Question

Find the general solution: $(x+y)\frac {dy}{dx}=1$

Answer 1

(x+y)$\frac {dx}{dy}$ = 1

⇒$$\frac {dx}{dy}= $\frac {1}{x+y}$

⇒$$\frac {dx}{dy} = x+y

⇒$$\frac {dx}{dy} - x = y

This is a linear differential equation of the form:

$\frac {dx}{dy}$ + px = Q (where p=-1 and Q=y)

Now, I.F. = e∫pdy= e-y.

The general solution of the given differential equation is given by the relation,

x(I.F.) = $\int$(Q×I.F.)dy + C

$⇒$xe-y = $\int$(y.e-y)dy + C

$⇒$xe-y = y.∫e-ydy-$\int$[$\frac {d}{dy}$(y)$\int$e-y dy]dy+C

$⇒$xe-y = y(-e-y)-$\int$(-e-y)dy + C

$⇒$xe-y = - ye-y+$\int$e-ydy+C

$⇒$xe-y = -ye-y-e-y+C

$⇒$x = - y-1+$\frac {C}{e^{-y}}$

$⇒$x+y+1 = Cey