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Solve the ODE assuming the solution to be a product of 2 unknown functions

$ x^2y''+14xy'+(x^2 +42)y=0$.

I will be gratefull for all hints. I don't know how to use the knowledge that "solution is a product of 2 unknow functions" to start solving the problem. Usually I solved ODE problems where 1st solution was given, then I applied Reduction of order or where I had any initial conditions to apply Power of series, or situation where I applied substitution $y=x^m$ where $m\in R \setminus \{0,1\}$, but all of this conditions were part of the problem.

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  • $\begingroup$ have you tried WA? $\endgroup$ – Dr. Sonnhard Graubner Jul 14 '17 at 16:11
  • $\begingroup$ Assume $y$ has the form $\sum_{n=0}^{\infty} a_n x^n$ ... This should give you a second order linear recurrence relation for$a_n$ ... don't worry that you do not know $a_0$ and $a_1$ they are the two arbitary constants ! $\endgroup$ – Donald Splutterwit Jul 14 '17 at 16:16
  • $\begingroup$ @Dr.SonnhardGraubner The shortcut WA is not familiar for me. $\endgroup$ – Vid Jul 14 '17 at 16:19
  • $\begingroup$ @DonaldSplutterwit I will try your hint. Thank you! $\endgroup$ – Vid Jul 14 '17 at 16:21
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Hint: Substitute $y(x) = x^{-7}u(x)$. I first assumed $y(x)=x^{n}u(x)$ (because the ODE looks alot like the Euler-Cauchy-ODE) and obtained: $$2(n+7)x^{n+1}u'(x)+(u(x)+u''(x))x^{n+2}+x^n(n+7)(n+6)u(x)=0.$$

Choosing $n=-7$ will eliminate the first and last term and we are left with: $\left[u(x)+u''(x)\right]x^{-7+2}=0$. The resulting ODE (assuming $x\neq 0$) will be $u''(x)+u(x)=0$ (harmonic oscillator) with $u(x)=c_1\cos(x)+c_2\sin(x)$.

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