x^2 y'' - (x^2 + 2x)y' + (x + 2) y = 0

i am supposed to find the 2 independent solutions for this equation but i'm not sure which method to use. i was going to use the methof of solving cauchy-euler equations but i got stuck

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- May 18th 2011, 03:35 PMhaneekNeed help solving (what method to use)
x^2 y'' - (x^2 + 2x)y' + (x + 2) y = 0

i am supposed to find the 2 independent solutions for this equation but i'm not sure which method to use. i was going to use the methof of solving cauchy-euler equations but i got stuck - May 18th 2011, 04:08 PMTheEmptySet
By inspection you can see (guess) that one solution is

$\displaystyle y=x$

Now we can use reduction of order

$\displaystyle y=xu(x) \quad y'=u(x)+xu'(x) \quad y''=xu''(x)+2u'(x)$

$\displaystyle x^2(xu''(x)+2u'(x))-(x^2+2x)(u(x)+xu'(x) )+(x+2)(xu(x))=0$

Simplifying gives

$\displaystyle x^3u''(x)+(2x^2-x^3-2x^2)u'(x)+(-x^2-2x+x^2+2x)u(x)=0$

$\displaystyle x^3(u''(x)-u'(x))=0 \iff \frac{du'}{dx}=u'$

$\displaystyle \frac{du'}{u'}=dx \implies \ln|u'|=x+c \implies u'(x)=Ae^{x}$

integrating once more gives

$\displaystyle u(x)=Ae^{x}+B$

Now picking A=1 B=0 gives

$\displaystyle u(x)=e^{x} \implies y=xe^{x}$

So the solution is

$\displaystyle y_c=c_1x+c_2xe^{x}$ - May 18th 2011, 11:54 PMbugatti79
- May 19th 2011, 08:58 AMbugatti79
- May 19th 2011, 09:12 AMTheEmptySet
- May 19th 2011, 09:30 AMbugatti79
- May 19th 2011, 09:41 AMTheEmptySet
It was mentioned (very briefly) in my first post that I was going to use reduction of order.

Reduction of order - Wikipedia, the free encyclopedia