solve 2 and 3 only Consider the following initial value prob

solve 2 and 3 only

Consider the following initial value problem: dy = (y - x - 1)dx + (x - y +2)-1 dx y(0) = 0 Write the equation in the form dy / dx = G(ax + by + c), where a, b, and c are constants and G is a function. Use the substitution u = ax + by + c to transfer teh equation into the variable u and x only. Solve the equation in (2).

Solution

Given:dy=(y-x-1)dx+(x-y+2)^-1dx        y(0)=0;

          dy=(y-x-1)dx+1/(x-y+2)dx    

          dy=[(y-x-1)+1/(x-y+2)]dx    

         dy/dx=(y-x-1)(x-y+2)+1/(x-y+2)

                 =xy-y²-x²+xy+2y-2x-x+y-2+1/(x-y+2)

                 =-x²-y²+2xy-3x+3y+1/(x-y+2)

                 =[(-x²-y²+2xy)/(x-y+2)]-3x+3y+1

     if u=(-3x+3y+1) then y=u+3x-1/3

                  =[(-x²-y²+2xy)/(x-y+2)]u

                  =[(-x²-((u+3x-1)/3)²+2x(u+3x-1)/3)]u.

        this is only in x and u only.

Let us solve above equation.

                            =[(-x²-((u+3x-1)/3)²+2x(u+3x-1)/3)]u.

                            =[-x²-(u²/9+x²+1/9+6xu-2u-6x)+2xu/3+6x²/3-2x/3]u

                          =-x²u-u³/9-ux²-u/9-6xu²+2u²+6xu+2/3xu²+2ux²-2/3xu

                            = -u³/9-16/3xu²-u/9+2u²+16/3xu