How Do You Solve the Differential Equation (x-1)y'=6y?

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goaliejoe35
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Homework Statement

[tex](x-1)y'=6y[/tex]

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

Here's my attempt...

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

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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ok so it would be like this...

[tex](x-1)dy/dx=6y[/tex]
 
ok now here's what i have...

[tex](x-1)dy/dx=6y[/tex]
[tex](x-1)dy=6ydx[/tex]
[tex]dy=6ydx/(x-1)[/tex]
[tex]dy/6y=dx/(x-1)[/tex]
[tex] \int dy/6y=\int dx/(x-1)[/tex]

is that correct?
 
goaliejoe35 said:
ok now here's what i have...

[tex](x-1)dy/dx=6y[/tex]
[tex](x-1)dy=6ydx[/tex]
[tex]dy=6ydx/(x-1)[/tex]
[tex]dy/6y=dx/(x-1)[/tex]
[tex]\ int dy/6y= \ int dx/(x-1)[/tex]

That's it. Good job.
 
ok but now I need to solve for y, so are these the right answers for the integral?

[tex] <br /> \int dy/6y=\int dx/(x-1)<br /> [/tex]
[tex](y^2/12)=ln(x-1)+c[/tex]
 
[tex]ln(y)/6[/tex] is that right? and is the other side correct?
 
[tex]ln(y)/6 = ln(x-1) + C[/tex]

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

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

y=(x-1)^6

is that correct now?