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
Open question
2022-08-13
Differential equation and the form of arbitrary constant in its general solution If I'm solving the differential equation in the title I will get to:
then I'll get eith arbitrary constant c. So I know I can write 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
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
but that assumes y is non-zero. Treating the case separately, you see that y≡0 is a solution which can be rewritten as
where d=0. As for the d negative case, note that is not an antiderivative for on because the former isn't even defined for . The correct statement is that is an antiderivative for on . Therefore, by antidifferentiating both sides of the equation
we obtain
Now if you rearrange for y, the presence of absolute values will produce a which allows for negative d.