- #1
J--me
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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
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