# Calculus: splitting variables

1. May 8, 2007

### t_n_p

1. The problem statement, all variables and given/known data

http://img291.imageshack.us/img291/4489/splittingvariablesgz9.jpg [Broken]

3. The attempt at a solution

I can't split the x and y variables in each of these three cases. Would somebody be able to help me get started?

Last edited by a moderator: May 2, 2017
2. May 8, 2007

### Hootenanny

Staff Emeritus
For (a).(i)

$$\frac{dy}{dx} = \frac{x(y^2+3)}{y}$$

Divide through by $(y^2+3)/y$ yielding;

$$\frac{y}{(y^2+3)}\cdot\frac{dy}{dx} = x$$

For (a).(ii)

It may be useful to note that;

$$e^{x-2y} = e^x\cdot e^{-2y}= \frac{e^x}{e^{2y}}$$

As for (b), I don't think it is seperable.

Last edited: May 8, 2007
3. May 8, 2007

### t_n_p

Regarding a) i) how would I differentiate y/(y²+3)?

and for a) ii) that's very helpful

Thanks!

4. May 8, 2007

### Hootenanny

Staff Emeritus
For (a)(i), I would use the quotient rule; but I don't think you mean differentiate, I think you mean integrate. If this is the case it may be useful to note that;

$$y = \frac{1}{2}\cdot\frac{d}{dx}(y^2+3)$$

and

$$\int\frac{f'(x)}{f(x)} = \ln|f(x)| + C$$

Last edited: May 8, 2007
5. May 8, 2007

### t_n_p

Silly me, of course I meant integrate! :yuck:

Thanks again!:tongue2:

Am I on the right track?
http://img150.imageshack.us/img150/2133/asdffy0.jpg [Broken]

Last edited by a moderator: May 2, 2017
6. May 8, 2007

### Hootenanny

Staff Emeritus
Yeah, you've got it basically correct, however, there are a few things I should say.

On line (5), you should only include one constant of integration (one one side), otherwise they cancel each other out and your left without a constant of integration!

Secondly, you should ask yourself is there any requirement for the modulus signs? Can y2+3 ever be negative?

Thirdly, a somewhat minor point, on line (6) your final term should be 2C, not just C.

7. May 8, 2007

### t_n_p

thanks for pointing that out, similiarly with (ii) should I only introduce the constant on one side (I'm assuming so, as the value (1,0) is given to find C...

Just double checking!

8. May 8, 2007

### Hootenanny

Staff Emeritus
Yes, well actually what your doing is;

$$\int f(y) dy = \int g(x) dx$$

$$F(y) + A = G(x) + B$$

$$F(y) = G(x) + C \hspace{1cm}\text{where }C = B-A$$

Does that make sense?

9. May 8, 2007

### t_n_p

Should I show likewise for a) i)?

10. May 8, 2007

### Hootenanny

Staff Emeritus
No not unless you want to , it isn't usually nesscary unless explicitly stated.

11. May 8, 2007

### t_n_p

Yeah, I don't see why not

Could you please double check my value of c for (ii).

I basically subbed in (1,0) as you do, but the answer seems a bit left of field, so I guess it's best to double check.

http://img501.imageshack.us/img501/7300/85694878mp9.jpg [Broken]

Last edited by a moderator: May 2, 2017
12. May 10, 2007

### t_n_p

*bump* can anybody confirm my value of c?

13. May 10, 2007

### t_n_p

actually I double checked, I now think c=-1-ln(2)

14. May 14, 2007

### t_n_p

Couldn't seem to edit my previous post, but my final answer is y=(x-1)/2 which sounds alot better!