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Variation of parameters 
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#1
Aug1511, 06:31 AM

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1. The problem statement, all variables and given/known data
Solve for general solution with variation of parameter [tex]y'''(x)  y'(x) = x[/tex] 3. The attempt at a solution I initially looked at [tex]y'''(x)  y'(x) = x[/tex] only and I foudn my answer to be [tex]y(x) = C_1e^{x} + C_2e^{x} + 1  x[/tex] Now i looked through my book and it says it works for ay'' + by' + c = f(t) only (second order). So I "integrated" [tex]y'''(x)  y'(x) = x[/tex] And I got [tex]y''(x)  y(x) = \frac{x^2}{2} + C[/tex], solving I got [tex]y(x) = C_1e^{x} + C_2e^{x} + C_3  \frac{x^2}{2}[/tex] Using the computer, it gave me http://www.wolframalpha.com/input/?i=Solve[y%27%27%27++y%27+%3D+x] Why does computer have negative sign?? 


#2
Aug1511, 08:22 AM

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I have no idea what you mean when you say you "looked at y''' y'= x only". That is the entire problem! Do you mean the associated homogeneous equation, y''' y'= 0? That has characteristic equation [itex]r^3 r= r(r^2 1)= r(r 1)(r+ 1)= 0[/itex] so the general solution to the associated homogeneous equation is [itex]y_h(x)= C_1e^x+ C_2e^{x}+ C_3[/itex]
Now, you have title this "variation of parameters". To use that method for this differential equation ("undetermined coefficients" would be simpler but I presume this is for practice of variation of parameters specifically), look for a solution of the form [tex]y(x)= u(x)e^x+ v(x)e^{x}+ w(x)[/tex] That is, we treat those constants (parameters) as variables that is the reason for the name "variation of parameters". Differentiating, [itex]y'= u'e^x+ ue^x+ v'e^{x} ve^{x}+ w'[/itex]. There will be many different possible functions that will work we limit our search by requiring that [itex]u'e^x+ v'e^x+ w'= 0[/itex] That means that we have [itex]y'= ue^x ve^{x}[/itex]. Differentiating again, [itex]y''= u'e^x+ ue^x v'e^{x}+ ve^{x}[/itex]. Once again, we "limit our search" (and simplify the equation) by requiring that [itex]u'e^x v'e^{x}= 0[/itex]. That gives, now, [itex]y''= ue^{x}+ ve^{x}[/itex] and, differentiating one more time, that [itex]y'''= u'e^{x}+ ue^{x}+ v'e^{x} ve^{x}[/itex]. Putting that and the formula for y' into the equation, [tex]u'e^x+ ue^x+ v'e^{x} ve^{x} (ue^x ve^{x})= u'e^x+ v'e^{x}= x[/tex] You see what has happened? We have no derivative of u, v, and w higher than first because of our "requirements" and we do not have u, v, and w themselves because they satify the homogenous equation and cancel. We have, instead the three equations [tex]u'e^x+ v'e^x+ w'= 0[/tex] [tex]u'e^x v'e^{x}= 0[/tex] and [tex]u'e^x+ v'e^{x}= x[/tex] which we can treat as three linear equations in u', v', and w'. Solve for those and integrate to find u, v, and w. 


#3
Aug1511, 10:52 AM

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#4
Aug1611, 02:12 AM

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Variation of parameters
HallsofIvy, my book did the same thing, they imposed on the requirement that
[tex]u'e^x+ v'e^x+ w'= 0[/tex] Why? Was I right though initially when I 'integrated' [tex]y'''  y' = x[/tex] 


#5
Aug1611, 06:57 AM

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Yes, because there is no "y" in the original equation you can integrate [itex]y''' y'= x[/itex] to get [itex]y'' y= (1/2)x^2+ C_1[/itex]. It will give you the same result.
However, you titled this "variation of parameters" which will be of about the same difficulty for either equation. 


#6
Aug1611, 07:11 AM

P: 2,568

But why this condition
[tex]u'e^x+ v'e^x+ w'= 0[/tex] My book threw this (similar, except they used y_1 and y_2 for general derivation) out of nowhere and I am left blank minded. How do they know it's going to work out? How did you know (don't say intuition...) the condition u'e^x+ v'e^x+ w'= 0 will enable to you solve the problem? 


#7
Aug1611, 07:14 AM

P: 2,568

Why does it make it equally correct that I "integrated" through [tex]y'''  y' = x[/tex] because y wasn't there? Why is it okay then that I don't need to know y', yet I would get the right answer? By the way, why do you use [/itex]? What does the i do? 


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