Thread: 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

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?

Originally Posted by mr fantastic

Do you need to make y the subject?

If so, then

where A = K + 1 is just as arbitrary as K.

Therefore



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 ........

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)

Originally Posted by mr fantastic

If so, then

where A = K + 1 is just as arbitrary as K.

Therefore



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 ........

where B is just as arbitrary as -1 + A.

ok got it thx.
i'll try another problem.