Q: Solve the differential equation :
(x+y+1)(dy/dx) = 1
Ans : I put (x+y)=t
(t+1)(d(t-x)/dx) = 1
(t+1)(dt/dx - 1) = 1
(t+1)(dt/dx) = t+2
[(t+1)/(t+2)]*(dt/dx) = dx
integrating both sides,
int[ t/(t+2) + 1/(t+2) ]dt = int.dx
now the second term in the LHS can be integrated directly, and so can the RHS.
but the term t/(t+2) has to be integrated separately by adding and subtracting 2 in numerator.
=> 1 - 2/(t+2)
and so this can now be solved too.
=> int.[ 1-2/(t+2) + 1/(t+2) ]dt = int.dx
=> int.[ 1 - 1/(t+2) ]dt = int.dx
=> t - log(t+2) = x +C
substituting t= x+y
x+y - log(x+y+2) = x +C
y= log(x+y+2) +C
which i think is the answer.
am I correct? I dont have the answers so if some1 can verify, It would be a great.
Post edited by PettyOfficer on