User:Egm6321.f12.team4.mishra/Homework6 R6.7

Homework 6 Problem R*6.7 – Comparison with the method in King 2003 p.8
sec35-4

Statement
Show that (1) p 34.6 agrees with King 2003 p.8 i.e.


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$$ y_P(x) = \int^{x} f(s) [\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}] ds $$ (6.7.1)
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with,


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$$ W(s) := u_1u_2' - u_2u_1' $$
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Discuss the feasibility of the following choices for the variation of parameters,


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$$ y(x) = U (x) \pm u_1(x) $$
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$$ y(x) = U (x) / u_1(x) $$
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$$ y(x) = u_1(x) / U(x) $$
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Solution
From (1)p 34.6,


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$$ y_P(x)= u_1(x)[\int\frac{1}{h(x)}[\int h(x)\frac{f(x)}{u_1(x)}dx]dx] $$     (6.7.2)
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From the hint, we observe that, (4)35.4,


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$$ (\frac{u_2}{u_1})'=\frac{1}{h(x)} $$     (6.7.3)
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Further differentiating,


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$$ (\frac{u_1u_2'-u_2u_1'}{u_1^2})=\frac{1}{h(x)} $$     (6.7.4)
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Integrating Eq. (6.7.2) by parts,


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$$ \int u dv = uv - \int vdu $$     (6.7.5)
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We select,


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$$ u = [\int h(x)\frac{f(x)}{u_1(x)}dx] $$     (6.7.6)
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$$ dv = \frac{1}{h(x)} = (\frac{u_2}{u_1})' $$     (6.7.7)
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Therefore,


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$$ y_P(x)= u_1(x)[\int\frac{1}{h(x)}dx[\int h(s)\frac{f(s)}{u_1(s)}ds] - \int[\int\frac{dx}{h(x)}][\frac{h(s)f(s)}{u_1(s)}]ds] $$     (6.7.8)
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As $$ \int\frac{1}{h(x)}dx = \frac{u_2}{u_1}$$ and using (6.7.4),


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$$ y_P(x)= \not{u_1(x)}[\frac{u_2(x)}{\not{u_1(x)}}[\int \frac{{u_1(s)}^{\not2}}{u_1(s)u_2'(s)-u_2(s)u_1'(s)}\frac{f(s)}{\not{u_1(s)}}ds]] - u_1(x)[\int[\frac{u_2(s)}{\not{u_1(s)}}][\frac{\not{u_1(s)^2} f(s)}{(u_1(s)u_2'(s)-u_2(s)u_1'(s){\not{u_1(s)}}}]ds]$$     (6.7.9)
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$$ y_P(x)= [\int \frac{{u_1(s)}u_2(x)}{u_1(s)u_2'(s)-u_2(s)u_1'(s)}f(s)ds] -[\int [\frac{u_2(s){u_1(x)}}{(u_1(s)u_2'(s)-u_2(s)u_1'(s)}f(s)]ds] $$     (6.7.10)
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which can be rewritten as,


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$$ y_P(x)= [\int \frac{{u_1(s)}u_2(x)-u_2(s)u_1(x)}{u_1(s)u_2'(s)-u_2(s)u_1'(s)}f(s)ds] $$     (6.7.11)
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which is in the form given in King 2003 p8.

Feasibility of Choices

1.)$$ y(x)= U(x) \pm u_1(x) $$


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$$ [y(x)= U(x) \pm u_1(x)] \times a_0(x) $$     (6.7.12)
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$$ [y'(x)= (U'(x) \pm u_1'(x))] \times a_1(x) $$     (6.7.13)
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$$ [y(x)= (U(x) \pm u_1''(x))] \times a_2(x) $$     (6.7.14)
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Adding (6.7.12), (6.7.13),(6.7.14)


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$$ a_0(x)y+a_1(x)y'+y= a_0(x) U(x) + a_1(x) U'(x) + a_2(x) U(x) + (a_0(x) u_1(x) + a_1(x) u_1'(x) + u_1''(x)) $$     (6.7.15)
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Since, $$(a_0(x) u_1(x) + a_1(x) u_1'(x) + u_1''(x) = 0 $$,


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$$ a_0(x)y+a_1(x)y'+y= a_0(x) U(x) + a_1(x) U'(x) + a_2(x) U(x) = f(x) $$     (6.7.15)
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2.)$$ y(x)= U(x) / u_1(x) $$


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$$ y'= \frac{u_1(x)U'(x)-U(x)u_1'(x)}{u_1(x)^2} $$     (6.7.16)
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$$ y= \frac{u_1^2(u_1(x)U(x)+U'(x)u_1'(x)-U'(x)u_1'(x)-U(x)u_1''(x))-(u_1(x)U'(x)-U(x)u_1'(x))2u_1(x)}{u_1(x)^4} $$     (6.7.17)
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3.) $$ y(x)= u_1(x) / U(x) $$


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$$ y'= \frac{u_1'(x)U(x)-U'(x)u_1(x)}{U(x)^2} $$     (6.7.18)
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$$ y= \frac{U(x)^2(U(x)u_1(x)+u_1'(x)U'(x)-u_1'(x)U'(x)-u_1(x)U''(x))-(U(x)u_1'(x)-u_1(x)U'(x))2U(x)}{U(x)^4} $$     (6.7.19)
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From the above, we can see that our choices of trial functions does not reduce the order of non-homogeneous L2-ODE VC, hence they are not feasible.

Author and References

 * Solved and Typed by -- Pushkar Mishra