A little help with calculus terms.

[sebasalekhine7]

The rate of change of y with respect to x is inversely proportional to the square root of y.
a)Write a differential equation for the given statement
b)Solve the differential equation in part a.

I don't know, but what I've done so far is:
({dy/dx}) k=y^{1/2}

 

[ToxicBug]

I'm not sure, but I think this is it:

a) (dy/dx)x=1/sqrt(y)

b)
(dy/dx)1/x²=y
-2x/x^4=y

 

[dextercioby]

Incorrect.It's INVERSE PROPORTIONALITY.We usually let the constant in the other side of the equality.

dy(x)/dx~y^{-\frac{1}{2}}=>\frac{dy(x)}{dx}=ky^{-\frac{1}{2}}

Daniel.

 

[dextercioby]

I'm not sure, but I think this is it:

a) (dy/dx)x=1/sqrt(y)

b)
(dy/dx)1/x²=y
-2x/x^4=y

Sorry,there's no "x" explicitely.Just "y" to a power & its first derivative of "y".


Daniel.

 

[sebasalekhine7]

why is the x beside the (dy/dx)?

 

[ToxicBug]

Nevermind, I think I misunderstood the whole point of the question :/

 

[sebasalekhine7]

ok, but why is the "x" in the left side of the equation, why is there an x at all?

 

[dextercioby]

It isn't.It shouldn't be.It was an error from the poster.

Daniel.

 

[sebasalekhine7]

so the solution should just be \frac{dy}{dx}=ky^{-\frac{1}{2}} ?

 

[dextercioby]

Yes.

Daniel.

 

[sebasalekhine7]

that would mean that y=?

 

[dextercioby]

You can find easier "x" as a function of "y".

Daniel.

 

[HallsofIvy]

\frac{dy}{dx}= ky^{\frac{-1}{2}}
is a "separable equation". Write it as
y^{\frac{1}{2}}dy= kdx
and integrate.