Question

(3x+3y-1)dx+(x+y+1)dy=0 solve the separable equation using linear substitution

Answer 1

$Solution: Given ~~ \\(3x+3y-1)dx+(x+y+1)dy=0\\Now,Substitute~\\ u=x+y~ \implies du=dx+dy ~\\y=u-x~\implies dy=du-dx \\Put~ in~ given ~ordinary ~~differential~ ~equation \\(3(u-x)+3x-1)dx+(u+1)(du-dx)=0 \\\implies (3u-3x+3x-1)dx+udu-udx-du-dx=0 \\Simplifying,we~ get \\(u+1)du=(2-2u)dx \\Divide ~by ~2-2u \\\frac{(-u-1)}{2(u-1)}du=dx \\Which ~is~ separable~ differential~ equation~form~ M(x)du=N(x)dx \\ \therefore Integrating~both~sides~ \\\int \frac{(-u-1)}{2(u-1)}du=\int dx \\\therefore \frac{-u}{2}-ln ~(u-1)=x+c \\back ~substitution~u=x+y \\-ln (x+y-1)-\frac{x+y}{2}=x+c \\Simplify, ~we ~get \\-ln (x+y-1)-\frac{y}{2}=frac{3x}{2}+c,x=1-y$