# Solving a system of 2 diffrential equations with 3 unknowns.

1. Jun 2, 2009

### torquerotates

1. The problem statement, all variables and given/known data
r''-(theta')^2=-g reads: second derivative of r minus the first derivative of theta squared equals negative g

2(r')theta'+(r)theta''=0 reads: 2 times the first derivative of r times the first derivative of theta plus r times the second derivative of theta.

both r and theta are a function of time. Oh and the derivatives of each variable are with respect to time.

2. Relevant equations

3. The attempt at a solution
I really have no clue as to how I'm going to solve this system of differential equations. I'm trying to solve for r'

2. Jun 3, 2009

### tiny-tim

Hi torquerotates!

(have a theta: θ and try using the X2 tag just above the Reply box )

(and it's ok using ' to mean derivative … you needn't explain it! )
Hint: trick is to multiply the second one by r …

that gives you 2rr'θ' + r2θ'' = 0,

which you can integrate to give … ?

3. Jun 3, 2009

### torquerotates

Im thinking about it but i just cant seem to solve it. At first I was thinking chain rule. But that clearly doesn't work.

4. Jun 3, 2009

### Cyosis

To make use of Tim's hint recognize the product rule.

5. Jun 3, 2009

### torquerotates

Oh I see. So if i integrate, I get (r^2 )theta'=C where C is a constant. Is this the right thinking?

6. Jun 3, 2009

### torquerotates

Ok so solving for theta' I get theta'=C/(r^2)

And plugging into the other equation, I get r''-(C^2)/(r^4)=-g

But I have no clue solve this one as well. Maybe if r^4 was in the numerator, I would get somewhere.

7. Jun 3, 2009

### tiny-tim

Yup!

(but what happened to that θ i gave you? )
Hint: this time, multiply by r'.