Seperation of Variables

1. Jun 23, 2008

goaliejoe35

The problem statement, all variables and given/known data

$$(x-1)y'=6y$$

Here's my attempt....

$$(x-1)y'=6y$$
$$(x-1)dy=6y$$
$$(x-1)=6y/dy$$
$$\int (x-1)dx=\int 6y/dy$$
$$((x^2,2)-x)= 3y^2+C$$

Then I try to get it to equal y and it comes out nothing like the answer. Am I even doing these steps correct?

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2. Jun 23, 2008

Dick

y'=dy/dx. Put that in and separate. Don't put dy or dx into the denominator. It's supposed to be in the numerator to integrate.

3. Jun 23, 2008

goaliejoe35

ok so it would be like this....

$$(x-1)dy/dx=6y$$

4. Jun 23, 2008

Dick

Yes. Now separate.

5. Jun 23, 2008

goaliejoe35

ok now heres what i have...

$$(x-1)dy/dx=6y$$
$$(x-1)dy=6ydx$$
$$dy=6ydx/(x-1)$$
$$dy/6y=dx/(x-1)$$
$$\int dy/6y=\int dx/(x-1)$$

is that correct?

6. Jun 23, 2008

Dick

That's it. Good job.

7. Jun 23, 2008

goaliejoe35

ok but now I need to solve for y, so are these the right answers for the integral?

$$\int dy/6y=\int dx/(x-1)$$
$$(y^2/12)=ln(x-1)+c$$

8. Jun 23, 2008

Dick

No, y is in the denominator on the left side. It should give you another log type integral.

9. Jun 23, 2008

goaliejoe35

$$ln(y)/6$$ is that right? and is the other side correct?

10. Jun 23, 2008

Dick

Yes. Now both sides are correct. Can you solve for y now?

11. Jun 23, 2008

goaliejoe35

$$ln(y)/6 = ln(x-1) + C$$

Now how do I solve for y? I am not sure what to do with the ln's.

12. Jun 23, 2008

TeTeC

**edit**

*erased*

13. Jun 23, 2008

Dick

Exponentiate both sides. exp(ln(A))=A. I'd move the 6 over to the x side first.

14. Jun 23, 2008

goaliejoe35

so it would be..

e^(ln(y))=6e^(ln(x-1)) + C

15. Jun 23, 2008

Dick

Noo. e^(6*ln(x-1)) is not equal to 6*e^(ln(x-1)). And e^(A+B) is not equal to e^(A)+e^(B). Be careful! Review log rules.

16. Jun 23, 2008

goaliejoe35

Ohhh my mistake. e^(ln(y))=e^6ln(x-1)

y=(x-1)^6

is that correct now?

17. Jun 23, 2008

Dick

Almost, where did the constant go?

18. Jun 26, 2008

goaliejoe35

y=C(x-1)^6

19. Jun 26, 2008

That's it.