Hello all! i have a doubt,consider a 2nd order ODE ,F[y]=y''+y'+y=0,it is said to be homogenous because F[ay]=aF[y],(where a is some constant) take this eqn where F[y]=y''+y'+y=3 ,it is said to be non homogenous because F[y] can be expressed as y''+y'+y-3=0 and thus F[ay] is not equal to aF[y],but if i take F[y] as y''+y'+y which is equal to 3 i.e then F[ay]=ay''+ay'+ay=a[y''+y'+y]=a*3=aF[y],where did i make a mistake here?

 

Hi,

Dont try to generalize right away as you have done so far. First, solve both equations then look at the actual aF(y) and compare it to the actual F(ay). See what you can find out after you look at both solutions.

 

Actually I mean, if F(y)=y"+y'+y=3 © then F(ay)=a[y"+y'+y] ,which is equal to a*3 (from eqn ©)
This is a homogeneous eqn as F(ay)=aF(y) but if I keep F(y) as y"+y'+y-3=0 then it is non homogenous because in this way F(ay) not equal to aF(y) "due to that constant 3". What iam asking is why the same eqn seems different in terms of homogeneity based on how I take F(y) as.Iam confused,how is that?

 

Actually I mean, if F(y)=y"+y'+y=3 © then F(ay)=a[y"+y'+y] ,which is equal to a*3 (from eqn ©)
This is a homogeneous eqn as F(ay)=aF(y) but if I keep F(y) as y"+y'+y-3=0 then it is non homogenous because in this way F(ay) not equal to aF(y) "due to that constant 3". What iam asking is why the same eqn seems different in terms of homogeneity based on how I take F(y) as.Iam confused,how is that?

Yeah i told you what to do and you basically said the same thing again.

Solve BOTH and then you can see if aF(y) is the same or different than F(ay).
That's the ONLY way you can tell unless you take it verbatim that the constant causes the non homogeneity which that way it does not help you.

 

Here is a wikipedia entry:
https://en.wikipedia.org/wiki/Homogeneous_differential_equation

 

Thanks, will refer to the entry u posted