.2.6 simplifies to:
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$$  \displaystyle (2c_4x^4) + (2c_3x^3 - 12c_4x^3) + (-9c_3x^2 + 2c_2x^2 + 12c_4x^2) + (2c_1x - 6c_2x + 6c_3x) + (2c_2 + c_0 - 3c_1) = \frac{1}{\ln(10)} (x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}) $$     (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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$$  \displaystyle c_0 = -1.7643 $$
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$$  \displaystyle c_1 = -1.7915 $$
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$$  \displaystyle c_2 = -0.9229 $$
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$$  \displaystyle c_3 = -0.2533 $$
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$$  \displaystyle c_4 = -0.0543 $$
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Finally the particular solution has been reached. The solution $$y_4(x)\,$$ is then:

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$$  \displaystyle y_{4} = y_{4,h}(x) -0.0543x^4 -0.2533x^3 -0.9229x^2 -1.7915x -1.7643 $$     (Eq.3.2.8)
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Where $$y_{n,h}(x)\,$$ is solved by setting Eq. 3.1 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 $$y_4(x)\,$$:

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$$  \displaystyle y_{4} = c_6e^{2x} + c_5e^x -0.0543x^4 -0.2533x^3 -0.9229x^2 -1.7915x -1.7643 $$     (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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$$  \displaystyle c_6 = -30.511 $$
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$$  \displaystyle c_5 = 23.777 $$
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$$  \displaystyle y_{4} = -30.511e^{2x} + 23.777e^x -0.0543x^4 -0.2533x^3 -0.9229x^2 -1.7915x -1.7643 $$     (Eq.3.2.15)
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$$  \displaystyle r_n(x) = \log(1+x) = \frac{1}{\ln(10)} \sum_{n=1}^n \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.16) Where n=7 \,:
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$$  \displaystyle r_7(x) = \frac{1}{\ln(10)} \sum_{n=1}^7 \frac{x^n}{n}(-1)^{n-1} = \frac{1}{10}(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7}) $$     (Eq.3.2.17)
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$$  \displaystyle y_{7,p} = \sum_{i=0}^7 c_i x_i = c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.18)
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Deriving Eq. 3.2.18 twice will yield the following:

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$$  \displaystyle y'_{7,p} = 7c_7x^6 + 6c_6x^5 + 5c_5x^4 + 4c_4x^3 + 3c_3x^2 + 2c_2x + c_1 $$     (Eq.3.2.19)
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$$  \displaystyle y''_{7,p} = 42c_7x^5 + 30c_6x^4 + 20c_5x^3 + 12c_4x^2 + 6c_3x + 2c_2 $$     (Eq.3.2.20)
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To solve for the particular solution y_7(x) \,, Eq. 3.2.19 and Eq. 3.2.20 will be substituted into the following particular equation:

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$$  \displaystyle y''_{7,p} - 3y'_{7,p} + 2y_{7,p} = $$
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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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$$  \displaystyle 2c_7x^7 + (2c_6x^6 -21c_7x^6) + (2c_5x^5 - 18c_6x^5 + 42c_7x^5) + (2c_4x^4- 15c_5x^4+30c_6x^4 ) + (2c_3x^3 - 12c_4x^3 + 20c_5x^3) + (2c_2x^2 - 9c_3x^2 + 12c_4x^2) $$
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$$  \displaystyle + (2c_1x - 6c_2x + 6c_3x) + (2c_0 - 3c_1 + 2c_2) = \frac{1}{\ln(10)}(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7}) $$     (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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$$  \displaystyle c_0 = 268.25 $$
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$$  \displaystyle c_1 = 267.25 $$
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$$  \displaystyle c_2 = 132.62 $$
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$$  \displaystyle c_3 = 43.610 $$
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$$  \displaystyle c_4 = 10.59 $$
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$$  \displaystyle c_5 = 1.998 $$
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$$  \displaystyle c_6 = 0.2895 $$
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$$  \displaystyle c_7 = 0.03102 $$
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Finally the particular solution has been reached. The solution y_7(x) \, is then:

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$$  \displaystyle y_{7} = y_{n,h}(x) + 0.03102x^7 + 0.2895x^6 + 1.998x^5 + 10.59x^4 + 43.610x^3 + 132.62x^2 + 267.25x + 268.25 $$     (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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$$  \displaystyle y_{7} = c_9e^{2x} + c_8e^x + 0.03102x^7 + 0.2895x^6 + 1.998x^5 + 10.59x^4 + 43.610x^3 + 132.62x^2 + 267.25x + 268.25 $$     (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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$$  \displaystyle c_9 = -4.002 $$
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$$  \displaystyle c_8 = -264.73 $$
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The final solution is then:
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$$  \displaystyle y_{7}(x) = -4.002e^{2x} -264.73e^x + 0.03102x^7 + 0.2895x^6 + 1.998x^5 + 10.59x^4 + 43.610x^3 + 132.62x^2 + 267.25x + 268.25 $$     (Eq.3.2.26)
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$$  \displaystyle r_{11}(x) = log(1+x) = \frac{1}{\ln(10)}\sum_{n=1}^{11} \frac{x^n}{n}(-1)^{n-1} $$     (Eq.3.2.27) Where n=11 \,:
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$$  \displaystyle r_{11}(x) = \frac{1}{\ln(10)} \sum_{n=1}^{11} \frac{x^n}{n}(-1)^{n-1} =\frac{1}{\ln(10)}(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7} - \frac{x^8}{8} + \frac{x^9}{9} - \frac{x^{10}}{10} + \frac{x^{11}}{11}) $$     (Eq.3.2.28)
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$$  \displaystyle y_{11,p} = \sum_{i=0}^{11} c_i x_i = c_{11}x^{11}+c_{10}x^{10}+c_9x^9+c_8x^8 + c_7x^7 + c_6x^6 + c_5x^5 + c_4x^4 + c_3x^3 + c_2x^2 + c_1x + c_0 $$     (Eq.3.2.29)
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Deriving Eq.3.2.28 twice will yield the following:

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

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$$  \displaystyle y''_{11,p} - 3y'_{11,p} + 2y_{11,p} = $$
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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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$$  \displaystyle 2c_{11}x^{11} + (2c_{10}x^{10}-33c_{11}x^{10}) + (2c_9x^9 - 30c_{10}x^9 + 110c_{11}x^9) + (2c_8x^8 - 27c_9x^8 + 90c_{10}x^8) $$
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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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$$  \displaystyle c_0 = 1.433640(10^6) $$
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$$  \displaystyle c_1 = 1.433318(10^6) $$
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$$  \displaystyle c_2 = 716337.8315 $$
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$$  \displaystyle c_3 = 238564.9257 $$
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$$  \displaystyle c_4 = 59534.0376 $$
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$$  \displaystyle c_5 = 11863.9372 $$
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$$  \displaystyle c_6 = 1963.0291 $$
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$$  \displaystyle c_7 = 276.3509 $$
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$$  \displaystyle c_8 = 33.5221 $$
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$$  \displaystyle c_9 = 3.4985 $$
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$$  \displaystyle c_{10} = 0.304006137 $$
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$$  \displaystyle c_{11} = 0.019740658 $$
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Finally the particular solution has been reached. The solution y_{11} \, is then:

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$$  \displaystyle y_{11} = y_{11,h}(x) +0.019740658x^{11}+0.304006137x^{10}+3.4985x^9+33.5221x^8+276.3509x^7 + $$ $$  \displaystyle 1963.0291x^6 +11863.9372x^5+59534.0376x^4+238564.9257x^3 +716337.8315x^2 +1433318.934x +1433640.5704 $$     (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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$$  \displaystyle y_{11}= c_{13}e^{2x} + c_{12}e^x +0.019740658x^{11}+0.304006137x^{10}+3.4985x^9+33.5221x^8+276.3509x^7 $$ $$  \displaystyle +1963.0291x^6 +11863.9372x^5+59534.0376x^4+238564.9257x^3 +716337.8315x^2 +1433318.934x +1433640.5704 $$     (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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$$  \displaystyle c_{13} = 315.5842 $$
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$$  \displaystyle c_{12} = -1433957.1455 $$
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The final solution is then:
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$$  \displaystyle y_{11}(x) = 315.5842e^{2x} -1433957.1455e^x +0.019740658x^{11}+0.304006137x^{10}+3.4985x^9+33.5221x^8+276.3509x^7 $$
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$$  \displaystyle +1963.0291x^6 +11863.9372x^5+59534.0376x^4+238564.9257x^3 +716337.8315x^2 +1433318.934x +1433640.5704 $$     (Eq.3.2.37)
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Below are the three corresponding graphs to the n=4,7,11

Part 3 Problem Statement
3. Use the matlab code ode45 to integrate numerically (5) p.7b-7 with (1)-(2) p.7-28 to obtain the numerical soln for y(x). Plot $$y_n(x)$$ in the same plot as y(x)

Part 3 Solution
M-file code used to set up the ODE

Command Window code to use the ode45 function and to plot the results

This is the graph from the above matlab code (ode45 function is green line)

The picture shows us that the solution using ode45 is exactly the same as the taylor series of log(1+x) where n = 7

Author & Proofreaders
Author:Egm4313.s12.team17.ying 09:13, 12 March 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.wheeler.tw 23:15, 12 March 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.Li 07:15, 13 March 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 14:05, 14 March 2012 (UTC)

=Problem R4.4 - Extend the Accuracy of the Solution beyond the Initial Point=

Given

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$$  \displaystyle \log(1+x) $$     (Eq.4.1)
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$$  \displaystyle x_1 = 0.9 $$     (Eq.4.1)
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Part 1 Statement
Extend the accuracy of the solution beyond \hat x=1 \,.

Consider x_1=0.9 \, and find the value n \, for, so that it doesn't differ from the numerical solution by more than 10^{-5} \,.

Part 2 Statement
2. Develop log(1+x) in taylor series about x^hat = 1 for n = 4,7,11 and plot these truncated series vs the exact function.

What is now the domain convergence? (by observation of you results.)

Part 2 Solution
The general function for a Taylor series function about
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$$  \displaystyle f(x) = \sum_{n=0}^\infty \frac{f^{n} (\hat x)}{n!} (x - \hat x)^n $$     (Eq.4.4.1)
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Therefore, the Taylor series for the function r(x) = log(1+x) can be approximated using equation 3.1.1 about $$\hat x = 1$$
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$$  \displaystyle r(x) = \log(1+x) = \sum_{n=0}^\infty \frac{f^{n}(1)}{n!}(x-1)^n = \frac{f^{0}(1)}{0!}(x-1)^0 + \frac{f^{1}(1)}{1!}(x-1)^1 + \frac{f^{2}(1)}{2!}(x-1)^2 + \frac{f^{3}(1)}{3!}(x-1)^3 + \frac{f^{4}(1)}{4!}(x-1)^4 + \frac{f^{5}(1)}{5!}(x-1)^5 + \ldots $$     (Eq.4.4.2)
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The derivative of $$\log_b(x)$$ is defined as
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$$  \displaystyle \frac{d}{dx}\log_b(x) = \frac{1}{(x)\ln(b)} $$     (Eq.4.4.3)
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Therefore the derivative of $$\log(1+x)$$ is

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$$  \displaystyle \frac{d}{dx}\log(1+x) = \frac{1}{(1+x)\ln(10)} $$     (Eq.4.4.4)
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Using Eq.4.4.4 in Eq.4.4.2 will then give
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$$  \displaystyle r(x) = \sum_{n=0}^\infty \frac{f^{n}(1)}{n!}(x-1)^n = \frac{1}{(2)\ln(10)}\frac{1}{0!}(x-1)^0 + \frac{1}{(2)\ln(10)}\frac{1}{1!}(x-1)^1 - \frac{1}{(2)\ln(10)}\frac{1}{2!}(x-1)^2 + \frac{1}{(2)\ln(10)}\frac{1}{3!}(x-1)^3 - \frac{1}{(2)\ln(10)}\frac{1}{4!}(x-1)^4 + \cdots $$     (Eq.4.4.5)
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Simplifying Eq. 4.4.5 will give:
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$$  \displaystyle r(x) = \frac{1}{2\ln(10)} \sum_{n=0}^\infty \frac{f^{0}(0)}{n!}(x-1)^n = \frac{1}{2\ln(10)}\Bigg(0 + \frac{1}{1!}(x-1)^1 - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 + \cdots\Bigg) $$     (Eq.4.4.6)
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The Taylor series equation for the excitation is then:
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$$  \displaystyle r(x) = \log(1+x) = \frac{1}{2\ln(10)} \sum_{n=0}^n \frac{(x-1)^n}{n!}(-1)^{n-1} $$     (Eq.4.4.7)
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For n = 4 the Taylor Series has already been solved above in Eq.4.4.6
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$$  \displaystyle \sum_{n=0}^4 \frac{(x-1)^n}{n!}(-1)^{n-1}= \frac{1}{2\ln(10)}\Bigg(\frac{1}{1!}(x-1)^1 - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4\Bigg) $$     (Eq.4.4.8.a)
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For n = 7 the Taylor Series Expansion is
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$$  \displaystyle \sum_{n=0}^4 \frac{(x-1)^n}{n!}(-1)^{n-1}= \frac{1}{2\ln(10)}\Bigg(\frac{1}{1!}(x-1)^1 - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 -\frac{1}{6}(x-1)^6+ \frac{1}{7}(x-1)^7\Bigg) $$     (Eq.4.4.8.b)
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And Finally for n=11 the Taylor Series Expansion is
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$$  \displaystyle \sum_{n=0}^4 \frac{(x-1)^n}{n!}(-1)^{n-1} = \frac{1}{2\ln(10)}\Bigg(\frac{1}{1!}(x-1)^1 - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 $$ $$  \displaystyle    -\frac{1}{6}(x-1)^6+ \frac{1}{7}(x-1)^7 - \frac{1}{8}(x-1)^8 + \frac{1}{9}(x-1)^9 -\frac{1}{10}(x-1)^{10} + \frac{1}{11}(x-1)^{11}\Bigg) $$     (Eq.4.4.8.c)
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Here is the graph of the taylor series of log(1+x) about center point 1 compared to the exact function:

The domain of convergence based on observation is [-0.5 1.5]

Part 3 Statement
Find $$y_n(x)$$, for n = 4,7,11 such that:

$$y''_{n} + ay'_{n} + by_{n} = r_n(x)$$

for x in [0.9,3] with the initial conditions found i.e

$$y_n(x_1), y'_n(x_1)$$

plot $$y_n(x)$$ for n = 4,7,11 for x in [0.9,3]

Part 4 Statement
4. Use the matlab command ode45 to integrate numerically (5) p.7b-7 ($$ y'' - 3y' + 2y$$) with (1) p.7-28 ($$log(1+x)$$) and the initial conditions $$y_n(x_1), y'_n(x_1)$$ to obtain the numerical soln for $$y(x)$$.

plot $$y_n(x)$$ in the same figure $$y_n(x)$$

Part 4 Solution
M-file code used to set up the ODE

Command Window code to use the ode45 function and to plot the results

Author & Proofreaders
Author :Egm4313.s12.team17.ying 14:10, 14 March 2012 (UTC)

Proofreader: Egm4313.s12.team17.wheeler.tw 14:12, 14 March 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 15:22, 14 March 2012 (UTC)

= Contributing Team Members =

title: User:Egm4313.s12.team17/Report 4


text:
Dominant group/Atmospheric sciences

Atmospheric sciences is an umbrella term for the study of the Earth's and other astronomical objects' atmospheres, its processes, the effects other systems have on the atmosphere, and the effects of the atmosphere on these other systems.

Any dominant group in the atmospheric sciences is expected to be either an entity, source, or object of these sciences or a group in some way associated with these sciences.

Notations
Notation: let the symbol VOC indicate Volatile Organic Compounds.

Sciences
Def. the disciplines or branches of learning, especially those "dealing with measurable or systematic principles ... the collective [disciplines] of study or learning acquired through the scientific method" are called the sciences.

Def. in the definite "[a] method of discovering knowledge about the natural world based in making falsifiable predictions (hypotheses), testing them empirically, & developing theories that match known data from repeatable physical experimentation" is called the scientific method.

Def. "[b]eing in a position of power, authority or ascendancy over others" is called dominance, or "[t]he state of being dominant; of prime importance".

Dominant group
Examples from primary sources are to be used to prove or disprove each hypothesis. These can be collected per subject or in general.

 * Accident hypothesis: dominant group is an accident of whatever processes are operating.
 * Artifact hypothesis: dominant group may be an artifact of human endeavor or may have preceded humanity.
 * Association hypothesis: dominant group is associated in some way with the original research.
 * Bad group hypothesis: dominant group is the group that engages in discrimination, abuse, punishment, and additional criminal activity against other groups. It often has an unfair advantage and uses it to express monopolistic practices.
 * Control group hypothesis: there is a control group that can be used to study dominant group.
 * Entity hypothesis: dominant group is an entity within each field where a primary author of original research uses the term.
 * Evolution hypothesis: dominant group is a product of evolutionary processes, such groups are the evolutionary process, produce evolutionary processes, or are independent of evolutionary processes.
 * Identifier hypothesis: dominant group is an identifier used by primary source authors of original research to identify an observation in the process of analysis.
 * Importance hypothesis: dominant group signifies original research results that usually need to be explained by theory and interpretation of experiments.
 * Indicator hypothesis: dominant group may be an indicator of something as yet not understood by the primary author of original research.
 * Influence hypothesis: dominant group is included in a primary source article containing original research to indicate influence or an influential phenomenon.
 * Interest hypothesis: dominant group is a theoretical entity used by scholarly authors of primary sources for phenomena of interest.
 * Metadefinition hypothesis: all uses of dominant group by all primary source authors of original research are included in the metadefinition for dominant group.
 * Null hypothesis: there is no significant or special meaning of dominant group in any sentence or figure caption in any refereed journal article.
 * Object hypothesis: dominant group is an object within each field where a primary author of original research uses the term.
 * Obvious hypothesis: the only meaning of dominant group is the one found in Mosby's Medical Dictionary.
 * Original research hypothesis: dominant group is included in a primary source article by the author to indicate that the article contains original research.
 * Primordial hypothesis: dominant group is a primordial concept inherent to humans such that every language or other form of communication no matter how old or whether extinct, on the verge of extinction, or not, has at least a synonym for dominant group.
 * Purpose hypothesis: dominant group is written into articles by authors for a purpose.
 * Regional hypothesis: dominant group, when it occurs, is only a manifestation of the limitations within a region. Variation of those limitations may result in the loss of a dominant group with the eventual appearance of a new one or none at all.
 * Source hypothesis: dominant group is a source within each field where a primary author of original research uses the term.
 * Term hypothesis: dominant group is a significant term that may require a 'rigorous definition' or application and verification of an empirical definition.

A humanistic approach to a dominant group in the atmospheric sciences likely involves individualization of the dominant group members so as to treat each member as an individual. The result of such individualist effort is the dissolution of the dominant group. The key is to show that the collective criteria of dominance that composes the group is only manifested individually rather than through conspiracy.

Each individual entity that has a position of power, authority or ascendancy over other entities is viewed without any more power, authority or ascendancy than any entity. In short, the hierarchy is dissolved and manifestations are viewed only as part of a spectrum of differences.

Atmospheres
Def. "[t]he gases surrounding the Earth or any astronomical body" are called an atmosphere.

Def. the invisible gaseous substance "above the surface of the earth" is called air.

Def. "[t]he air in a particular place" is called an atmosphere.

Aerology
Def. "[t]he observation of the atmosphere" is called aerology.

"From these data there is no longer any doubt that dominant group microseisms are generated by various types of meteorological disturbances over the oceans."

Aeronomy
Def. "the atmosphere of the Earth and the other planets with reference to their chemical composition, physical properties, relative motion, and responses to radiation from space" is called aeronomy.

Aeronomy is the study of the upper layers of the atmosphere, where dissociation and ionization are important.

Aeronomy is the science of the upper region of the atmosphere, where dissociation and ionization are important.

Compounds
“Aromatic VOCs is the dominant group both for new and in-use 4-stroke engines. Alkane VOCs is the dominant group for 2-stroke engines.”

“2-Decenal and 2-undecanal were the main components in this class of aldehydes and alkenals were the dominant group in all three cooking oils at both temperatures studied, except in canola oil at 180 °C.”

“Additionally, both species fall into the 10 same functional group (ie reactive aromatics) (Zhang et al., 2004), and thus their ratio will be most relevant to this most dominant group.”

Meteors
“These values yield a dominant group velocity consistent with the Eady lee wave extension along the y axis in Fig. 5f:

 * Cg ≈ (6,-20) m s-1”

“Silicate particle types of probable crustal origin were the dominant group by volume through most of the period.”

“The 73 “silicate group” PM 2.5 analyses from the Lizard mostly comprise felsic particles (84%), with the Al-felsic (52%) being by far the most dominant group, followed by AlSi-felsic (20%) and relatively little quartz-rich material (Fig. 10).”

“In contrast, on weekends, gasoline vehicles were the dominant group, except at midnight.”

“When gasoline and LPG vehicles were the dominant types of running vehicles, OC/EC ratios were larger than 1.0.”

“Gasoline- and LPG-vehicle exhaust had similar carbon profiles; i.e., OC3 was the dominant species, followed by OC2.”

“Days with liquid precipitation (rain and drizzle) are a dominant group and account for 58.1% of all precipitation days.”

Atmochemistry
"The biogenic volatile organic compounds (BVOC) include the quantitatively dominant group of the isoprenoids [isoprene, monoterpenes (e.g. α-pinene), sesquiterpenes] and oxygen-containing species (OVOC), such as alcohols, carbonyls and organic acids."

"Group 1 was the dominant group and accounted for 45.00%, 33.04%, 48.36% of the total trajectories in 2004–2005, 2006–2007, and 2007–2008, respectively."

Tropospheric ozone: “The dominant groups were chosen from the dendrogram based on three criteria: (1) minimize the scatter around the mean curve for each dominant group; (2) the number of classification days within each group had to approach or exceed the number of variables in the subsequent discriminant analysis (24 in our case), and (3) retain the smallest number of groups satisfying criteria (1) and (2).”

Atmophysics
"There is a dominant group of periodicities around 4±7 years, which may be associated with the ENSO phenomena."

"It is the speed of the dominant group that remains because of the self-canceling rapid oscillations away from the point of stationary phase."

Climatology
Climatology "is the study of climate, scientifically defined as weather conditions averaged over a period of time, and is a branch of the atmospheric sciences."

Climate encompasses the statistics of temperature, humidity, atmospheric pressure, wind, rainfall, atmospheric particle count and numerous other meteorological elements in a given region over long periods of time.

"From the outset WG1, chaired by Sir John Houghton of the UK Meteorological Office, was the dominant group."

"A dominant group size of at least 50% might be a useful criterion since a size smaller than 50% would mean that the other two groups combined would outnumber the dominant group. Ten of the 20 circulation types shown in Fig. 3 pass this test."

"Relevant amongst our findings is that sinuous tracks are the minority group from May to August, but then become the dominant group from September through to December, suggesting a major shift in the degree of synoptic control on typhoon movement between the first and second half of the typhoon season"

"The results of the factor analysis show that the rainfall characteristics (amount, intensity, duration) are the dominant group of factors, representing the highest weight for explaining the amount of throughfall, with high positive correlations (r = 0.99; r = 0.90 yr = 0.83, respectively)."

Meteorology
Def. "[t]he study of the atmosphere and its phenomena, especially with weather and weather forecasting" is called meteorology.

“The dashed lines mark a distance of 500 m, and the arrow gives the travel direction of the dominant group in the images.”

Weather astronomy
"A dominant group size of at least 50% might be a useful criterion since a size smaller than 50% would mean that the other two groups combined would outnumber the dominant group."

Weather forecasting astronomy
"In the currently proposed method, that is, if no information about the land use distribution for a grid cell is available, we shall use only the A type station as it belongs to the dominant group."

Hypotheses

 * 1) A dominant group in the atmospheric sciences may prevent acceptance of the effects the natural electric field of the Earth has on the Earth's atmosphere.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Dominant group/Atmospheric sciences


text:
University of Florida/Egm4313/s12.team11.perez.gp/R3.2

Problem Statement
Developing the second homogeneous solution for the case of double real root as a limiting case of distinct roots.

Given
Consider two distinct roots of the form:

$$ \lambda_1 = x \! $$ and $$ \lambda_2 = x + \epsilon \! $$

(where $$ \epsilon \! $$ is perturbation).

Given
Find the homogeneous L2-ODE-CC having the above distinct roots.

Solution
$$ (\lambda - \lambda_1)(\lambda - (\lambda_1))=0 \! $$

$$ (\lambda - x)(\lambda - (x + \epsilon))=0 \! $$

$$ \lambda^2 - \lambda x - \lambda \epsilon - \lambda x + x^2 + x \epsilon =0 \! $$

$$ \lambda^2 - \lambda (2x+\epsilon) + x (x+\epsilon )=0 \! $$

$$ \therefore y'' - y'(2\lambda +\epsilon) + y \lambda (\lambda + \epsilon) = 0 \! $$ (1)

Given
Show that $$ \frac{e^{(\lambda +\epsilon) x} - e^{\lambda x}}{\epsilon } \! $$ is a homogeneous solution. (2)

Solution
Let's find the corresponding derivatives:

$$ y = \frac{e^{(\lambda+\epsilon) x} - e^{\lambda x}}{\epsilon} \! $$

$$ y' = \frac{(\lambda +\epsilon) e^{\lambda + \epsilon x}-\lambda e^{\lambda x}}{\epsilon } \! $$

$$ y'' = \frac{(\lambda +\epsilon)^2 e^{(\lambda + \epsilon) x}-\lambda^2 e^{\lambda x}}{\epsilon} \! $$

If we now take these three equations and plug them into the homogeneous L2-ODE-CC (1), we get:

$$ e^{(\lambda + \epsilon)x} (\lambda^2+2\lambda \epsilon + \epsilon^2 - 2\lambda^2-2\lambda \epsilon - \lambda \epsilon -\epsilon^2 + \lambda^2 + \epsilon \lambda)+e^{\lambda x}(- \lambda^2+2\lambda^2+\lambda \epsilon - \lambda ^2 - \lambda \epsilon) = 0 \! $$

$$ \therefore 0 \equiv 0. \! $$

Since the left and right hand sides of the equation are zero, the solution is in fact a homogeneous equation.

Given
Find the limit of the homogeneous solution in (2) as epsilon approaches zero (think l'Hopital's Rule).

Solution
Using l'Hopital's Rule,

$$ \lim_{\epsilon \rightarrow 0}\frac{e^{(\lambda +\epsilon) x }-e^{\lambda x}}{\epsilon} = \frac{0}{0} \! $$

(this is an indeterminate form).

L'Hopital's Rule states that we can divide this function into two functions, $$ f(\epsilon) \! $$ and $$ g(\epsilon) \! $$, and then find their derivatives and attempt to find the limit of $$ \frac{f'(\epsilon)}{g'(\epsilon)} \! $$. If a limit exists for this, then a limit exists for our original function.

$$ \lim_{\epsilon \rightarrow 0} \frac{f'(\epsilon)}{g'(\epsilon)} = \lim_{\epsilon \rightarrow 0} \frac{xe^{x(\lambda + \epsilon)}}{1}\! $$

$$ = \frac{x e^{(\lambda+0)x}}{1} \! $$

$$ = xe^{x \lambda}. \! $$

Given
Take the derivative of $$ e^{\lambda x} $$ with respect to lambda.

Solution
Taking the derivative with respect to lambda, we find that:

$$ \frac{d(e^{\lambda x})}{d\lambda} = xe^{\lambda x} $$.

It is important to remember that we must hold $$ x \! $$ as a constant when finding this derivative.

Given
Compare the results in parts (3) and (4), and relate to the result by using variation of parameters

Solution
Taking a closer look at Parts 3 and 4 of this problem, we discover that they're in fact equal:

$$ \frac{d(e^{\lambda x})}{d\lambda }=\lim_{e\rightarrow 0}\frac{e^{(\lambda +\epsilon )x} - e^{\lambda x}} {\epsilon} = xe^{\lambda x} \! $$

Given
Numerical experiment: Compute (2) setting lambda equal to 5 and epsilon equal to 0.001, and compare to the value obtained from the exact second homogeneous solution.

Solution
After performing these calculations, from (2) we get 148.478.

And from the exact second homogeneous solution, we get 200.05.


text:
University of Florida/Egm4313/s12.team11.perez.gp/R3.9

Problem Statement
Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail.

(K 2011 pg.85 #13)
$$ 8y'' - 6y' + y = 6 \cosh x \! $$              (1)

Initial conditions are:

$$ y(0) = 0.2, y'(0) = 0.05 \! $$

Solution
The general solution of the homogeneous ordinary differential equation is

$$ 8y'' - 6y' + y = 0 \! $$

We can use this information to determine the characteristic equation:

$$ 8 \lambda^2 - 6 \lambda + 1 = 0 \! $$

And proceeding to find the roots,

$$ 4\lambda(2 \lambda - 1) - 1(2 \lambda - 1) = 0 \! $$

Thus, $$ (4 \lambda - 1)(2 \lambda - 1) = 0 \! $$.

Solving for the roots, we find that $$ \lambda = \frac{1}{4}, \frac{1}{2}, \! $$

where the general solution is

$$ y_k = c_1 e^{\frac{1}{4}x} + c_2 e^{\frac{1}{2}x} \! $$.

The solution of $$ y_p \! $$ of the non-homogeneous ordinary differential equation is

$$ x = \frac{e^x+e^{-x}}{2} \! $$.

Using the Sum rule as described in Section 2.7, the above function translates into the following:

$$ y_p = y_{p1} + y_{p2} \! $$, where Table 2.1 tells us that:

$$ y_{p1} = Ae^x \! $$ and $$ y_{p2} = Be^x \! $$.

Therefore, $$ y_{p} = Ae^x + Be^{-x} \! $$.

Now, we can substitute the values ($$ y_p, y_p', y_p'' \! $$) into (1) to get:

$$ 8(Ae^x+Be^{-x}) - 6(Ae^x - Be^{-x}) + Ae^x + Be^{-x} = 6(\frac{e^x + e^{-x}}{2}) \! $$

$$ = 3Ae^x + 15Be^{-x} \! $$

$$ = 3(e^x + e^{-x}) \! $$

Now that we have this equation, we can equate coefficients to find that:

$$ 3A = 3 \! $$

$$ \therefore A = 1 \! $$

$$ B = \frac{3}{15} = \frac{1}{5} \ $$

and thus, $$ y_p = e^x + \frac{1}{5}e^{-x} \! $$

We find that the general solution is in fact:

$$ y = y_k + y_p \! $$

$$ y = c_1 e^{\frac{1}{4}}x + c_2 e^{\frac{1}{2}}x + e^x + 3e^{-x} \! $$

whereas the general solution of the given ordinary differential equation is actually:

$$ y = c_1 e^{\frac{1}{4}}x + c_2 e^{\frac{1}{2}}x + e^x + \frac{1}{5} e^{-x} \! $$

Solving for the initial conditions given and first plugging in $$ y(0) = 0.2 \! $$, we get that:

$$ 0.2 = c_1 e^{\frac{1}{4}(0)} + c_2 e^{\frac{1}{2}(0)} + e^0 + 3e^{(0)} \! $$

$$ 0.2 = c_1 e^{(0)} + c_2 e^{(0)} + e^{(0)} + 3e^{(0)} \! $$

$$ 0.2 = c_1 + c_2 + 1 + \frac{1}{5} \! $$

$$ \therefore c_1 + c_2 = - 1 \! $$.             (2)

And now we can determine the first order ODE :

$$ y' = \frac{1}{4}c_1 e^{\frac{1}{4}(x)} + \frac{1}{2}c_2 e^{\frac{1}{2}(x)} + e^{x} - \frac{1}{5}e^{x} \! $$

The second initial condition that was given to us, $$ y'(0) = 0.05 \! $$ can now be plugged in:

$$ 0.05 = \frac{1}{4}c_1 e^{\frac{1}{4}(0)} + \frac{1}{2}c_2 e^{\frac{1}{2}(0)} + e^{(0)} - \frac{1}{5}e^{(0)} \! $$

$$ 0.05 = \frac{1}{4}c_1 e^{(0)} + \frac{1}{2}c_2 e^{(0)} + e^{(0)} - \frac{1}{5}e^{(0)} \! $$

$$ 0.05 = \frac{1}{4}c_1 + \frac{1}{2}c_2 + 1 - \frac{1}{5} \! $$

$$ \frac{1}{4}c_1 + \frac{1}{2}c_2 = -0.75 \! $$

$$ \therefore c_1 + 2 c_2 = -3 \! $$             (3)

Once we solve (2) and (3), we can get the values:

$$ c_1 = 1, c_2 = - 2 \! $$.

And once we substitute these values, we get the following solution for this IVP:

$$ y = e^{\frac{1}{4}x} - 2 e^{\frac{1}{2}x} + e^{x} + \frac{1}{5} e^{-x} \! $$

(K 2011 pg.85 #14)
$$ y'' + 4y' + 4y = e^{-2x} sin2x \! $$              (1)

Initial conditions are:

$$ y(0)=1, y'(0) = -1.5 \! $$

Solution
The general solution of the homogeneous ordinary differential equation is

$$ y'' + 4y' + 4y = 0 \! $$

We can use this information to determine the characteristic equation:

$$ \lambda^2 + 4 \lambda + 4 = 0 \! $$

And proceeding to find the roots,

$$ (\lambda + 2) (\lambda + 2) = 0 \! $$

Solving for the roots, we find that $$ \lambda = -2, -2 \! $$

where the general solution is:

$$ y_k = c_1 e^{-2x} + c_2 e^{-2x} x \! $$, or:

$$ y_k = (c_1 + c_2 x) e^{-2x} \! $$

Now, according to the Modification Rule and Table 2.1 in Section 2.7, we know that we have to multiply by x to get:

$$ y_p = e^{-2x}(K x cos2x + Mx sin2x) \! $$, since the solution of $$ y_k \! $$ is a double root of the characteristic equation.

We can then derive to get $$ y_p' \! $$:

$$ y_p' = -2 e ^{-2x} (K x cos2x + M x sin2x) + e^{-2x} (Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x) \! $$

$$ y_p' = (-2K+2M) e^{-2x} x cos2x + (-2M-2K) e^-2x xsin2x + Ke^{-2x} cos2x + Me^{-2x} sinx \! $$

Deriving once again to solve for $$ y_p'' \! $$, we get the following:

$$ y_p'' = 4e^{-2x} (Kxcos2x + Mxsin2x) - 2e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) - 2e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) \! $$

$$ y_p'' = 4e^{-2x} (Kxcos2x + Mxsin2x) - 4e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) + e^{-2x} (-4Ksin2x - 4Kxcos2x + 4Mcos2x - 4Mxsin2x) \! $$

$$ y_p'' = (-4K + 4M) e^{-2x} cos2x + (-4K - 4M)e^{-2x} sin2x \ $$

Now, we can substitute the values ($$ y_p, y_p', y_p'' \! $$) into (1) to get:

$$ (-4K + 4M) e^{-2x}cos2x+(-4M-4K)e^{-2x}sin2x+4((-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-2x}xsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx) + 4(e^{-2x}(Kxcos2x + Mxsin2x)) = e^{2x}sin2x \! $$

$$ \therefore (-3K+4M)e^{-2x}cos2x + (-3M-4K)e^{-2x}sin2x = e^{-2x}sin2x \! $$

Now that we have this equation, we can equate coefficients to find that:

$$ -3K + 4M =0 \! $$ and $$ -4K - 3M = 1 \! $$

and finally discover that:

$$ M = - \frac{3}{25} \! $$ and $$ K = - \frac{4}{25} \! $$.

Plugging in these values in $$ y_p \! $$, we find that:

$$ y_p = e^{-2x}( - \frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$

And finally, we arrive at the general solution of the given ordinary differential equation:

$$ y = y_k + y_p \! $$

$$ y = (c_1 + c_2 x) e^{-2x}+e^{-2x} (-\frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$

Solving for the initial conditions given and first plugging in $$ y(0) = 1 \! $$, we get that:

$$ 1 = (c_1 + c_2 (0)) e^{-2(0)}+e^{-2(0)} (-\frac{4}{25}(0)cos2(0) - \frac{3}{25}(0)sin2(0)) \! $$

$$ 1 = (c_1 + c_2(0))e^0 \! $$

$$ \therefore c_1 = 1 \! $$

The second initial condition that was given to us, $$ y'(0) = -1.5 \! $$ can now be plugged in:

$$ y' = - \frac{1}{5} e^-2x (10c_1 + 10c_2x - 5c_2 + (3-14x)sin2x+(4-2x)cos(2x)) \! $$

$$ -1.5 = -\frac{1}{5}(10c_1 - 5c_2 + 4) \! $$

$$ \therefore c_2 = -3.5 \! $$

And once we substitute these values, we get the following solution for this IVP:

$$ y = (1 - 3.5 x) e^{-2x}+e^{-2x} (-\frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$


text:
University of Florida/Egm4313/s12.team8.dupre/R3.4

Problem Statement
Use the Basic Rule and the Sum Rule to show that the appropriate particular solution to $$\displaystyle y''-3y'+2y=4x^2-6x^5 $$      (4.1) is of the form $$\displaystyle y_{p}(x)=\sum_{j=0}^{n}c_{j}x^{j} $$ with n=5, i.e, $$\displaystyle y_{p}(x)=\sum_{j=0}^{5}c_{j}x^{j} $$.

Solution
The Basic rule and Sum rule allow us to choose our particular y forms from table 2.1 on page 82 in the Kreyszig Advanced Engineering Mathematics book. These rules state that, using your given $$\displaystyle r(x)$$, you can find what $$\displaystyle y_{p}(x) $$ you should choose to use. From (4.1), we know that: $$\displaystyle r(x)=4x^2-6x^5 $$     (4.2) Referring to table 2.1, and knowing that the Basic rule tells us that we can match up the form of our $$\displaystyle r(x)$$ to one in the table, and use the $$\displaystyle y_{p}(x) $$ accordingly, we find the following two equations: The $$\displaystyle y_{p}(x) $$ for $$\displaystyle 4x^{2} $$ is: $$\displaystyle k_{2}x^{2}+k_{1}x^{1}+k_{0} $$     (4.3) And the $$\displaystyle y_{p}(x) $$ for $$\displaystyle 6x^{5} $$ is: $$\displaystyle K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x^{1}+K_{0} $$     (4.4) The sum rule tells us that we are able to add these equations, (4.3) and (4.4), to obtain a final solution for $$\displaystyle y_{p}(x) $$. The solution for this is as follows: $$\displaystyle k_{2}x^{2}+k_{1}x^{1}+k_{0} +K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x^{1}+K_{0} $$ Adding similar variables gives us: $$\displaystyle K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+(K_{2}+k_{2})x^2+(K_{1}+k_{1})x^1+(K_{1}+k_{0}) $$    (4.5) Since these k and K values are just constants, we can set: $$\displaystyle K_{5}=c_{5} $$ $$\displaystyle K_{4}=c_{4} $$ $$\displaystyle K_{3}=c_{3} $$ $$\displaystyle (K_{2}+k_{2})=c_{2} $$ $$\displaystyle (K_{1}+k_{1})=c_{1} $$ $$\displaystyle (K_{1}+k_{0})=c_{0} $$ Our FINAL solution for the $$\displaystyle y_{p}(x) $$ of (4.1) is: $$\displaystyle y_{p}(x)=c_{5}x^{5}+c_{4}x^{4}+c_{3}x^{3}+c_{2}x^{2}+c_{1}x^{1}+c_{0} $$    (4.6) And, as was the original problem statement, equation (4.6) is of the form: $$\displaystyle y_{p}(x)=\sum_{j=0}^{5}c_{j}x^{j} $$


text:
User:Egm4313.s12.team8.koester/R3.5

=R3.5 - Particular solution to L2-ODE-CC from series=

Problem Statement
Complete the solution for equation (2) p 7-11 as follows:

$$ \displaystyle {y}''-3{y}'+2y=4x^{2}-6x^{5} $$

Step 1
Obtain equations for 2, 3, 4, and 6 from p 7-14

-coefficients of $$ \displaystyle x $$

-coefficients of $$ \displaystyle x^{2} $$

-coefficients of $$ \displaystyle x^{3} $$

-coefficients of $$ \displaystyle x^{5} $$

Use equation (2) from p 7-13:

$$ \ \sum_{j = 0}^{3} \left [ c_{j+2} (j+2) (j+1) - 3c_{j+1} (j+1) + 2c_j\right] x^j - 3c_5(5)x^4 + 2 \left [c_4x^4 + c_5x^5 \right ] = 4x^2 - 6x^5\ $$

For the coefficients of x, let j=1 to get:

$$ \displaystyle (c_3(3)(2)-3c_2(2)+2c_1)x=0 $$

Reducing to:

$$ \displaystyle 6c_3-6c_2+2c_1=0 $$

For the coefficients of x^{2}, let j=2 to get:

$$ \displaystyle (c_4(4)(3)-3c_3(3)+2c_2)x^{2}=4x^{2} $$

Reducing to:

$$ \displaystyle 12c_4-9c_3+2c_2=4 $$

For the coefficients of x^{3}, let j=3 to get:

$$ \displaystyle (c_5(5)(4)-3c_4(4)+2c_3)x^{3}=0 $$

Reducing to:

$$ \displaystyle 20c_5-12c_4+2c_3=0 $$

For the coefficients of x^{5}, let j=5 to get:

$$ \displaystyle 2c_5x^{5}=-6x^{5} $$

Reducing to:

$$ \displaystyle 2c_5=-6 $$

Step 2
Verify all equations by long-hand expansion of the series in equation (4) p 7-12:

$$ \ \sum_{j = 2}^{5} c_j(j)(j-1)(x^{j-2}) - 3\sum_{j = 1}^{5} c_j(j)(x^{j-1}) + 2\sum_{j = 0}^{5} c_jx^j = 4x^2 - 6x^5\ $$

Expanding these series the following terms are obtained:

$$ \displaystyle c_2(2)(1)+c_3(3)(2)x+c_4(4)(3)x^2+c_5(5)(4)x^3-3c_1(1)-3c_2(2)x-3c_3(3)x^2-3c_4(4)x^3-3c_5(5)x^4 $$

$$ \displaystyle +2c_0+2c_1x+2c_2x^2+2c_3x^3+2c_4x^4+2c_5x^5= 4x^2-6x^5  $$

Collecting like coefficients the equation can be rearranged to:

$$ \displaystyle (2c_0-3c_1+2c_2)+(2c_1-6c_2+6c_3)x+(2c_2-9c_3+12c_4)x^2+(2c_3-12c_4+20c_5)x^3+(2c_4-15c_5)x^4+(2c_5)x^5=4x^2-6x^5  $$

Examining this equation it can be seen that the coefficients match those previously found so the equations are verified.

Step 3
Put the system of equations into matrix form:

$$ \ \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0 \\ 0 & 2 & -6 & 6 & 0 & 0 \\ 0 & 0 & 2 & -9 & 12 & 0 \\ 0 & 0 & 0 & 2 & -12 & 20 \\ 0 & 0 & 0 & 0 & 2 & -15 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 4 \\ 0 \\ 0 \\ -6 \end{bmatrix} $$

Step 4
Solve for the coefficients by back substitution.

This is done by first solving for $$ \displaystyle c_5 $$ and then $$ \displaystyle c_4 $$ and eventually all the way back to $$ \displaystyle c_0 $$

$$ \displaystyle 2c_5=-6 $$

$$ \displaystyle c_5=-3 $$

$$ \displaystyle -15c_5+2c_4=0 $$

$$ \displaystyle 2c_4=-45 $$

$$ \displaystyle c_4=-22.5 $$

$$ \displaystyle 20c_5-12c_4+2c_3=0 $$

$$ \displaystyle 2c_3=-210 $$

$$ \displaystyle c_3=-105 $$

$$ \displaystyle 12c_4-9c_3+2c_2=4 $$

$$ \displaystyle 2c_2=-671 $$

$$ \displaystyle c_2=-335.5 $$

$$ \displaystyle 6c_3-6c_2+2c_1=0 $$

$$ \displaystyle 2c_1=-1383 $$

$$ \displaystyle c_1=-691.5 $$

$$ \displaystyle 2c_2-3c_1+2c_0=0 $$

$$ \displaystyle 2c_0=-1403 $$

$$ \displaystyle c_0=-701.75 $$

Step 5
Consider the initial conditions:

$$ \displaystyle y(0)=1 $$

$$ \displaystyle y'(0)=0 $$

Find the solution y(x) and plot it

$$ \displaystyle y(x)= y_p + y_h $$

$$ \displaystyle y_p= c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + c_5x^5 $$

So for our equation:

$$ \displaystyle y_p=-701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5 $$

$$ \displaystyle y_h= C_1e^{\lambda_1x}+C_2e^{\lambda_2x} $$

where:

$$ \displaystyle \lambda_1=1 $$ and $$ \displaystyle \lambda_2=2 $$ from quadratic equation.

So the overall solution becomes:

$$ \displaystyle y(x)= -701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5+C_1e^{x}+C_2e^{2x} $$

Now use the given initial conditions to solve for the constants:

$$ \displaystyle y(0)=1=-701.75 + C_1 + C_2 $$

$$ \displaystyle y'(0)=0=-691.5 + C_1 + 2C_2 $$

Cleaning up to:

$$ \displaystyle 702.75 = C_1 + C_2 $$

$$ \displaystyle 691.5 = C_1 + 2C_2 $$

Subtracting the first equation from the second the constants can then be solved for.

$$ \displaystyle C_2=-11.25 $$

$$ \displaystyle C_1=714 $$

Yielding the final complete solution of:

$$ \displaystyle y(x)= -701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5+714e^{x}-11.25e^{2x} $$

title: User:Egm4313.s12.team8.koester/R3.5


text:
Talk:University of Canberra/Critical Care Nursing/Resources

How to contribute?
--Holly Northam 07:24, 22 February 2012 (UTC)
 * 1) Explore and read the material on the site and question...
 * 2) Is it correct?
 * 3) Can I improve it?
 * 4) What resources can I add that will help meet the information needs of patients,  families and other staff in critical care environments?

Did you really mean this? I really did! We have a really powerful role in making a difference in the outcomes experienced by patients and their families...but more than that we have a responsibility in advocating for change and improvement when evidence shows that there are better ways of doing things. An interesting recent example is an insightful critical care student recently shared with me her indignation at the waste she observes in the way we use resources in the critical care environment. She commented on the equipment, linen and and other material that is discarded without being used because it is placed in excess in rooms where patients with infectious conditions are being nursed. It is an important aspect of nursing care that contaminated material should not be returned for use in the ward, however, thoughtful stocking and judicious management of resources can reduce the cost of critical care whilst still maintaining excellence in patient care. Some innovative work in this area is being carried out by the Institute of Health Care []. A number of intiatives in other centres are also working to reduce the environmental impact of caring for patients with critical illness. The Austin is targeting their ICU staff with education and have provided facilities to sort waste []. Please let me know what other initiaves are around and consider what you can do personally to decease the waste in providing care.Holly Northam (talk) 00:32, 5 June 2012 (UTC)


title: File:Problem5Graph.jpg