User:Egm4313.s12.team1.wyattling/r3

=Problems for Report 3=

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

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$.

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, 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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$$\displaystyle{y_p(x) = Cx^2e^{5x} - [K_2x^2 + K_1x + K_0]}$$ (1.5)
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$$\displaystyle{y_p(x)' = 2Cxe^{5x} + 5Cx^2e^{5x} - [2K_2x + K_1]}$$ (1.6)
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$$\displaystyle{y_p(x)'' = 2Ce^{5x} + 20Cxe^{5x} + 25Cx^2e^{5x} - [2K_2]}$$ (1.7)
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Now, plug these into the homogeneous solution (1.2) and simplify:


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$$\displaystyle{2Ce^{5x} - 2K_2 + 20K_2x + 10K_1 - 25K_2x^2 - 25K_1x - 25K_0 = 7e^{5x} - 2x^2}$$ (1.8)
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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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$$\displaystyle{e^{5x}: 2C = 7 \rightarrow C = 7/2}$$ (1.9)
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$$\displaystyle{x^2: -25K_2 = -2 \rightarrow K_2 = 0.08}$$ (1.10)
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$$\displaystyle{x: 20K_2 - 25K_1 = 0 \rightarrow K_1 = 0.064}$$ (1.11)
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$$\displaystyle{-2K_2 + 10K_1 - 25K_0 = 0 \rightarrow K_0 = 0.0192}$$ (1.12)
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Plug into $$ y_p(x)\!$$:


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$$\displaystyle{y_p(x) = \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192]}$$ (1.13)
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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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$$\displaystyle{y(x) = C_1e^{5x} + C_2xe^{5x} + \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192]}$$ (1.14)
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Take derivative:


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$$\displaystyle{y(x)' = 5C_1e^{5x} + C_2e^{5x} +5C_2xe^{5x} + 7xe^{5x} + \frac{35}{2}x^2e^{5x} - [0.16x + 0.064]}$$ (1.15)
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Using the given initial conditions:


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$$\displaystyle{y(0) = 4 \rightarrow 4 = C_1 - 0.0192 \rightarrow C_1 = 4.0192}$$ (1.16)
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$$\displaystyle{y'(0) = 5 \rightarrow 5 = 5C_1 + C_2 - 0.064 \rightarrow C_2 = -15.032}$$ (1.17)
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Plug these constants back into the solution to obtain the final solution:

$$  \displaystyle y(x) = 4.0192e^{5x} - 15.032xe^{5x} + \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192] $$     (1.18)
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Author
Solved and Typed By -

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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:


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$$\displaystyle{\lambda^2 + 4\lambda + 4\lambda = 0 \rightarrow (\lambda + 2)^2}$$ (8.0)
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Which means that the characteristic equation has a double root of $$ \lambda = -2 \!$$.

Based on the double root, then, the general solution is:


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$$\displaystyle{y_g(x) = (C_1 + C_2x)e^{-\frac{ax}{2}} \rightarrow y_g(x) = (C_1 + C_2x)e^{-2x}}$$ (8.1)
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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:


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$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-x}Kcosx + e^{-x}Msinx}$$ (8.2)
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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 (8.2):


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$$\displaystyle{y_p(x) = e^{-x}Kcosx + e^{-x}Msinx}$$ (8.2)
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$$\displaystyle{y_p(x)' = -e^{-x}Ksinx - e^{-x}Kcosx + e^{-x}Mcosx - e^{-x}Msinx}$$ (8.3)
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$$\displaystyle{y_p(x)'' = -e^{-x}Kcosx + e^{-x}Ksinx + e^{-x}Ksinx + e^{-x}Kcosx - e^{-x}Msinx - e^{-x}Mcosx - e^{-x}Mcosx + e^{-x}Msinx}$$ (8.4)
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Now, plug these into the homogeneous solution and simplify:


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$$\displaystyle{-e^{-x}(Kcosx - Ksinx - Ksinx - Kcosx + Msinx + Mcosx + Mcosx - Msinx) + 4[-e^{-x}(Ksinx + Kcosx - Mcosx + Msinx)] + 4[e^{-x}(Kcosx + Msinx)] = e^{-x}cosx}$$ (8.5)
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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
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