1
$\begingroup$

Solve the following ODE:

$x \cdot y'' - y' - x^3 \cdot y = 0$

I have made the following substitution: $y = u \cdot z$. And have found $u$ s.t. the coefficient at $z'$ became zero.

$u = \sqrt{x}$

And I've got the following equation: $z'' - (x^2 + \frac{3}{4} \cdot x^{-2}) \cdot z = 0$ But can't figure out how to solve it

  • 1
    The solution is a linear combination of $\sinh(x^2/2)$ and $\cosh(x^2/2)$. – 2017-01-20
  • 0
    I know the answer. I want to know how to get it – 2017-01-20
  • 1
    Wolfram Alpha says set $$t=\frac{ix^2}{2}$$ and we obtain $$x=(-1+i)\sqrt{t}$$ – 2017-01-20
  • 0
    Can you double-check your $z^{\prime\prime}$ equation? I think you left out a $z$. – 2017-01-20
  • 0
    Yes, thank you! – 2017-01-20

0 Answers 0