1. ## [SOLVED ] Reduction of Order

$x(1+2x)y'' + (1+6x)y' + 2y = \frac{1}{x^2}$

$y_1=\frac{1}{1+2x}$ is a solution to do related homogeneous equation, and I need to find a second linearly independent solution of the DE using reduction of order.

So, $y_2 = V(x)\frac{1}{1+2x}$

$y_2' = \frac{V'}{1+2x} - \frac{2V}{(1+2x)^2}$

$y_2'' = \frac{V''}{1+2x} - \frac{4V'}{(1+2x)^2} + \frac{8V}{(1+2x)^3}$

Plugging $y_2$ into the homogeneous DE:

$x(1+2x)[\frac{V''}{1+2x} - \frac{4V'}{(1+2x)^2} + \frac{8V}{(1+2x)^3}] + (1+6x)[\frac{V'}{1+2x} - \frac{2V}{(1+2x)^2}] + 2[V(x)\frac{1}{1+2x}] = 0$

After some simplification, I get

$\frac{(4x^3+4x^2+x)V'' + (12x^3+3x)V' + (8x^2+8x)V}{(1+2x)^2} = 0$

I know that for reduction of order to work, the terms involving V need to go away, but I can't see a way to make that happen.

2. why not make it generally at once?

consider $a(x)y''(x)+b(x)y'(x)+c(x)y=0,$ by knowing that $y_1(x)$ is a solution of the homogeneous equation, then substitute $y_2(x)=y_1(x)u$ and prove that $u(x)=\displaystyle\int\frac1{y_1^2}e^{-\int\frac{b(x)}{a(x)}\,dx}\,dx.$

3. My teacher doesn't allow the use of that formula.

4. completely useless, it's worth of time to prove that formula and then apply it.

well, there's nothing for me to say here.

5. I found an error in my simplification.