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.

  • $\begingroup$ 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

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 $u(x)=c_1\cos(x)+c_2\sin(x)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.