# Need help solving (what method to use)

• May 18th 2011, 03:35 PM
haneek
Need 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 PM
TheEmptySet
Quote:

Originally Posted by haneek
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

By inspection you can see (guess) that one solution is

$y=x$

Now we can use reduction of order

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

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

Simplifying gives

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

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

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

integrating once more gives

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

Now picking A=1 B=0 gives

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

So the solution is

$y_c=c_1x+c_2xe^{x}$
• May 18th 2011, 11:54 PM
bugatti79
Quote:

Originally Posted by TheEmptySet
By inspection you can see (guess) that one solution is

$y=x$

Now we can use reduction of order

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

Just in case the OP doesnt follow, I think TheEmptySet has a typo in the last equation, ie y should be y''
• May 19th 2011, 08:58 AM
bugatti79
Quote:

Originally Posted by TheEmptySet
By inspection you can see (guess) that one solution is

$y=x$

Now we can use reduction of order

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

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

Simplifying gives

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

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

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

integrating once more gives

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

Now picking A=1 B=0 gives

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

So the solution is

$y_c=c_1x+c_2xe^{x}$

Your welcome. I was meant to ask you why/how you set A=1 and B=0?

Thanks
• May 19th 2011, 09:12 AM
TheEmptySet
Quote:

Originally Posted by bugatti79
Your welcome. I was meant to ask you why/how you set A=1 and B=0?

Thanks

We are free to pick them because when we form the general solution to the homogenous solution it has the form

$y=c_1x+c_2xu(x)$

So if we left them we would just get new constants in the final solution
• May 19th 2011, 09:30 AM
bugatti79
Quote:

Originally Posted by TheEmptySet
We are free to pick them because when we form the general solution to the homogenous solution it has the form

$y=c_1x+c_2xu(x)$

So if we left them we would just get new constants in the final solution

Ok, just one more question. How did you arrive at the general solution to be in the form you have above? I dont see this form anywhere in the thread?
• May 19th 2011, 09:41 AM
TheEmptySet
Quote:

Originally Posted by bugatti79
Ok, just one more question. How did you arrive at the general solution to be in the form you have above? I dont see this form anywhere in the thread?

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