Reduction of Order

**staevobr** · October 17th 2010, 10:59 AM

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

$y_1=\frac{1}{1+2x}$ 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.

**Krizalid** · October 17th 2010, 01:31 PM

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 $u(x)=\displaystyle\int\frac1{y_1^2}e^{-\int\frac{b(x)}{a(x)}\,dx}\,dx.$

**staevobr** · October 17th 2010, 01:34 PM

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

**Krizalid** · October 17th 2010, 01:58 PM

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

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

**staevobr** · October 17th 2010, 09:07 PM

I found an error in my simplification.

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