Series solution of xy″+2y′−xy=0I get r(r+1)=0,(r+1)(r+2)c1=0 andcn+1=cn−1(n+1+r)(n+r+2)The first equation gives the indicial roots r=−1 and r=0. The case for r=0 is fine.For r=−1, I don't see how c1=0 is implied by the second equation. The way I see it, since 1+r=0 when r=−1,c1 is not necessarily zero here, but this leads to a different solution to what is in the text book. What am I missing? Why is c1 forced to be zero?

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Series solution of xy″+2y′−xy=0I get r(r+1)=0,(r+1)(r+2)c1=0 andcn+1=cn−1(n+1+r)(n+r+2)The first equation gives the indicial roots r=−1 and r=0. The case for r=0 is fine.For r=−1, I don&#039;t see how c1=0 is implied by the second equation. The way I see it, since 1+r=0 when r=−1,c1 is not necessarily zero here, but this leads to a different solution to what is in the text book. What am I missing? Why is c1 forced to be zero?

Karsyn Wu · Accepted Answer

As you already found a basis solution for r=0, you need just one other. Setting c1 in the case r=−1 to some non-zero value will just add a multiple of the first, r=0, basis solution, so nothing new gets found.Note that your DE can be written as (xy)″−(xy)=0,so that the solution you find are y1(x)=sinh(x)x and y2(x)=cosh(x)x

Series solution of \(\displaystyle{x}{y}^{{''}}+{2}{y}^{{'}}-{x}{y}={0}\) I get \(\displaystyle{r}{\left({r}+{1}\right)}={0},{\left({r}+{1}\right)}{\left({r}+{2}\right)}{c}_{{1}}={0}\)

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