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Question: Verify that $y=c_1x^{-1}+c_2x^5$ is a solution of $x^2y''-3xy-5y=0$ on any interval $[a,b]$ that does not contain the origin. If $x_o \ne 0$, and if $y_o$ and $y_o'$ are arbitrary, show directly that $c_1$ and $c_2$ can be chosen in one and only one way so that $y(x_0) = y_o$ and $y'(x_o) = y_o'$.

Approach: I found $y''$ and substituted it in the given equation to get $\frac{-3c_1}{x}+15c_2x^5-3c_2x^6-3c_1=0$. How do I solve this and where should I go from here? The first part of the question states that the solution does not contain origin which I think can be clearly seen from this equation I got. For the second part of the question, I am confused what the question wants me to do. Isn't $y'(x_o) = y_o'$ obvious?

Edit: I found a link to similar thread asking same question here: verification of solution of differential equation. So apparently there is a typo in the question in book. It should be $3*x*y'$ instead of $3*x*y$. I can solve the first part now but can someone give me a hint about second part?

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    $\begingroup$ Are you sure there isn't a typo in the question? The function $y = c_1/x + c_2 x^5$ is not a solution to the ODE as written. I suspect that it's supposed to be $x^2 y'' - 3 x y' - 5 y = 0$; if it is, then the given $y$ is indeed a solution of the ODE. $\endgroup$ – Michael Seifert Jul 10 '18 at 13:01
  • $\begingroup$ Yes. It was a mistake in book. I made an edit for the same. $\endgroup$ – Seth Rollins Jul 10 '18 at 13:05
  • $\begingroup$ @user539887 Yes $\endgroup$ – Seth Rollins Jul 10 '18 at 13:12
  • $\begingroup$ Regarding the second question: Can you show that $x^{-1}$ and $x^5$ form a fundamental system of solutions of the given linear homogeneous equation on an interval not containing zero? If yes, then use the fact that any solution can be written as $c_1 x^{-1}+c_2 x^5$. $\endgroup$ – user539887 Jul 10 '18 at 13:16

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