Solve the differential equation 2x y 1ySolution 2xy1y 1 l

Solution

(2x+y+1)y\' = 1 let 2x+y+1 = t 2+dy/dx = dt dy/dx = dt/dx -2 hence t(dt/dx - 2) = 1 tdt/dx - 2t = 1 dt/dx - 2 = 1/t dt/dx = 1/t +2 dt/dx = (2t+1)/t tdt/(2t+1) = dx 1/2[t/(t+1/2)]dt = dx [t/(t+1/2)]dt = 2dx [1 - 1/2(t+1/2)]dt = 2dx integrating wrt x t - 1/2 ln(t+1/2) = 2x + C putting the value of t 2x+y+1 - 1/2 ln(2x+y+1+1/2) = 2x + C y+1 - 1/2 ln|2x+y+3/2| = C y = 1/2 ln|2x+y+3/2| + C1