# Homework Help: A little help with calculus terms.

1. Mar 11, 2005

### 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:
$$({dy/dx}) k=y^{1/2}$$

Last edited: Mar 11, 2005
2. Mar 11, 2005

### 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

3. Mar 11, 2005

### 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.

Last edited: Mar 11, 2005
4. Mar 11, 2005

### dextercioby

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

Daniel.

5. Mar 11, 2005

### sebasalekhine7

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

6. Mar 11, 2005

### ToxicBug

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

7. Mar 11, 2005

### sebasalekhine7

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

8. Mar 11, 2005

### dextercioby

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

Daniel.

9. Mar 11, 2005

### sebasalekhine7

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

10. Mar 11, 2005

### dextercioby

Yes.

Daniel.

11. Mar 11, 2005

### sebasalekhine7

that would mean that y=?

12. Mar 11, 2005

### dextercioby

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

Daniel.

13. Mar 12, 2005

### HallsofIvy

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