Question

how come y'' + 4y=0 is homogeneous?

Answer 1

dy/dx+4y=0
dydx=-4y-------(1)
substitutig ky
=>y'=-4ky which is not equal to (1)
therefore its not a homogeneous equation

Answer 2

Well go back to the previous section and take a look at Example 7 and Example 8. In those two examples we solved homogeneous (and that’s important!) BVP’s in the form,
search for this at http://tutorial.math.lamar.edu/Classes/DE/BVPEvals.aspx

Answer 3

it says that it is a homogeneous

Answer 4

isn't because this diff eq has set to zero i.e. g(t)=0

Answer 5

oopps !!!!!
its double differentiated equation
y''=-4y
dy/y=-4 . dx
integrating on both side
logy =-4x
but still it isnt a homogeneous eq

Answer 6

A DE is said homogeneous if there is a homogeneous function F such that
\[y \prime = F(x,y)\]
i.e., I think that you can write this as:
y'' = -4y
y' = -4yx
and F(x,y) = -4yx. Only have to prove that F is homogeneous now.

Answer 7

I was thinking the same as @bmp but somewhere I remembered, if the function can be defined as F(y/x) then it is called homogeneous differential equation.

Answer 8

though I am not quite sure about either.

Answer 9

@experimentX I think this proof suffices, since
F(tx, ty) = t*(-4yx) = t*F(x,y), t being a scalar. So F is homogeneous

Answer 10

I meant, t^2*F(x,y)

Answer 11

not really sure ... i'm quite illiterate on these stuffs.