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Homework Statement
[tex]x^2\frac{d^2y}{dx^2}+ax\frac{dy}{dx}+by=0[/tex]
show that if there is one real double root of the aux. eq'n show that the G.S. is given by
[tex]y=c_1x^{n_1}+c_2x^{n_1}ln(x)[/tex]
Homework Equations
Assume the trial solution [itex]y=x^n[/itex]
The Attempt at a Solution
[tex] y=x^n
\frac{dy}{dx}=nx^{n-1}
\frac{d^2y}{dx^2}=n(n-1)x^{n-2}
=> n(n-1)x^n + anx^n +bx^n =n^2-(a-1)n+b=0 (Aux. Eq'n)[/tex]
for equal roots [tex](a-1)^2-4b=0[/tex]
therefore [tex]n=\frac{1-a}{2}[/tex]
hence one solution is [tex]y=x^\frac{1-a}{2}[/tex]
so the other solution is [tex]y=vx^\frac{1-a}{2}
\frac{dy}{dx}=\frac{dv}[dx}x^\frac{1-a}+ \frac{1-a}{2}Vx^\frac{-a-1}{2}
\frac{d^2y}{dx^2}=\frac{d^2v}{dx^2}x^\frac{-1-a}{2} + (1-a)\frac{dv}{dx}x^\frac{-1-a}{2} -\frac{1-a^2}{4}vx^\frac{-a-3}{2}[/tex]
sub. into the diff. eq'n...(this is where I think I am wrong)
(This is what I get)
[tex]\frac{d^2v}{dx^2}x^\frac{3-a}{2} + \frac{dv}{dx}x^\frac{3-a}{2} +\frac{vx^\frac{1-a}{2}((a-1)^2+4b)}{4}[/tex]
Now I was hoping to get [itex](a-1)^2-4b[/itex] in the last term there so that it would be zero but I apparently did something wrong..where did I go wrong?