# Seperation of Variables Method

• Jul 17th 2010, 02:59 PM
Benji123
Seperation of Variables Method
Hi All, new to the forum!

Having a few problems with this question: "Use the separation of variables method to solve the first ordinary differential equation dy/dx = x(y-1) given that when x = -, y = 4

I came up with two answers... 12 and 1.099 and I'm not sure which one is correct, if any of them are correct! Any help will be deeply appreciated, thanks!
• Jul 17th 2010, 03:15 PM
Also sprach Zarathustra
dy/dx = x(y-1)

==>

dy/(y-1) = xdx

ln(y-1)=x^2/2 +C

correct?
• Jul 17th 2010, 03:18 PM
mr fantastic
Quote:

Originally Posted by Benji123
Hi All, new to the forum!

Having a few problems with this question: "Use the separation of variables method to solve the first ordinary differential equation dy/dx = x(y-1) given that when x = -, y = 4

I came up with two answers... 12 and 1.099 and I'm not sure which one is correct, if any of them are correct! Any help will be deeply appreciated, thanks!

In order to best help you, please post all your work - show every step.
• Jul 17th 2010, 03:29 PM
Benji123
Thanks for the quick replies, I have 2 answers which I got not sure if there correct...

∫(y-1) dy = ∫x dx

y^2 /2 - y = x^2 /2 + C1

Multiplied both sides by 2

y^2 - y + x^2 = C

= 12

Do I need to find 'C' in order to answer this question?

My second answer is,

∫1/y-1 dy = ∫x dx

ln(y-1) = x^2 /2 + C1

Mutliplied both sides by 2

ln(y-1) + x^2 = C

= 1.099

Again do I need to find 'C' to answer this question or just leave it with at the integrated solution?

Thanks
• Jul 17th 2010, 03:56 PM
mr fantastic
Quote:

Originally Posted by Benji123
Thanks for the quick replies, I have 2 answers which I got not sure if there correct...

∫(y-1) dy = ∫x dx Mr F says: This is plainly wrong (at this level I do hope you understand why) and so all that follows from it will be wrong.

y^2 /2 - y = x^2 /2 + C1

Multiplied both sides by 2

y^2 - y + x^2 = C

= 12

Do I need to find 'C' in order to answer this question?

-----------------------------------------------------------------------------------------

My second answer is,

∫1/y-1 dy = ∫x dx Mr F says: Post #2 tells you that this is correct.

ln(y-1) = x^2 /2 + C1 Mr F says: Wrong. It is ln|y - 1| NOT ln(y - 1). There is a big difference that you should be aware of from basic calculus.

Mutliplied both sides by 2

ln(y-1) + x^2 = C Mr F says: If you multiply both sides by 2, and then subtract x^2, then surely you get 2 ln|y - 1| - x^2 = C. Although I don't know why you moved the x^2. It is better left as 2 ln|y - 1| = x^2 + C.

= 1.099 Mr F says: This has come out of nowhere and is meaningless without explanation.

Again do I need to find 'C' to answer this question or just leave it with at the integrated solution?

Thanks

Obviously you need to find the value of C, why do you think you were "given that when x = -, y = 4" ? I assume you are meant to get y as a function of x, so I don't know why you are giving a single number as an answer.

Sorry, but from what I can see your problem is mainly algebra not calculus. You would be well-advised to take much greater care with the algebra. And it wouldn't hurt to review your class notes and textbook on solving differential equations.