Question

Help!!! Find the solution of the given initial value problem: 4x^2y'' + 8xy' + 17y = 0.

Answer 1

Let:
\[y=x^{\lambda} \implies y'=\lambda x^{\lambda - 1} \implies y'' = \lambda(\lambda - 1)x^{\lambda - 2} \]\[\implies 4x^2 \lambda(\lambda-1)x^{\lambda -2}+8x \lambda x^{\lambda-1}+17x^{\lambda}=0\]
\[\rightarrow 4\lambda^2-4\lambda+8\lambda+17=\lambda^2+\lambda+\frac{17}{4}=0 \implies \lambda = -\frac{1}{2} \pm 2 i\]
\[\implies y=C_1 x^{\alpha} \cos(\beta \ln(x))+C_2 x^{\alpha} \sin(\beta \ln(x)); \alpha = \Re(m);\beta = \Im(m)\]

Answer 2

\(\lambda=m,\qquad\)right?

Answer 3

Yeah, sorry I was double checking wikipedia to make sure I remembered it right and they used an m. :/

Answer 4

that was a good solution ı solve it using x=e^t and t=lnx but this is more easier than mine thanks :)