Could someone explain steps in this differentian equation?

Question

I have this notebook that i copied from a college long time ago with solutions and steps...but i have trouble understanding this particular example if someone could explain to me in plain english what's going on i would be more than grateful. I copied it in full and it goes like this :

$\ddot y + \frac{2}{x} \dot y + y =0 $

$ y(x)=a(x)*Z(x)$

$\dot y = \dot a(x) \cdot Z(x) + a(x) \cdot \dot Z(x)$

$ \ddot y= \ddot a(x)\cdot Z(x) + \dot a(x) \cdot \dot Z(x)+ \dot a(x) \cdot \dot Z(x)+ a(x) \cdot \ddot Z(x)$

$ 2\dot a+ \frac{2}{x}a=0$

.... $\frac{da}{a} = -\frac{dx}{x}$

$\log_e|a|=- \log_e|x|+C /\cdot e$

$|a|=\frac 1x \cdot c1 => a= \frac cx= \frac 1x$

$\frac 1x \ddot Z +(\frac 2{x^3}+ \frac 2x \cdot \frac {-1}x+ \frac 1x )\cdot Z=0$

...

I would appreciate if someone could explain the method used here or point to some reference to learn more about it

Answer 1

$\ddot y + \frac{2}{x} \dot y + y =0 $

Our foe the ODE.

$y(x)=a(x)*Z(x)$

Looks for a solution where $y$ consists of two factors.

$\dot y = \dot a(x) \cdot Z(x) + a(x) \cdot \dot Z(x)$

$ \ddot y= \ddot a(x)\cdot Z(x) + \dot a(x) \cdot \dot Z(x)+ \dot a(x) \cdot \dot Z(x)+ a(x) \cdot \ddot Z(x)$

Taking the derivatives for such an $y$. Then the ODE is applied and sorted to $Z^{(k)}$: \begin{align} 0 &= \ddot a(x)\cdot Z(x) + 2\dot a(x) \cdot \dot Z(x)+ a(x) \cdot \ddot Z(x) + \\ & \quad \,\, (2/x) \left( \dot a(x) \cdot Z(x) + a(x) \cdot \dot Z(x) \right) + a(x) \,Z(x) \\ &= a(x) \cdot \ddot Z(x) + \left( \ddot a(x) + (2/x) \dot a(x) + a(x) \right) Z(x) + \left( 2\dot a(x) + (2/x) a(x) \right) \dot Z(x) \end{align}

$ 2\dot a+ \frac{2}{x}a=0$

Such an $a$ would make the term with $\dot{Z}$ in the above ODE vanish.

.... $\frac{da}{a} = -\frac{dx}{x}$

$\log_e|a|=- \log_e|x|+C /\cdot e$

$|a|=\frac 1x \cdot c1 => a= \frac cx= \frac 1x$

Here separation of variables is used to solve the ODE for $a$.

$\frac 1x \ddot Z +(\frac 2{x^3}+ \frac 2x \cdot \frac {-1}x+ \frac 1x )\cdot Z=0$

We still have $$ a(x) \cdot \ddot Z(x) + \left( \ddot a(x) + (2/x) \dot a(x) + a(x) \right) Z(x) = 0 $$ to solve, now we insert $a(x) = 1/x$ and get: $$ (1/x) \cdot \ddot Z(x) + \left( (2/x^3) + (2/x) (-1/x^2) + (1/x) \right) Z(x) = 0 $$ Note the difference to your version in the middle term.