Rewriting the given diff. eqn. (DE) as #dy/dx-y=2x#, we find
that it is a linear DE of the form : #dy/dx+yP(x)=q(x)#.
To find its gen. soln. (GS), we need to multiply it by the
integrating factor (IF) #e^(intP(x)dx#.
Since,
#P(x)=-1, intP(x)dx=int-1dx=-x :." IF is "e^-x.#
Multiplying the DE by IF, we get,
#e^-xdy/dx-ye^-x=2xe^-x#.
#:. e^-x*d/dx(y)+y*d/dx(e^-x)=2xe^-x, or, #
# d/dx(y*e^-x)=2xe^-x#.
#:. y*e^-x=int2xe^-xdx+C#,
#=2[x*inte^-xdx-int{d/dx(x)inte^-xdx}dx]+C......[because," Integration by Parts]"#,
#=2[x(-e^-x)-int(-e^-x)dx]+C#,
#=2[-xe^-x-e^-x]+C#.
# rArr" The GS is, "y*e^-x+2(x+1)e^-x=C, or, #
# y+2(x+1)=Ce^x#.
Feel the Joy of Maths.!