MHB Understanding the Reduction of Order Method for Solving LODEs

[Carla1985]

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 don't 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]

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 don't 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:

$$y(x)=xv(t)$$

Using the product rule we obtain:

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

Now, using the product rule again, we find:

$$y''(x)=xv''(x)+v'(x)+v'(x)=xv''(x)+2v'(x)$$

 

[Carla1985]

Aaaah I see. That makes perfect sense now. Thank you ever so much :)