Question

Find all solutions of y'=e^(2x)+(1-2e^x)y+y^2.

Answer 1

\[y ^{-2} y'=e ^{2x} y ^{-2}+(1-2e ^{x})y ^{-1}+1\]

Answer 2

\[v=y ^{-1}\]

Answer 3

\[v'=-y ^{-2} y'\]

Answer 4

\[-v'=e ^{2x} v ^{2}+(1-2e ^{x})v+1\]

Answer 5

\[-v'=e ^{2x}v ^{2}+v-2e ^{x}v+1\]

Answer 6

@ganeshie8

Answer 7

You can complete the square on the v on the right side,\[\Large\rm -v'=\color{orangered}{(v e^x)^2-2(v e^x)+1}+v\]\[\Large\rm -v'=\color{orangered}{(ve^x-1)^2}+v\]I'm not sure how that would help :c
Just a thought...