Question

x^2y"+y=0
\(\alpha =0; \beta=1\) therefore, characteristic equation is \(r^2-r+1=0\) and \(r_{1,2} = \frac{1}{2}\pm\frac {\sqrt{3}}{2}i\) which leads to solution \[y=C_1x^{\frac{1}{2}}cos(\frac{\sqrt{3}}{2}lnx)+C_2x^{\frac{1}{2}sin(\frac{\sqrt{3}}{2}lnx)\]
Then, I don't know how to do next. Please, help

Answer 1

\[r^2-r+1=0\\r=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\]

Answer 2

\[y=C_1x^{\frac{1}{2}}cos(\frac{\sqrt{3}}{2}lnx) +C_2 x^{\frac{1}2}sin(\frac{\sqrt{3}}{2}lnx)\]

Answer 3

then, I am stuck, @oldrin.bataku help me, please

Answer 4

try putting y(x) into your DE and see if it is solution or not

Answer 5

I shortcut a step which tests whether x =0 is regular singular point or not. It passes that test, that mean that y is solution.

Answer 6

\[lim_{x\rightarrow 0}x^2\frac{1}{x^2} =1\]

Answer 7

whats the problem?

Answer 8

I need a solution under the form of series \(\sum\)

Answer 9

the series method is a mess, this is just Euler-Cauchy equation, that it is best dealt with.
if you want final answer in series then just expand the final answer.

Answer 10

how?

Answer 11

assume the solution be of the form y=x^n,
n(n-1) + 1 = 0 , let n1 and n2 be the roots of this quadratic equation
your solution should be of the form
y(x) = C1x^(n1) + C2 x^(n2)

Answer 12

http://en.wikipedia.org/wiki/Frobenius_method
since 1/x^2 is not analytic at x=0, no power series solution exist.

\[ \lim_{x\to 0} x^{1/2} \sin( k \log(x)) = 0\]
http://www.wolframalpha.com/input/?i=limit+x-%3E0+x%5E%281%2F2%29+sin%28log%28x%29%29
y[x] = 0 satisfield eqn for any x, if y[x] is singular at x=0, then probably that's the case of removable singularity.