User:Egm6321.f12.team4.harris/Homework6 R*6.8

Problem R*6.8 – Proof of an Invalid Root
sec35-4

Statement
Explain why


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$$ r_2(x)=\frac{1}{x-1} $$     (6.8.1)
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is not a valid root for the equation


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$$ (x-1)y''-xy'+y=0 $$     (6.8.2)
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Solution
If $$ y_2=e^{xr_2(x)} $$ then, with equation 6.8.1


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$$ y_2=exp(\frac{x}{x-1}) $$     (6.8.3)
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One must take the first and second derivatives of equation 6.8.3, and plug these and equation 6.8.3 back into equation 6.8.2. The derivatives are as follows:
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$$ y'=\frac{d}{dx}[exp(\frac{x}{x-1})] $$     (6.8.4) Using the rules
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$$ \frac{d}{dx}(e^{f(x)})=f'(x)e^{f(x)} $$ and
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$$ \frac{d}{dx}(\frac{g(x)}{h(x)})=\frac{h(x)g'(x)-h'(x)g(x)}{(h(x))^2} $$ then
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$$ y'=[\frac{(x-1)(1)-x(1)}{(x-1)^2}]exp(\frac{x}{x-1})=\frac{-1}{(x-1)^2}exp(\frac{x}{x-1}) $$     (6.8.5)
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$$ y''=(\frac{-1}{(x-1)^2})(\frac{-1}{(x-1)^2})exp(\frac{x}{x-1})+[\frac{(x-1)^2(0)-(-1)(2(x-1)(1))}{(x-1)^4}](\frac{-1}{(x-1)^2})exp(\frac{x}{x-1})=(\frac{2x-1}{(x-1)^4})exp(\frac{x}{x-1}) $$     (6.8.6)
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Substituting equations 6.8.3, 6.8.5, and 6.8.6 back into equation 6.8.2 gives
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$$ (x-1)[\frac{2x-1}{(x-1)^4}]exp(\frac{x}{x-1})-x[\frac{-1}{(x-1)^2}]exp(\frac{x}{x-1})+exp(\frac{x}{x-1})=0 $$     (6.8.7)
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One can divide out the $$ exp(\frac{x}{x-1}) $$ term and simplify equation 6.8.7 to
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$$ \frac{2x-1}{(x-1)^3}+\frac{x}{(x-1)^2}+1=0 $$     (6.8.8)
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This equation is not valid, as there are no values of $$ x $$ that will make the left hand side of the equation equal to zero. Since $$ r_2(x) \ne constant $$, it is not a valid solution to the homogeneous equation.