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!