# Thread: Need help solving (what method to use)

1. ## 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

2. 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

$y_c=c_1x+c_2xe^{x}$

3. 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''

4. 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

5. 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

6. 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?

7. 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