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 = Ce^y