Question

how to solve the exact 2nd order differential equation, (x^2)y''+xy'-y=0?

Answer 1

there is a method of solving the characteristic quadratic ....

Answer 2

or, power series are fun

Answer 3

\[Let\\~\\y=\sum_0a_nx^n\]\[y'=\sum_0a_n~nx^{n-1}\]\[y''=\sum_0a_n~n(n-1)x^{n-2}\]

Answer 4

forgot to up my indexes ... y' from 1, and y'' from 2 :)

Answer 5

\[(x^2)y''+xy'-y=0\]
\[(x^2)\sum_2a_n~n(n-1)x^{n-2}+x\sum_1a_n~nx^{n-1}-\sum_0a_nx^{n}=0\]
\[\sum_2a_n~n(n-1)x^{n}+\sum_1a_n~nx^{n}+\sum_0-a_nx^{n}=0\]
\[\sum_2a_n~n(n-1)x^{n}+\sum_1(a_n~n-a_n)x^{n}-a_1x=0\]
\[\sum_2(a_n~n(n-1)+a_n~n-a_n)x^{n}+(a_1-a_1)x-a_1x=0\]
\[\sum_2~a_n~(n(n-1)+n-1)x^{n}-a_1x=0\]
this is identically zero when a1 = 0, and the recurrsion of the coeffs of the summation are zero