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