Reduction of Order ODE - Stuck on question

[J--me]

Reduction of Order ODE - Stuck on question! Help Please!

The question says that y₁= x is a solution to:

x^3 y'' + x y' - y = 0

It then says to use y² = y₁ f(x)

So I can do it this far and then I just get lost and my notes don't seem to clear anything!

I'm just going to say y(2) = y₂ , f = f(x) and y = y₁ just to make it easier!

So:

y(2) = f x
y'(2) = f' x + f
y''(2) = f'' x + 2f'

then sub it in:

x^3 (f'' x + 2f') + x (f' x + f) -f x = 0
f'' x^4 + f' ( 2x^3 + x^2) + f (x - x) = 0
Therefore: f'' x^4 + f' (2x^3 + x^2) = 0
Then my notes say that f' = g
So: g' x^4 + g (2x^3 + x^2) = 0
Then rearrange and integrate:
INT(1/g) dg = -INT(x^-2 + 2x^-1) dx
Ln(g) = 1/x - 2ln(x)
So g = exp(1/x) - x^2
As f' = g
f = INT (exp(1/x) - x^2) dx
So f = x exp(1/x) - 1/3x^3
I then put the final answer for y(2) into the equation but it doesn't equal 0 therefore its not a solution and wrong =S! So I've gone wrong somewhere. The question then goes onto say obtain a general solution and I am not sure how to do that either!
Any help would be good Thanks

 

[tiny-tim]

Hi J--me!

Ln(g) = 1/x - 2ln(x)
So g = exp(1/x) - x^2

nooo

 

[J--me]

Hey!
Lol! Thanks!
Should g = -x^2 exp(1/x)?

 

[HallsofIvy]

Hey!
Lol! Thanks!
Should g = -x^2 exp(1/x)?

Again no! ln(a-b)= ln(a/b).

 

[tiny-tim]

nooo!

third try? ​

 

[J--me]

>.<! I may need a fourth try..

Ok so ln(a-b) = ln(a/b)
sooo: 0 = 1/x - ln(x^2) - ln(g)
0 = 1/x - ln(x^2/g)
Therefore: g = x^2 exp(-1/x) ? =S but if i use ln(a+b) = ln(a*b) then i get:
g = x^-2 exp(1/x) ? =S! (sorry if I am doing something really silly!)

 

[tiny-tim]

g = x^-2 exp(1/x) ? =S!

At last!

 

[J--me]

Cool! Thanks! =D! So the integration of that is g = f' so:
f = INT (g) dt = -exp(1/x) what do i do now to get the general solution?? =S

 

[tiny-tim]

x^3 y'' + x y' - y = 0

It then says to use y₂ = y₁ f(x)

so your general solution is y = xf

(btw, you lost a solution of f' = g = 0 earlier on when you divided by g … did you notice that?)

 

[J--me]

Oh right so the general solution is just y(2) = -x exp(1/x) Thanks!

(Where did i lose a solution?? =S)

 

[tiny-tim]

you divided by g, which you can't do if g = 0 …

So: g' x^4 + g (2x^3 + x^2) = 0
Then rearrange and integrate:
INT(1/g) dg = -INT(x^-2 + 2x^-1) dx

that means you lost the solution g = 0 (so f = constant, so y = Cx)