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Reduction of order method

Carla1985

Member
Feb 14, 2013
93
Hi, I'm practicing some of the methods for solving LODE's for my exam tomorrow afternoon and am struggling to understand one of the steps, I'm hoping someone will be able to explain it for me :)

So I have the equation

$(x-1)y''-xy+y=0$

with x being a solution. Then:

$y(x)=xv(t)$
$y'(x)=v(t)+xv'(x)$
$y''(x)=2v'(x)+xv''(x)$

I dont see where the 2 is coming from in the last line as my previous understanding of it was that:

$y''=x''v(x)+x'v'(x)+xv(x)$

so I thought it would just be one v'(x) with the v(x) disappearing?

Thanks
Carla
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hi, I'm practicing some of the methods for solving LODE's for my exam tomorrow afternoon and am struggling to understand one of the steps, I'm hoping someone will be able to explain it for me :)

So I have the equation

$(x-1)y''-xy+y=0$

with x being a solution. Then:

$y(x)=xv(t)$
$y'(x)=v(t)+xv'(x)$
$y''(x)=2v'(x)+xv''(x)$

I dont see where the 2 is coming from in the last line as my previous understanding of it was that:

$y''=x''v(x)+x'v'(x)+xv(x)$

so I thought it would just be one v'(x) with the v(x) disappearing?

Thanks
Carla
I assume the ODE is actually:

$(x-1)y''-xy'+y=0$

However, that is just an aside. You have assumed:

\(\displaystyle y(x)=xv(t)\)

Using the product rule we obtain:

\(\displaystyle y'(x)=xv'(x)+v(x)\)

Now, using the product rule again, we find:

\(\displaystyle y''(x)=xv''(x)+v'(x)+v'(x)=xv''(x)+2v'(x)\)
 

Carla1985

Member
Feb 14, 2013
93
Aaaah I see. That makes perfect sense now. Thank you ever so much :)