User:Egm6322.s12.team2.steele.m2/Mtg36

=EGM6321 - Principles of Engineering Analysis 1, Fall 2011=

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Plan: $$\Rightarrow$$Application: Obtaining L2-ODE-VC from characteristic equation $$\Rightarrow$$Application: Trial solution with singularity $$\Rightarrow$$WolframAlpha (Mathematica) syntax $$\Rightarrow$$Obtian L2-ODE-VC from known trial solution and characteristic equation

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=Application: Obtaining L2-ODE-VC from Characteristic Equation= Reverse engineering. Consider two roots: Hence:

Since $$r_2(x)$$ is not a constant, there is only one valid root, i.e., $$r_1 = 2$$. Consider the following L2-ODE-VC:

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=With the trial solution:= $$y = e^{rx}$$, r = constant		[[media:Pea1.f11.mtg25.djvu|Eqn(4) p.25-4]] We recover the same characteristic equation (3) p.36-1. So if there is at least one constant root, the above inverse procedure can be used to generate L2-ODE-VC given the two roots, or equivalently the characteristic equation. Consider another example: Given

Characteristic equation:

(3)	Cannot be used in the above inverse procedure to generate a L2-ODE-VC having the trial solution $$e^{xr_3(x)}$$	or $$ex^{xr_4(x)}$$		Why?

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R*6.5: Find $$u_1(x), u_2(x)$$	(2 homogeneous solutions) of (5) p.36-1 using the trial solution $$y(x) = e^{rx}$$ Application: Trial solution with singularity Consider the L2-ODE-VC: (see also F09 p.22-2) Given the 1st homogeneous solution:

1)	Verify exactness of (1)	HW* 2)	Can you use the iFM? HW* 3)	Trial solutions

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Show that  is as shown in (2) p.36-3. 4)	Find $$u_2(x)$$ using variation of parameters. Substitute (1) in (1) p.36-3 to find (by hand or use WA):

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WolframAlpha (Mathematica) syntax Substitute (1) p.36-4 in (1) p.36-3 using WA. 1)	For $$\frac{d^2}{dx^2} \frac{e^{rx}}{x}$$ use $$D^2[e^{rx}/x] / dx^2$$ Note: You can use either square brackets[] or parentheses. 2)	For $$x\frac{d^2}{dx^2} \frac{e^{rs}}{x}-2\frac{d}{dx} \frac{e^{rx}}{x}$$ Xd^2(e^{rx}/x)/dx^2 – [2d(e^{rx}/x)/dx] Note the square brackets[] surrounding the 2nd term; try to remove the square brackets, then you would get a completely different expression. The reason is because of the confusion in interpreting d^2 and dx^2.

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So brackets [] deleted Xd^2(e^{rx}/x)/dx^2 – 2d(e^{rx}/x)/dx Leads to: $$xd^2 \frac{\displaystyle \frac{e^{rx}}{x}}{\displaystyle \frac{\partial}{\partial x} x^{\displaystyle 2-2d \frac{e^{rx}}{x}}}$$ Which is the same as Xd^2(e^{rx}/x)/[d[x^(2 – 2d(e^{rx}/x))]/dx] Added Alternatively, use the “derivative” function: X*derivative(derivative(e^(rx)/x, x), x) -2*derivative(e^(rx)/x, x)

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3)	For $$x\frac{d^2}{dx^2} \frac{e^{rx}}{x}-2\frac{d}{dx} \frac{e^{rx}}{x} = x\frac{e^{rx}}{x}$$ Xd^2(e^{rx}/x)/dx^2-[2d(e^{rx}/x)/dx]+[x(d^[rx}/x)] Or bracket only the 2nd term: Xd^2(e^{rx}/x)/dx^2 + [2d(e^{rx}/x)/dx] + x(e^{rx}/x) But bracket only the 1st term does not work. When in doubt, bracket! For more details 	$$\Rightarrow$$http://en.wikiversity.org/wiki/User:Egm6321.f11/WolframAlpha_syntax 	$$\Rightarrow$$-> FE1 S11 Team 7 HW6.7: Syntax of Wolfram Alpha 	$$\Rightarrow$$http://en.wikiversity.org/w/index.php?title=User:Eml5526.s11.team7/HW6&oldid=732398#Part_3:_Syntax_of_WolframAlpha

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Obtain L2-ODE-VC from known trial solution and characteristic equation

Add the following expressions:

And equate the result to the product of the trial solution by the characteristic equation:

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3 equations for $$a_2, a_1, a_0$$

Thus:

Which then leads to the following L2-ODE-VC:

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Which is the same as (1) p.36-3. R*6.6: Find a homogeneous L2-ODE-VC that accepts

(1)-(3) p.36-1				F11