Seperation of Variables

[goaliejoe35]

Homework Statement

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

Could someone please help explain to me how to do this problem.

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?

 

[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.

 

[goaliejoe35]

ok so it would be like this...

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

 

[Dick]

ok so it would be like this...

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

Yes. Now separate.

 

[goaliejoe35]

ok now here's 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?

 

[Dick]

ok now here's what i have...

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

That's it. Good job.

 

[goaliejoe35]

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

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

 

[Dick]

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

 

[goaliejoe35]

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

 

[Dick]

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

 

[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.

 

[TeTeC]

**edit**

*erased*

 

[Dick]

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

 

[goaliejoe35]

so it would be..

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

 

[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.

 

[goaliejoe35]

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

y=(x-1)^6

is that correct now?

 

[Dick]

Almost, where did the constant go?

 

[goaliejoe35]

y=C(x-1)^6

 

[Dick]

y=C(x-1)^6

That's it.