$dy/ dx = (4x + y + 1)$ ..(i)
Put 4x + y + 1 = t (ii)
Differentiating w.r.t. x, we get
$4 + dy/dx = dt/ dx$
$dy/dx = dt/ dx - 4$ .... (iii)
Substituting (ii) and (iii) in (i), we get
$dt/ dx 4 = t$
$dt/ dx = t + 4$
$dt/ (t + 4) = dx$
Integrating on both sides, we get
$intdt/(t+4) = int dx$
log | t + 4 | = x + c
log |(4x + y + 1) + 4 | = x + c
log | 4x + y + 5 | = x + c (iv)
When y = 1, x = 0, we have
log |4(0) + 1 + 5| = 0 + c
c = log |6|
Substituting c = log |6| in (iv), we get
log |4x + y + 5| = x + log |6|
log |4x + y + 5 | - log |6| = x
log $|(4x+y+ 5) /6 | = x$,
which is the required particular solution.
