User:Egm4313.s12.team1.armanious/team1/R3

=Report 3=

Statement
Given the double root $$ \lambda = 5\!$$ and the excitation $$ r(x)=7e^{5x}-2 x^2 \!$$, with the initial conditions $$ y(0) = 4, y'(0) = 5 \!$$ find the solution $$ y(x)\!$$. Plot this solution and the solution to the same problem except with the excitation $$ r(x) = 7e^{5x} \!$$.

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$. Using the given double root, we can find the equation for the characteristic equation which leads to the homogeneous solution:


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$$\displaystyle{(\lambda - 5)^2 = \lambda^2 - 10\lambda + 25 = 0}$$ (1.0)
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Homogeneous solution:
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$$\displaystyle{y'' - 10y' + 25y = r(x)}$$ (1.1)
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$$\displaystyle{y'' - 10y' + 25y = 7e^{5x} - 2x^2}$$ (1.2)
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First, by using the Modification Rule, we find that the general equation associated with the given double root is:


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$$\displaystyle{y_g(x) = C_1e^{5x} + C_2xe^{5x}}$$ (1.3)
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We need to find the particular solution to the excitation $$ r(x) = 7e^{5x} - 2x^2\!$$. In analyzing the excitation, it is found that the particular solution looks like this:


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$$\displaystyle{y_p(x) = Cx^2e^{5x} - [K_2x^2 + K_1x + K_0]}$$ (1.4)
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Now, we need to find the values for the constants $$ C, K_2, K_1, K_0\!$$, by taking the first and second derivatives of the particular solution:


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$$\displaystyle{y_p(x) = Cx^2e^{5x} - [K_2x^2 + K_1x + K_0]}$$ (1.4)
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$$\displaystyle{y_p(x)' = 2Cxe^{5x} + 5Cx^2e^{5x} - [2K_2x + K_1]}$$ (1.5)
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$$\displaystyle{y_p(x)'' = 2Ce^{5x} + 20Cxe^{5x} + 25Cx^2e^{5x} - [2K_2]}$$ (1.6)
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Now, plug these into the homogeneous solution (1.2) and simplify:


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$$\displaystyle{2Ce^{5x} - 2K_2 + 20K_2x + 10K_1 - 25K_2x^2 - 25K_1x - 25K_0 = 7e^{5x} - 2x^2}$$ (1.7)
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Comparing the coefficients for $$ e^{5x}, x^2 \!$$ and $$ x \!$$ allows us to solve for the unknown coefficients:


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$$\displaystyle{e^{5x}: 2C = 7 \rightarrow C = 7/2}$$ (1.8)
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$$\displaystyle{x^2: -25K_2 = -2 \rightarrow K_2 = 0.08}$$ (1.9)
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$$\displaystyle{x: 20K_2 - 25K_1 = 0 \rightarrow K_1 = 0.064}$$ (1.10)
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$$\displaystyle{-2K_2 + 10K_1 - 25K_0 = 0 \rightarrow K_0 = 0.0192}$$ (1.11)
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Plug into $$ y_p(x)\!$$ (1.4):


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$$\displaystyle{y_p(x) = \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192]}$$ (1.12)
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Now that we have the particular solution, we can use the initial conditions to solve for the unknown constants in the general solution, $$ y_g(x)\!$$, and then we will be able to solve for the final solution $$ y(x)\!$$:


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$$\displaystyle{y(x) = C_1e^{5x} + C_2xe^{5x} + \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192]}$$ (1.13)
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Take derivative of (1.13):


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$$\displaystyle {y(x)' = 5C_1e^{5x} + C_2e^{5x} +5C_2xe^{5x} + 7xe^{5x} + \frac{35}{2}x^2e^{5x} - [0.16x + 0.064]}$$ (1.14)
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Using the given initial conditions and equations (1.13) and (1.14):


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$$\displaystyle{y(0) = 4 \rightarrow 4 = C_1 - 0.0192 \rightarrow C_1 = 4.0192}$$ (1.15)
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$$\displaystyle{y'(0) = 5 \rightarrow 5 = 5C_1 + C_2 - 0.064 \rightarrow C_2 = -15.032}$$ (1.16)
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Plug these constants back into the solution (1.13) to obtain the final solution:

$$  \displaystyle y(x) = 4.0192e^{5x} - 15.032xe^{5x} + \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192] $$     (1.17)
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The solution to the same problem but with excitation $$ r(x) = 7e^{5x} \! $$ is:


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$$\displaystyle{y(x) = 4e^{5x}-25xe^{5x}+\frac{7}{2}x^{2}e^{5x}}$$ (1.18)
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This is the plot for (1.17) and (1.18):



Author
Solved and Typed By - --Egm4313.s12.team1.wyattling 22:00, 20 February 2012 (UTC)

Reviewed By - Egm4313.s12.team1.armanious 22:39, 21 February 2012 (UTC)

Statement
Perturbation method for double real root: Developing the 2nd homogeneous solution for the case of double real root as a limiting case of distinct roots (see [[media:iea.s12.sec7.djvu| Sec7 p. 7-5 ]]). Consider two distinct real roots of the form: $$\lambda_1= \lambda, \lambda_2= \lambda + \epsilon \!$$

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

2) Show that the following is a homogeneous solution:

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

The fraction in (3) p.7-5, for small $$\epsilon\!$$, is a finite difference formula that approximates the derivative

$$\frac{e^{(\lambda + \epsilon)x} - e^{\lambda x}}{\epsilon}\approx \frac{d}{d \lambda} e^{\lambda x} \!$$

In fact, $$ \frac{d}{d \lambda} e^{\lambda x} = \lim_{\epsilon \to 0}\frac{e^{(\lambda + \epsilon)x} - e^{\lambda x}}{\epsilon} \!$$

3)Find the limit of the homogeneous solution in (3) p.7-5 as [epsilon goes to 0] (think l'Hopital's rule)

4)Take the derivative of $$ e^{\lambda x}\!$$ with respect to $$\lambda\!$$

5)Compare the results in Parts (3) and (4), and relate to the result by variation of parameters.

6)Numerical experiment: Compute (3) p.7-5 using at $$\lambda=5\!$$ and with $$\epsilon = 0.001\!$$, and compare to get the value obtained from the exact 2nd homogeneous solution.

Solution
In order to find the corresponding L2-ODE-CC, the characteristic equation corresponding to the given solution must be found:
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$$  \displaystyle (\lambda'-\lambda)(\lambda'-\lambda-\epsilon)=0 $$     (2.0)
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$$  \displaystyle (\lambda')^2-2\lambda' \lambda + \lambda^2 - \lambda' \epsilon + \lambda \epsilon=(\lambda')^2-(2\lambda+\epsilon)\lambda'+(\lambda^2+\lambda\epsilon)=0 $$     (2.1)
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Therefore the corresponding L2-ODE-CC is:
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$$  \displaystyle y''-(2\lambda+\epsilon)y'+(\lambda^2+\lambda\epsilon)y=0 $$     (2.2) Note: if ε is equal to zero, the characteristic equation of 2.2 has a double real root at λ.
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To show that the following is a homogeneous solution, the first and second derivatives must be taken:
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$$  \displaystyle y(x)=\frac{e^{(\lambda + \epsilon)x}-e^{\lambda x}}{\epsilon} $$     (2.3)
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$$  \displaystyle y'(x)=\frac{(\lambda+\epsilon)e^{(\lambda + \epsilon)x}-\lambda e^{\lambda x}}{\epsilon} $$     (2.4)
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$$  \displaystyle y''(x)=\frac{(\lambda+\epsilon)^2e^{(\lambda + \epsilon)x}-\lambda^2 e^{\lambda x}}{\epsilon} $$     (2.5) Using these values in 2.2 yields:
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$$  \displaystyle \frac{(\lambda+\epsilon)^2e^{(\lambda + \epsilon)x}-\lambda^2 e^{\lambda x}}{\epsilon}-\frac{(2\lambda+\epsilon)[(\lambda+\epsilon)e^{(\lambda + \epsilon)x}-\lambda e^{\lambda x}]}{\epsilon}+\frac{(\lambda^2+\lambda\epsilon)[e^{(\lambda + \epsilon)x}-e^{\lambda x}]}{\epsilon}=0 $$     (2.6) Rearranging and simplifying yields:
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$$  \displaystyle \frac{[\lambda^2 +2\lambda\epsilon+\epsilon^2-2\lambda^2-3\lambda\epsilon-\epsilon^2+\lambda^2+\lambda\epsilon]e^{(\lambda + \epsilon)x}+[-\lambda^2+2\lambda^2+\lambda\epsilon-\lambda^2-\lambda\epsilon]e^{\lambda x}}{\epsilon}=0 $$     (2.7) All of the terms in the above equation cancel to yield:
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$$  \displaystyle \frac{0e^{(\lambda + \epsilon)x}+0e^{\lambda x}}{\epsilon}=0 $$     (2.8) Using l'Hopital's rule:
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$$  \displaystyle \lim_{\epsilon \to 0}\frac{e^{(\lambda+\epsilon)x}-e^{\lambda x}}{\epsilon}=\lim_{\epsilon \to 0}\frac{\frac{d}{d\epsilon}(e^{(\lambda+\epsilon)x}-e^{\lambda x})}{\frac{d}{d\epsilon} (\epsilon)}=\lim_{\epsilon \to 0}\frac{\frac{d}{d\epsilon}(e^{(\lambda+\epsilon)x})-0}{1} $$     (2.9)
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Simplifying further:
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$$  \displaystyle \lim_{\epsilon \to 0}\frac{d}{d\epsilon}(e^{(\lambda+\epsilon)x})=\lim_{\epsilon \to 0}xe^{(\lambda+\epsilon)x}=x\lim_{\epsilon \to 0}e^{(\lambda+\epsilon)x} $$     (2.10) This ultimately yields:
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$$  \displaystyle \lim_{\epsilon \to 0}\frac{e^{(\lambda+\epsilon)x}-e^{\lambda x}}{\epsilon}=xe^{\lambda x} $$ (2.11) The following should also be taken into consideration:
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$$  \displaystyle \frac{d}{d\lambda}(e^{\lambda x})=xe^{\lambda x} $$ (2.12) Clearly, the results of 2.11 and 2.12 are equivalent. This shows that $$y(x)=xe^{\lambda x} \!$$ is an appropriate solution to a homogeneous L2-ODE-CC having one double root.
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To test this, test values will be used to solve for the approximate (2.12) and exact (2.13) solutions of the ODE. For this test, $$\lambda=5\!$$ and $$\epsilon = 0.001\!$$
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$$  \displaystyle y_{approx}(x)=\frac{e^{(5 + 0.001)x}-e^{5 x}}{0.001}=1000(e^{0.001x}-1)e^{5x} $$     (2.13)
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$$  \displaystyle y_{exact}(x)=xe^{5x} $$     (2.14) If the above derivations are true, then the following approximation must also be true:
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$$  \displaystyle 1000(e^{0.001x}-1)\approx x $$ (2.15) Plotting both sides of 2.15 as a function of x shows that this is a valid approximation for most values of x. Figure 3.2-1
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Author
Solved and Typed By - Egm4313.s12.team1.armanious 05:51, 21 February 2012 (UTC)

Reviewed By - --Egm4313.s12.team1.durrance 02:53, 22 February 2012 (UTC)

Statement
Find the complete solution for $$ y'' - 3y' + 2y = 4x^2\!$$, with the initial conditions $$ y(0) = 1, y'(0) = 0\!$$ Plot the solution $$y(x).\!$$

Solution
First we create the characteristic equation in standard form:
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$$\displaystyle{\lambda^2-3 \lambda +2=0}$$ (3.0) Then, by setting it equal to zero, we can find what $$ \lambda \!$$ equals:
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$$\displaystyle{(\lambda - 2)(\lambda -1) = 0}$$ (3.1)
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$$\displaystyle{\lambda = 2,\lambda =1}$$ (3.2)
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Given two, distinct, real roots, the general solution looks like this:
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$$  \displaystyle y_g(x)=C_1e^{2x} + C_2e^{x} $$     (3.3) By using the method of undetermined coefficients, the excitation $$ 4x^2\!$$ is analyzed to yield a particular solution: In assessing a polynomial with a second power, the form of the particular solution will look like this:
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$$  \displaystyle y_p (x)= A_2 x^2 + A_1 x + A_0 $$     (3.5)
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It's derivative would look like this:
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$$  \displaystyle y_p '(x)= 2 A_2 x + A_1 $$     (3.6)
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And the second derivative to follow would then become:
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$$  \displaystyle y_p ''(x)= 2 A_2 $$     (3.7)
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Based on the coefficients, the following system of equations exists:
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$$  \displaystyle 2A_2 =4 $$     (3.8)
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$$  \displaystyle -6A_2 + A_1=0 $$     (3.9)
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$$  \displaystyle 2A_2 -3 A_1 + A_0 =0 $$     (3.10)
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The results of this set of equations make the coefficients of A's:
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$$  \displaystyle A_2 =2 $$
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$$  \displaystyle A_1 =12 $$
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$$  \displaystyle A_0 =32 $$ The resulting particular equation looks like this:
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$$  \displaystyle y_p (x)= 2 x^2 + 12 x + 32 $$     (3.11) By adding the particular and general solutions, we get the complete solution:
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$$  \displaystyle 2 x^2 + 12 x + 32 + C_1e^{2x} + C_2e^{x} = y $$ (3.12)
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We consider the initial conditions by taking the first derivative of the complete solution:


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$$  \displaystyle 4 x + 12 + 2 C_1e^{2x} + C_2e^{x} = y' $$ (3.13)
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By plugging in 0 for x, 1 for y, and 0 for y', we can solve for the constants $$C_1, C_2 \!$$:
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$$  \displaystyle y(0)=1 =2 * 0^2 + 12 *0 + 32 + C_1e^{2*0} + C_2e^{0}=32 +C_1 +C_2=1 $$     (3.14)
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$$  \displaystyle y'(0)=0=4 *0 + 12 + 2 C_1e^{2*0} + C_2e^{0} = 12+ 2 C_1 + C_2=0 $$     (3.15)
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Solving the equations proves that $$C_1 = 19, C_2 = -50 \!$$: The resulting complete solution with consideration for initial conditions then becomes:
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$$  \displaystyle 2 x^2 + 12 x + 32 + 19 e^{2x} - 50e^{x} = y $$ (3.12)
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y plotted looks like this:

Author
Solved and Typed By -Egm4313.s12.team1.silvestri 15:56, 19 February 2012 (UTC)

Reviewed By -Egm4313.s12.team1.armanious 03:02, 22 February 2012 (UTC)

Statement
From [[media:iea.s12.sec7b.djvu|R3.4 in Sec 3 p. 7-11]]; Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for
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$$  \displaystyle y''-3y'+2y=4x^{2}-6x^{5} $$     (4.1) is of the form
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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{n}c_{j}x_{j} $$     (4.2) i.e.,
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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{5}c_{j}x_{j} $$     (4.3)
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Basic Rule: Select $$y_{p}(x)\!$$ from the table and determine the coefficients by substituting $$y_{p}(x)\!$$ in


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$$  \displaystyle y''+ay'+by=r(x) $$     (4.4) Sum Rule: If $$r(x)\!$$ is the sum of the terms in the 1st column of table 2.1 then $$y_{p}(x)\!$$ is the sum of the corresponding terms in the 2nd column of this table. Table 2.1
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Solution
According to the table the Homogeneous equation has two $$r(x)\!$$ values of the form $$kx^{n}(n=0,1,2...)$$. Using the Basic Rule this means that there is a particular solution of the form $$K_{n}x^{n}+K_{n-1}x^{n-1}+...+K_{1}x+K_{0}\!$$ for each $$r(x)\!$$.


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$$  \displaystyle r_{1}(x)=4x^{2}\rightarrow r_{1}(x)=Kx^{n}(n=2) $$     (4.5)
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So the particular solution for this $$r_{1}(x)\!$$ should be:


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$$  \displaystyle y_{p1}= K_{2}x^{2}+K_{1}x^{1}+K_{0} $$     (4.6)
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Where $$K=4\!$$. Which simplifies to:


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$$  \displaystyle y_{p1}= \sum_{j=0}^{2}K_{j}x^{j} $$     (4.7)
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For the second $$r_{2}(x)\!$$:


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$$  \displaystyle r_{2}(x)=-6x^{5}\rightarrow r_{2}(x)=Cx^{n}(n=5) $$     (4.8)
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So the particular solution for this $$r_{2}(x)\!$$ should be:


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$$  \displaystyle y_{p2}= C_{5}x^{5}+C_{4}x^{4}+C_{3}x^{3}+C_{2}x^{2}+C_{1}x+C_{0} $$     (4.9)
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Where $$C=4\!$$. Which simplifies to:


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$$  \displaystyle y_{p1}= \sum_{j=0}^{5}C_{j}x^{j} $$     (4.10)
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Using the sum rule:


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$$  \displaystyle y_{p}=\sum_{j=0}^{2}K_{j}x^{j}+ \sum_{j=0}^{5}Cx^{j} $$     (4.11) Now using the Sum Rule which just states if there are two $$r(x)\!$$ values in any form on the left side of the table, then the particular solution is the sum of the solutions for $$r(x)\!$$ on the right side of the table.
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So the particular solution for $$y''-3y'+2y=4x^{2}-6x^{5}\!$$ where $$c=K+C$$


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$$  \displaystyle y_{p}=\sum_{j=0}^{5}c_{j}x^{j} $$     (4.12)
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Author
Solved and Typed By - User:Egm4313.s12.team1.stewart 22:47, 21 February 2012 (UTC) Reviewed By - Egm4313.s12.team1.silvestri 04:57, 22 February 2012 (UTC)

Statement
Given:
 * {| style="width:100%" border="0"

y''-3y'+2y=4x^2-6x^5 $$     (5.0)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

y_p(x)=\sum_{j=0}^5c_jx^j=4x^2-6x^5 $$     (5.1)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Find: -Coefficients of particular solution using series expansion and matrix back-substitution -Solution $$\displaystyle y(x)$$ using initial conditions $$\displaystyle y(0)=1, y'(0)=0$$ Plot solution.

Solution
Using [http://upload.wikimedia.org/wikiversity/en/a/a6/Iea.s12.sec7b.djvu Sec7b-1 p.7-13 Eq. (1)]:


 * {| style="width:100%" border="0"

\sum_{j=0}^3c_{j+2}(j+2)(j+1)x^j-3\sum_{j=0}^4c_{j+1}(j+1)x^j+2\sum_{j=0}^5c_jx^j=4x^2-6x^5 $$     (5.2) Which is a combined series expansion for the general coefficient series expansion of [http://upload.wikimedia.org/wikiversity/en/a/a6/Iea.s12.sec7b.djvu Sec7b-1 p.7-12 Eq. (4)]:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \sum_{j=2}^5c_jj(j-1)x^{j-2}-3\sum_{j=1}^5c_jjx^{j-1}+2\sum_{j=0}^5c_jx^j=4x^2-6x^5 $$     (5.3)
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 * <p style="text-align:right">
 * }

We can find a simultaneous linear system of equations to solve for the coefficients of $$\displaystyle y_p(x)$$.

Note: by combining the summations in Eq. (5.2) by setting the upper bounds of the summations to a common value 3, the equation can be simplified to:
 * {| style="width:100%" border="0"

$$\displaystyle \sum_{j=0}^3[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_j]x^j+(2c_4-15c_5)x^4+2c_5x^5=4x^2-6x^5 $$     (5.4)
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 * <p style="text-align:right">
 * }

Therefore providing a system of equations by equating coefficients from the summations to the $$\displaystyle y_p(x)$$:
 * {| style="width:100%" border="0"

j=0: [c_2(2)(1)-3c_1(1)+2c_0]x^0=0x^0=[2c_0-3c_1+2c_2]x^0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=1: [c_3(3)(2)-3c_2(2)+2c_1]x^1=0x^1=[2c_1-6c_2+6c_3]x^1 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=2: [c_4(4)(3)-3c_3(3)+2c_2]x^2=4x^2=[2c_2-9c_3+12c_4]x^2 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=3: [c_5(5)(4)-3c_4(4)+2c_3]x^3=0x^3=[2c_3-12c_4+20c_5]x^3 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

[2c_4-15c_5]x^4=0x^4 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

2c_5x^5=-6x^5 $$     (5.5) Eqs. (5.5) can be verified by solving for the coefficients by Eq. (5.2) to prove that the summations were combined correctly:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

j=0: 2c_0x^0=2c_0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=1: 2c_1x^1-3c_1(1)x^0=2c_1x-3c_1 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=2: 2c_2x^2-3c_2(2)x^1+c_2(2)(1)x^0=2c_2x^2-6c_2x+3c_2 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=3: 2c_3x^3-3c_3(3)x^2+c_3(3)(2)x^1=2c_3x^3-9c_3x^2+6c_3x $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=4: 2c_4x^4-3c_4(4)x^3+c_4(4)(3)x^2=2c_4x^4-12c_4x^3+12c_4x^2 $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0"

j=5: 2c_5x^5-3c_5(5)x^4+c_5(5)(4)x^3=2c_5x^5-15c_5x^4+20c_5x^3 $$     (5.6) By summing the terms in (5.6) and grouping like terms, then equating it to the $$\displaystyle y_p(x)$$, we find the coefficient equations are the same as Eqs. (5.5):
 * $$\displaystyle
 * $$\displaystyle
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 * }
 * {| style="width:100%" border="0"

[2c_0-3c_1+2c_2]x^0+[2c_1-6c_2+6c_3]x^1+[2c_2-9c_3+12c_4]x^2+[2c_3-12c_4+20c_5]x^3+[2c_4-15c_5]x^4+2c_5x^5=0x^0+0x^1+4x^2+0x^3+0x^4-6x^5 $$     (5.7)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

The linear system of equations in Eqs. (5.5) can be placed in matrix form:
 * {| style="width:100%" border="0"

$$ \left[ \begin{array}{cccccc} 2 & -3 & 2 & 0 & 0 & 0\\ 0 & 2 & -6 & 6 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0\\ 0 & 0 & 0 & 2 & -12 & 20\\ 0 & 0 & 0 & 0 & 2 & -15\\ 0 & 0 & 0 & 0 & 0 & 2\end{array} \right]\ \left( \begin{array}{cccccc} {c}_{0}\\ {c}_{1}\\ {c}_{2}\\ {c}_{3}\\ {c}_{4}\\ {c}_{5}\end{array} \right)\ = \left[ \begin{array}{cccccc} 0\\ 0\\ 4\\ 0\\ 0\\ -6\end{array} \right]\ $$ (5.8)
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 * <p style="text-align:right">
 * }

Solving this system of equations by back-substitution yields the values of the coefficients for the $$\displaystyle y_p(x)$$:
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c=\{-701.75, -691.5, -335.5, -105, -22.5, -3\} $$     (5.9)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Substituting these coefficients into Eq. (5.1):
 * {| style="width:100%" border="0"

y_p(x)=-701.75x^0-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5 $$     (5.10)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

The general solution is the sum of the homogeneous (found in R3.3) and particular solutions:
 * {| style="width:100%" border="0"

y(x)=-701.75-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5+C_1e^{2x}+C_2e^x $$     (5.11)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Considering the initial value condition $$\displaystyle y(0)=1$$ in the general solution:
 * {| style="width:100%" border="0"

y(0)=1=-701.75+C_1e^0+C_2e^0=-701.75+C_1+C_2 $$     (5.12) Taking the first derivative of Eq. (5.11) to consider the initial value condition $$\displaystyle y'(0)=0$$:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

y'(x)=-691.5-671x-315x^2-90x^3-15x^4+2C_1e^{2x}+C_2e^x $$     (5.13)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

y'(0)=0=-691.5+2C_1+C_2 $$     (5.14)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:
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C_1=-11.25, C_2=714 $$     (5.15)
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right">
 * }

This yields our final solution by substituting the coefficients into Eq. (5.11): Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:
 * {| style="width:100%" border="1"

$$\displaystyle y(x)=-701.75-691.5x^1-335.5x^2-105x^3-22.5x^4-3x^5-11.25e^{2x}+714e^x $$     (5.16)
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 * <p style="text-align:right">
 * }

A plot of the solution: Figure 3.5-1

Author
Solved and Typed By - --Egm4313.s12.team1.durrance 22:56, 21 February 2012 (UTC)--

Reviewed By - Egm4313.s12.team1.silvestri 03:28, 22 February 2012 (UTC)

Statement
Solve the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6 differently as follows. Consider the following two L2-ODEs-CC (see p.7-2b):
 * {| style="width:100%" border="0"

$$\displaystyle{y}_{p,1}''-3{y}_{p,1}'+2{{y}_{p,1}}={{r}_{1}}(x):=4{{x}^{2}}$$ (6.0)
 * style="width:95%" |
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle{y}_{p,2}''-3{y}_{p,2}'+2{{y}_{p,2}}={{r}_{2}}(x):=-6{{x}^{5}}$$ (6.1) The particular solution $$\displaystyle{y}_{p,1}$$ had been found in R3.3 p.7-11. Find the particular solution $$\displaystyle{y}_{p,2}$$, and then obtain the solution $$\displaystyle{y}$$ for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.
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 * <p style="text-align:right">
 * }

Compare the result with that obtained in R3.5.

Solution
Beginning with equation (6.1), we find that a particular solution has the form
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$$\displaystyle{{y}_{p,2}}(x)=\sum\limits_{j=0}^{n}$$ (6.2) where $$\displaystyle{n=5}$$. That is,
 * style="width:95%" |
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 * }
 * {| style="width:100%" border="0"

$$\displaystyle {{y}_{p,2}}(x)={{K}_{0}}+{{K}_{1}}x+{{K}_{2}}{{x}^{2}}+{{K}_{3}}{{x}^{3}}+{{K}_{4}}{{x}^{4}}+{{K}_{5}}{{x}^{5}}$$
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 * style="width:95%" |

(6.3) Differentiating twice:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle{y}_{p,2}'(x)={{K}_{1}}+2{{K}_{2}}x+3{{K}_{3}}{{x}^{2}}+4{{K}_{4}}{{x}^{3}}+5{{K}_{5}}{{x}^{4}}$$ (6.4)
 * style="width:95%" |
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle{y}_{p,2}''(x)=2{{K}_{2}}+6{{K}_{3}}x+12{{K}_{4}}{{x}^{2}}+20{{K}_{5}}{{x}^{3}}$$ (6.5) Substitute (6.3-5) into (6.1) to obtain
 * style="width:95%" |
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle (2{{K}_{2}}+6{{K}_{3}}x+12{{K}_{4}}{{x}^{2}}+20{{K}_{5}}{{x}^{3}})-3({{K}_{1}}+2{{K}_{2}}x+3{{K}_{3}}{{x}^{2}}+4{{K}_{4}}{{x}^{3}}+5{{K}_{5}}{{x}^{4}})$$ $$\displaystyle+2({{K}_{0}}+{{K}_{1}}x+{{K}_{2}}{{x}^{2}}+{{K}_{3}}{{x}^{3}}+{{K}_{4}}{{x}^{4}}+{{K}_{5}}{{x}^{5}})=-6{{x}^{5}}$$
 * style="width:95%" |
 * style="width:95%" |

(6.6) Rearranging terms with respect to $$\displaystyle{x}$$ power:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle (2{{K}_{2}}-3{{K}_{1}}+2{{K}_{0}})+x(6{{K}_{3}}-6{{K}_{2}}+2{{K}_{1}})+{{x}^{2}}(12{{K}_{4}}-9{{K}_{3}}+2{{K}_{2}})$$ $$\displaystyle+{{x}^{3}}(20{{K}_{5}}-12{{K}_{4}}+2{{K}_{3}})+{{x}^{4}}(-15{{K}_{5}}+2{{K}_{4}})+{{x}^{5}}(2{{K}_{5}})=-6{{x}^{5}}$$ (6.7) In matrix form:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \left[ \begin{array}{cccccc} 2 & -3 & 2 & 0 & 0 & 0\\ 0 & 2 & -6 & 6 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0\\ 0 & 0 & 0 & 2 & -12 & 20\\ 0 & 0 & 0 & 0 & 2 & -15\\ 0 & 0 & 0 & 0 & 0 & 2\end{array} \right]\ \left( \begin{array}{cccccc} {K}_{0}\\ {K}_{1}\\ {K}_{2}\\ {K}_{3}\\ {K}_{4}\\ {K}_{5}\end{array} \right)\ = \left[ \begin{array}{cccccc} 0\\ 0\\ 0\\ 0\\ 0\\ -6\end{array} \right]\ $$ (6.8) Solving for $$\displaystyle K$$, using MATLAB, yields
 * style="width:95%" |
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle K=\{-708.75,-697.5,-337.5,-105,-22.5,-3\}$$ (6.9) which means that
 * style="width:95%" |
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle {y}_{p,2}=-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}}$$ (6.10) From R3.3:
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle {{y}_{p,1}}=32+12x+2{{x}^{2}}$$ (6.11) Summing for the final solution:
 * style="width:95%" |
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y={{C}_{1}}{{y}_{p,1}}+{{C}_{2}}{{y}_{p,2}}$$ (6.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y={{C}_{1}}(32+12x+2{{x}^{2}})$$ $$\displaystyle +{{C}_{2}}(-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}})$$ (6.13) For initial condition $$\displaystyle y(0)=-5$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y(0)=-5={{C}_{1}}(32)+{{C}_{2}}(-708.75)$$ (6.14) Similarly, for initial condition $$\displaystyle y'(0)=2$$
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y'(0)=2={{C}_{1}}(12)+{{C}_{2}}(-697.5)$$ (6.15) Equations (6.14) and (6.15) yield the following matrix equation:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \left[ \begin{array}{cc} 32 & -708.75\\ 12 & -697.5\end{array} \right]\ \left( \begin{array}{cc} {C}_{1}\\ {C}_{2}\end{array} \right)\ = \left[ \begin{array}{cc} -5\\ 2\end{array} \right]\ $$ (6.16) Solving (6.16) in MATLAB yields
 * style="width:95%" |
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle C=\{-0.355,-0.009\}$$ (6.17) Therefore the combined solution of $$\displaystyle y$$ is
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=-0.355(32+12x+2{{x}^{2}})$$ $$\displaystyle -0.009(-708.75-697.5x-337.5{{x}^{2}}-105{{x}^{3}}-22.5{{x}^{4}}-3{{x}^{5}})$$ (6.18) Which can be simplified to Final Equation
 * style="width:95%" |
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 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="1"

$$  \displaystyle y(x)=-4.98125+2.0175x+2.3275{{x}^{2}}+0.945{{x}^{3}}+0.2025{{x}^{4}}+0.027{{x}^{5}}$$ (6.19)
 * style="width:95%" |
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 * <p style="text-align:right">
 * }

Author
Solved and Typed By - Egm4313.s12.team1.essenwein 00:40, 18 February 2012 (UTC)

Reviewed By - Egm4313.s12.team1.silvestri 04:29, 22 February 2012 (UTC)

Statement
Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities. These equalities are:
 * {| style="width:100%" border="0"


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$$ \sum_{j=2}^{5} C_{j} j(j-1) x^{^{j-2}}=\sum_{j=0}^{3} C_{j+2} (j+2)(j+1) x^{^{j}} \!$$ (7.0)
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{j=1}^{5} C_{j} jx^{^{j-1}}=\sum_{j=0}^{4} C_{j+1} (j+1) x^{^{j}} \!$$ (7.1)
 * <p style="text-align:right">
 * }

Solution
This transition occurs through a mid-level variable change. In this particular case, for (7.0), j-2 is represented by k:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{k=0}^{3} C_{k+2} (k+2)(k+1) x^{^{k}} \!$$ (7.3)
 * <p style="text-align:right">
 * }

We can then represent the variable k with a j. This new summation looks like this:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{j=0}^{3} C_{j+2} (j+2)(j+1) x^{^{j}} \!$$ (7.4)
 * <p style="text-align:right">
 * }

Expanding both sides, (7.0 and 7.4) yields the same result:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$C_{5}(5)(4)x^{^{3}} + C_{4}(4)(3)x^{^{2}}+C_{3}(3)(2)x^{^{1}} +C_{2}(2)(1)x^{^{0}}\!$$ (7.5)
 * <p style="text-align:right">
 * }

We follow the same process for equality (7.1): This transition occurs through a mid-level variable change. In this particular case, for (7.1), j-1 is represented by k:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{k=0}^{4} C_{k+1} (k+1) x^{^{k}} \!$$ (7.6)
 * <p style="text-align:right">
 * }

We can then represent the variable k with a j. This new summation looks like this:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$ \sum_{j=0}^{4} C_{j+1} (j+1) x^{^{j}} \!$$ (7.7)
 * <p style="text-align:right">
 * }

Expanding both sides, (7.1 and 7.7)yields the same result:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$C_{5}(5)x^{^{4}} + C_{4}(4)x^{^{3}}+C_{3}(3)x^{^{2}} +C_{2}(2)x^{^{1}}+C_{1}\!$$ (7.8)
 * <p style="text-align:right">
 * }

Author
Solved and Typed By - Egm4313.s12.team1.silvestri 17:20, 19 February 2012 (UTC)

Reviewed By - --128.227.113.77 17:05, 22 February 2012 (UTC)

Statement
Find a general solution for the following two problems:

Given
Homogeneous solution: $$ y'' + 4y' + 4y = e^{-x}cosx\!$$

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$. First, we will find the general solution, $$ y_g(x) \!$$, by finding the roots of the characteristic equation:


 * {| style="width:100%" border="0"

$$\displaystyle{\lambda^2 + 4\lambda + 4\lambda = 0 \rightarrow (\lambda + 2)^2}$$ (8.0)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Which means that the characteristic equation has a double root of $$ \lambda = -2 \!$$. Based on the double root, then, the general solution is:


 * {| style="width:100%" border="0"

$$\displaystyle{y_g(x) = (C_1 + C_2x)e^{-\frac{ax}{2}} \rightarrow y_g(x) = (C_1 + C_2x)e^{-2x}}$$ (8.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Next, we need to find the particular solution, and based on analysis of the excitation, $$ r(x) = e^{-x}cosx \!$$, we find that the particular solution is the following:


 * {| style="width:100%" border="0"

$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-x}Kcosx + e^{-x}Msinx}$$ (8.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now, we need to find the values for the constants $$ K, M\!$$, by taking the first and second derivatives of the particular solution (8.2):


 * {| style="width:100%" border="0"

$$\displaystyle{y_p(x) = e^{-x}Kcosx + e^{-x}Msinx}$$ (8.2)
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 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\displaystyle{y_p(x)' = -e^{-x}Ksinx - e^{-x}Kcosx + e^{-x}Mcosx - e^{-x}Msinx}$$ (8.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\displaystyle{y_p(x)'' = -e^{-x}Kcosx + e^{-x}Ksinx + e^{-x}Ksinx + e^{-x}Kcosx - e^{-x}Msinx - e^{-x}Mcosx - e^{-x}Mcosx + e^{-x}Msinx}$$ (8.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now, plug these into the homogeneous solution and simplify:


 * {| style="width:100%" border="0"

$$\displaystyle{-e^{-x}(Kcosx - Ksinx - Ksinx - Kcosx + Msinx + Mcosx + Mcosx - Msinx) + 4[-e^{-x}(Ksinx + Kcosx - Mcosx + Msinx)] + 4[e^{-x}(Kcosx + Msinx)] = e^{-x}cosx}$$ (8.5)
 * style="width:95%" |
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 * <p style="text-align:right">
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$$\displaystyle{-e^{-x}[-2Ksinx + 2Mcosx] - e^{-x}[4Ksinx + 4Kcosx - 4Mcosx + 4Msinx] + e^{-x}[4Kcosx + 4Msinx] = e^{-x}cosx}$$ (8.6)
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Pull $$ e^{-x} \! $$ out of equation (8.6) and simplify:


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$$\displaystyle{e^{-x}[4Kcosx + 4Msinx - 4Ksinx - 4Kcosx + 4Mcosx - 4Msinx + 2Ksinx - 2Mcosx] = e^{-x}cosx}$$ (8.7)
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$$\displaystyle{e^{-x}[-2Ksinx + 2Mcosx] = e^{-x}cosx}$$ (8.8)
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Now, solve for the unknown coefficients:


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$$\displaystyle{2M = 1 \rightarrow M = \frac{1}{2}}$$ (8.9)
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$$\displaystyle{-2K = 0 \rightarrow K = 0}$$ (8.10)
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Plug into $$ y_p(x)\!$$ (8.2):


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$$\displaystyle{y_p(x) = \frac{1}{2}e^{-x}sinx}$$ (8.11)
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Now, we have both the particular solution and the general solution, which allows for us to solve for the final solution, $$ y(x) \! $$

$$  \displaystyle y(x) = (C_1 + C_2x)e^{-2x} + \frac{1}{2}e^{-x}sinx $$     (8.12)
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Given
Homogeneous solution: $$ y'' + y' (\pi^2 + \frac{1}{4}) = e^{-\frac{x}{2}}sin{\pi x}\!$$

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$. First, we will find the general solution, $$ y_g(x) \!$$, by finding the roots of the characteristic equation, using the quadratic equation:


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$$\displaystyle{\lambda^2 + \lambda + (\pi^2 + \frac{1}{4}) = 0 \rightarrow \lambda = \frac{-1 \pm \sqrt{1-4(1)(\pi^2 + \frac{1}{4})}}{2}}$$ (8.13)
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$$\displaystyle{\lambda = -\frac{1}{2} + i\pi}$$ (8.14)
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The roots of the characteristic equation are complex conjugates, meaning that the general solution, $$ y_g(x) \!$$, is the following:


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$$\displaystyle{y_g(x) = e^{-\frac{x}{2}}(Acos{\pi x} + Bsin{\pi x})}$$ (8.15)
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Next, we need to find the particular solution, and based on analysis of the excitation, $$ r(x) = e^{-\frac{x}{2}}sin{\pi x} \!$$, we find that the particular solution is the following:


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$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-\frac{x}{2}}Kcos{\pi x} + e^{-\frac{x}{2}}Msin{\pi x}}$$ (8.16)
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Now, we need to find the values for the constants $$ K, M \!$$, by taking the first and second derivatives of the particular solution (8.16):


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$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-\frac{x}{2}}Kcos{\pi x} + e^{-\frac{x}{2}}Msin{\pi x}}$$ (8.16)
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$$\displaystyle{y_p(x)' = {-\pi e^{\frac{-x}{2}}Ksin{\pi x} - \frac{{}e^{\frac{-x}{2}}}{2}Kcos{\pi x}} + {\pi e^{\frac{-x}{2}}Mcos{\pi x} + \frac{{}e^{\frac{-x}{2}}}{2}Msin{\pi x}}}$$ (8.17)
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$$\displaystyle{y_p(x)'' = {-\pi^2 e^{\frac{-x}{2}}Kcos{\pi x} +  \pi \frac{e^{\frac{-x}{2}}}{2}Ksin{\pi x}}  +  \pi \frac{e^{\frac{-x}{2}}}{2}Ksin{\pi x}  + \frac{e^{\frac{-x}{2}}}{4}Kcos{\pi x}  - \pi^2 {e^{\frac{-x}{2}}}Msin{\pi x} - \pi \frac{e^{\frac{-x}{2}}}{2}{}Mcos{\pi x} - \pi \frac{e^{\frac{-x}{2}}}{2}{}Mcos{\pi x} + \frac{e^{\frac{-x}{2}}}{4}{}Msin{\pi x}}$$ (8.18)
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Now, we plug these into the homogeneous solution and simplify but we find that everything cancels out, meaning the unknown constants, $$ M, K \!$$ are both equal to $$ 0 \! $$. This means that the final solution is equal to just the general solution:

$$  \displaystyle y(x) = e^{\frac{-x}{2}}(Acos{\pi x} + Bsin{\pi x}) $$     (8.19)
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Author
Solved and Typed By ---Egm4313.s12.team1.wyattling 22:10, 20 February 2012 (UTC)

Reviewed By - Egm4313.s12.team1.rosenberg 03:07, 22 February 2012 (UTC)

Statement
K 2011 page 85 problems 13 and 14 Problem 13: Find the complete solution for $$ 8y'' - 6y' + y = 6\cosh x\!$$, with the initial conditions $$ y(0) = 0.2, y'(0) = 0.05\!$$ Problem 14: Find the complete solution for $$ y'' + 4y' + 4y = e^{-2x}\sin 2x\!$$, with the initial conditions $$ y(0) = 1, y'(0) = -1.5\!$$

Solution
Problem 13: In order to solve the equation we must use the definition $$ y=y_{h} + y_{p}\!$$. We start with the given equation:
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$$\displaystyle{8y''-6y'+y=6\cosh x}$$ (9.1) From this we get the homogeneous characteristic equation:
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$$\displaystyle{8 \lambda ^2-6 \lambda +1=0 x}$$ (9.2) Factoring the characteristic equation gives us:
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$$\displaystyle{(4 \lambda -1)(2 \lambda -1)=0}$$ (9.3) This give us the roots:
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$$\displaystyle{\lambda = \frac {1}{4}, \frac {1}{2}}$$ (9.4) Plugging in the roots gives us the general homogeneous solution:
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$$\displaystyle{y_{h} = c_{1} e^{\frac {1}{4} x}+c_{2} e^{\frac {1}{2} x}}$$ (9.5) Now we must solve for the particular solution. So far we know:
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$$\displaystyle{\cosh x = \frac {e^x +e^{-x}}{2}}$$ (9.6) Now using the sum rule, $$\displaystyle{y_{p} = y_{p1} + y_{p2}}$$, gives us:
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$$\displaystyle{y_{p} = Ce^x + Ke^{-x}}$$ (9.7) Now we must take the first and second derivatives of equation 9.7:
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$$\displaystyle{y_{p}' = Ce^x - Ke^{-x}}$$ (9.8)
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$$\displaystyle{y_{p}'' = Ce^x + Ke^{-x}}$$ (9.9) Substituting 9.8 and 9.9 back into our original equation gives us:
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$$\displaystyle{8(Ce^x+Ke^{-x})-6(Ce^x-Ke^{-x})+Ce^x+Ke^{-x}=6(\frac {e^x+e^{-x}}{2})}$$ (9.10) Simplifying equation 9.10 give us:
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$$\displaystyle{3Ce^x+15Ke^{-x}=3(e^x+e^{-x})}$$ (9.11) From this we can deduce that:
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$$\displaystyle{3C=3, 15K=3}$$ (9.12) Thus we get that:
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$$\displaystyle{C=1, K= \frac{1}{5}}$$ (9.13) We now have our particular equation:
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$$\displaystyle{y_{p} = e^x + \frac{1}{5} e^{-x}}$$ (9.14) Thus:
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$$\displaystyle{y= c_{1} e^{\frac {1}{4} x}+c_{2} e^{\frac {1}{2} x}+e^x + \frac{1}{5} e^{-x}}$$ (9.15) Now we can solve for $$ c_{1}\!$$ and $$ c_{2}\!$$ using the given initial values. We are given that at $$ y(0)=0.2\!$$, we can now plug this into our equation to get:
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$$\displaystyle{0.2= c_{1} e^{\frac {1}{4} (0)}+c_{2} e^{\frac {1}{2} (0)}+e^{(0)} + \frac{1}{5} e^{-(0)}}$$ (9.16) This simplifies to:
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$$\displaystyle{0.2= c_{1}+c_{2}+1+ \frac{1}{5}}$$ (9.17) Giving us that:
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$$\displaystyle{c_{1}+c_{2} = -1}$$ (9.18) For our second condition, $$y'(0)=0.05\!$$, we must we must take the derivative of our general solution:
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$$\displaystyle{y'= \frac {1}{4} c_{1} e^{\frac {1}{4} x}+ \frac {1}{2} c_{2} e^{\frac {1}{2} x}+e^x - \frac{1}{5} e^{-x}}$$ (9.19) At our initial condition gives us:
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$$\displaystyle{0.05= \frac {1}{4} c_{1} + \frac {1}{2} c_{2} +1 - \frac{1}{5}}$$ (9.20) Thus:
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$$\displaystyle{c_{1}+2c_{2}=-3}$$ (9.21) Combining 9.18 and 9.21 gives us that:
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$$\displaystyle{c_{1}=1, c_{2}=-2}$$ (9.22) Plugging these coefficients in gives us our final solution:
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$$\displaystyle{y= e^{\frac {1}{4} x}-2e^{\frac {1}{2} x}+e^x + \frac{1}{5} e^{-x}}$$ (9.23)
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Problem 14: In order to solve the equation we must use the definition $$ y=y_{h} + y_{p}\!$$. We start with the given equation:
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$$\displaystyle{y''+4y'+4y=e^{-2x} \sin 2x}$$ (9.24) From this we get the homogeneous characteristic equation:
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$$\displaystyle{\lambda ^2 +4 \lambda +4 = 0}$$ (9.25) Factoring the characteristic equation gives us:
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$$\displaystyle{(\lambda +2)(\lambda +2)=0}$$ (9.26) This give us the double root:
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$$\displaystyle{\lambda = -2}$$ (9.27) Plugging in the roots gives us the general homogeneous solution:
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$$\displaystyle{y_{h} = c_{1} e^{-2x}+c_{2} xe^{-2x}}$$ (9.28) Now we must solve for the particular solution. So far we know:
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$$\displaystyle{e^{-2x}\sin 2x = Cxe^{-2x}\cos 2x+Kxe^{-2x}\sin 2x}$$ (9.29) Now we must take the first and second derivatives of equation 9.29:
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$$\displaystyle{y_{p}' = -2e^{-2x}(Cx\cos 2x+Kx\sin 2x)+e^{-2x}(C\cos 2x-2Cx\sin 2x+K\sin 2x+2Kx\cos 2x)}$$ (9.30)
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$$\displaystyle{y_{p}'' = (-4C+4K)e^{-2x}\cos 2x+(-4k-4C)e^{-2x}\sin 2x}$$ (9.31) Substituting 9.30 and 9.31 back into our original equation gives us:
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$$\displaystyle{(-3C+4K)e^{-2x}\cos 2x+(-3k-4C)e^{-2x}\sin 2x=e^{-2x}\sin 2x}$$ (9.32) From this we can deduce that:
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$$\displaystyle{-3C+4K=0,-3K-4C=1}$$ (9.33) Thus we get that:
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$$\displaystyle{C=\frac{-4}{25}, K= \frac{-3}{25}}$$ (9.34) We now have our particular equation:
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$$\displaystyle{y_{p} = e^{-2x}(\frac {-4}{25}x\cos 2x-\frac {3}{25}x\sin 2x)}$$ (9.35) Thus:
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$$\displaystyle{y=c_{1} e^{-2x}+c_{2} xe^{-2x}+e^{-2x}(\frac {-4}{25}x\cos 2x-\frac {3}{25}x\sin 2x)}$$ (9.36) Now we can solve for $$ c_{1}\!$$ and $$ c_{2}\!$$ using the given initial values. We are given that at $$ y(0)=1\!$$, we can now plug this into our equation to get:
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$$\displaystyle{y(0)=(c_{1} +c_{2} x)e^{-2(0)}+0=1}$$ (9.37) Giving us that:
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$$\displaystyle{c_{1} = 1}$$ (9.38) For our second condition, $$y'(0)=-1.5\!$$, we must we must take the derivative of our general solution:
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$$\displaystyle{y'= -2c_{1}e^{-2x}+c_{2} (-2xe^{-2x}+e^{-2x})+\frac {-4}{25} e^{-2x}\cos 2x - \frac {3}{25} e^{-2x}\sin 2x+\frac {2}{25}e^{-2x}x\cos 2x + \frac {14}{25}e^{-2x}x\sin 2x}$$ (9.39) At our initial condition we have:
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$$\displaystyle{y'(0)=-2+c_{2} -\frac{4}{25} -0 +0 +0 =0}$$ (9.40) Thus:
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$$\displaystyle{c_{2}=\frac {54}{25} }$$ (9.41) Plugging these coefficients in gives us our final solution:
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$$\displaystyle{y=1 e^{-2x}+\frac {54}{25} xe^{-2x}+e^{-2x}(\frac {-4}{25}x\cos 2x-\frac {3}{25}x\sin 2x)}$$ (9.42)
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Author
Solved and Typed By - Egm4313.s12.team1.rosenberg 03:04, 22 February 2012 (UTC)

Reviewed By - Egm4313.s12.team1.essenwein 02:39, 22 February 2012 (UTC)