1. ## [SOLVED ] Reduction of Order

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

$\displaystyle y_1=\frac{1}{1+2x}$ 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, $\displaystyle y_2 = V(x)\frac{1}{1+2x}$

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

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

Plugging $\displaystyle y_2$ into the homogeneous DE:

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

After some simplification, I get

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

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

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.