# integration

• February 7th 2008, 03:33 PM
kid funky fried
integration
i am having some trouble with the following problem.

solve:

dy/dx= (x+1)/y-1), y>1

here is what is have done so far-

(y-1) dy/dx=(x+1)

S[(y-1)dy/dx]dx=S(x+1)dx

y^2/2 -y=x^2/2+x <-i feel this is where i made a mistake

Any help would be appreciated.
thx
• February 7th 2008, 04:49 PM
mr fantastic
Quote:

Originally Posted by kid funky fried
i am having some trouble with the following problem.

solve:

dy/dx= (x+1)/y-1), y>1

here is what is have done so far-

(y-1) dy/dx=(x+1)

S[(y-1)dy/dx]dx=S(x+1)dx

y^2/2 -y=x^2/2+x <-i feel this is where i made a mistake Mr F says: Add an arbitrary constant and your answers fine ...... y^2/2 -y=x^2/2+x + C. Personally, I'd multiply through by 2:

y^2 - 2y = x^2 + 2x + K, where K is just as arbitrary as 2C ......

Any help would be appreciated.
thx

Do you need to make y the subject?
• February 7th 2008, 04:53 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Do you need to make y the subject?

If so, then

$y^2 - 2y = x^2 + 2x + K \Rightarrow (y - 1)^2 - 1 = x^2 + 2x + K \Rightarrow (y - 1)^2 = x^2 + 2x + A$ where A = K + 1 is just as arbitrary as K.

Therefore

$y - 1 = \pm \sqrt{x^2 + 2x + A} \Rightarrow y = \pm \sqrt{x^2 + 2x + A} + 1.$

The value of A and which of the + or - root solutions are kept will depend on the given boundary conditions. Since y > 1 is given ........
• February 7th 2008, 04:57 PM
kid funky fried
well, i did that but then i checked the answer in the back of the book and it reads:

y=1+ sqrt((x+1)^2+c)
• February 7th 2008, 05:02 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
If so, then

$y^2 - 2y = x^2 + 2x + K \Rightarrow (y - 1)^2 - 1 = x^2 + 2x + K \Rightarrow (y - 1)^2 = x^2 + 2x + A$ where A = K + 1 is just as arbitrary as K.

Therefore

$y - 1 = \pm \sqrt{x^2 + 2x + A} \Rightarrow y = \pm \sqrt{x^2 + 2x + A} + 1.$

The value of A and which of the + or - root solutions are kept will depend on the given boundary conditions. Since y > 1 is given ........

$y = \sqrt{x^2 + 2x + A} + 1 = \sqrt{(x + 1)^2 - 1 + A} + 1 = \sqrt{(x + 1)^2 + B} + 1$ $= 1 + \sqrt{(x + 1)^2 + B}$ where B is just as arbitrary as -1 + A.
• February 7th 2008, 05:03 PM
kid funky fried
ok got it thx.
i'll try another problem.