# Bernoulli's Equation

• Feb 22nd 2007, 08:03 PM
Sebastian
Bernoulli's Equation
I also have a bernouli's equation problem that I wonder if you guys would check the work on. Here goes:
Bernouli's equation:
y'+p(x)y=Q(x)y^n

dv/du=(u-v)^2-2(u-v)-2 w=u-v dw/du=1-dv/du dv/du=1-dw/du
1-dw/du=w^2-2w-2
dw/du=-w^2+2w+3
dw/du-2w=-w^2+3
p=-2, Q=-W^2+3
Linear form:
dz/dw+(1-n)PZ=(1-n)Q
dz/dw=(1-2)(-2)(Z)=(1-2)(-W^2+3)
dz/dw+2z=w^2-3
u=e∫2dw=e^2w
z=1/u(
∫uq+c)
Z=1/e^2w(
∫(e^2w)(w^2-3) + C Now where I go from here is a mystery to me. Anyone have some idea of how to intergrate that function?

• Feb 22nd 2007, 08:16 PM
ThePerfectHacker
Quote:

Originally Posted by Sebastian

dv/du=(u-v)^2-2(u-v)-2 w=u-v dw/du=1-dv/du dv/du=1-dw/du
1-dw/du=w^2-2w-2
dw/du=-w^2+2w+3
dw/du-2w=-w^2+3
p=-2, Q=-W^2+3

Wait.
Why is Q(u)=-W^2+3 :confused:

Remember it must be of the form,
Say for example,
Q(u)w^2
Now if you say that,
Q(u)=-w^2+2
Then you imply that,
Q(u)w^2=-w^2+2
Which is clearly false.
• Feb 22nd 2007, 10:21 PM
Sebastian
Quote:

Originally Posted by ThePerfectHacker
Wait.
Why is Q(u)=-W^2+3 :confused:

Remember it must be of the form,
Say for example,
Q(u)w^2
Now if you say that,
Q(u)=-w^2+2
Then you imply that,
Q(u)w^2=-w^2+2
Which is clearly false.

I don't understand what to do from here then. Does my w become my z function? If not, how do I make it so?
• Feb 23rd 2007, 07:54 AM
ThePerfectHacker
Quote:

Originally Posted by Sebastian
I don't understand what to do from here then. Does my w become my z function? If not, how do I make it so?

Here are two ways of writing it (assuming it is non-zero),

-w^2+2=(-w+2/w)w^1 thus Q=-w+2/w

Or,

-w^2+2=(-1+2/w^2)w^2 thus Q=-1+2/w^2