Using the indicated substitutions x^2=z, find a general solution of x^2y\'\'+ xy
ID: 1813475 • Letter: U
Question
Using the indicated substitutions x^2=z, find a general solution of x^2y''+ xy'+4(x^4-v^2)y=0 in terms of Jv and J-v.
A.) y=c1J2(x)+c2J-2(x)
B.) y=c1Jv(x) + c2J-v(x)
C.) y=c1J1/4(x)-c2J-1/4(x)
D.) y=c1Jv(x^2)+c2J-v(x^2)
E.) None of the above; see Problem Work
Explanation / Answer
x^2 y''(x)+x y'(x)+4 (x^4-v^2) y(x) = 0 alternative forms.... 4 (v^2-x^4) y(x) = x (x y''(x)+y'(x)) x (x y''(x)+y'(x))-4 (v^2-x^4) y(x) = 0 -4 v^2 y(x)+4 x^4 y(x)+x^2 y''(x)+x y'(x) = 0 y(x) = c_1 Gamma(1-v) J_(-v)(x^2)+c_2 Gamma(v+1) J_v(x^2) so the correcct answer is "D"
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