Differential equation xy′+2y=0 and the form of arbitrary constant in its general solution If I'm solving the differential equation in the title I will get to: log(y)=−2log(x)+c then I'll get y=e^c/x^2 eith arbitrary constant c. So I know I can write y=d/x^2 where d is an arbitrary positive constant. But now it's easy to see that d can be negative or zero as well. How can I see that from the derivation? Can I reason from the derivation that d can be arbitrary? (I know I can just show d can be arbitrary e.g. by calculating y′ and see that d cancels out.)

Leypoldon

Leypoldon

Open question

2022-08-13

Differential equation x y + 2 y = 0 and the form of arbitrary constant in its general solution
If I'm solving the differential equation in the title I will get to:
log ( y ) = 2 log ( x ) + c
then I'll get y = e c / x 2 eith arbitrary constant c.
So I know I can write y = d / x 2 where d is an arbitrary positive constant. But now it's easy to see that d can be negative or zero as well. How can I see that from the derivation? Can I reason from the derivation that d can be arbitrary? (I know I can just show d can be arbitrary e.g. by calculating y′ and see that d cancels out.)

Answer & Explanation

Lisa Acevedo

Lisa Acevedo

Beginner2022-08-14Added 18 answers

If you're more careful, you can see that d can be arbitrary.
First of all, to solve xy′+2y=0 as you have done, you need to rewrite it as
y y = 2 x ,
but that assumes y is non-zero. Treating the y 0 case separately, you see that y≡0 is a solution which can be rewritten as
y = d x 2
where d=0.
As for the d negative case, note that log x is not an antiderivative for 1 x on R { 0 } because the former isn't even defined for x < 0. The correct statement is that log | x | is an antiderivative for 1 x on R { 0 }. Therefore, by antidifferentiating both sides of the equation
y y = 2 x ,
we obtain
log | y | = 2 log | x | + c .
Now if you rearrange for y, the presence of absolute values will produce a ± which allows for negative d.

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