Obtain the general solution of the following two Cauchy-Euler equations
(i) y''(x) +(1/x)y'(x) −(16/x^2)y(x) = 0,
(ii) x^2y''(x) + xy'(x) + y(x) = 0,
for x 0. Obtain the general solution of the following two Cauchy-Euler equations
(i) y''(x) +(1/x)y'(x) −(16/x^2)y(x) = 0,
(ii) x^2y''(x) + xy'(x) + y(x) = 0,
for x 0. @Mathematics

Question

Obtain the general solution of the following two Cauchy-Euler equations
(i) y''(x) +(1/x)y'(x) −(16/x^2)y(x) = 0, 
(ii) x^2y''(x) + xy'(x) + y(x) = 0,
for x > 0. Obtain the general solution of the following two Cauchy-Euler equations
(i) y''(x) +(1/x)y'(x) −(16/x^2)y(x) = 0, 
(ii) x^2y''(x) + xy'(x) + y(x) = 0,
for x > 0. @Mathematics

amistre64 · Answer

x^2y'' + xy' + y = 0

y'' + y'/x + y/x^2 = 0

i dont know to much about these, but this is what i recall from previous stuff.  might be applicable

amistre64 · Answer

y = c1 e^(r1 x) +c2 e^(r2 x)  depending on how the roots go i spose

amistre64 · Answer

http://www.wolframalpha.com/input/?i=+x%5E2y%27%27%28x%29+%2B+xy%27%28x%29+%2B+y%28x%29+%3D+0%2C ewww, those look nasty

mendicant_bias · Answer

I'm particularly confused by any of these where the right-hand function f(x) is already equal to zero; if f(x) is equal to zero already (multiplying across both LHS and RHS will get in proper Cauchy-Euler form, but will not change the RHS to have a nonzero f(x)---doesn't that mean your W_1 and W_2 Wronskian matrices will be equal to zero?