Question

Solve y'+y=y^2

Answer 1

diff. eq?

Answer 2

Yes, linear dff eq

Answer 3

dy/dx =y^2 - y

dx/dy = 1/y^2 -y
dx= 1/y^2-y dy
integrate both side

Answer 4

he is correct

Answer 5

After simplifying I get 1/y=1-e^x but it doesnt match the book answer: 1/y=Ce^x + 1

Answer 6

\[x=\ln (y-1)-\ln (y) +c\]
\[x+c=\ln (y-1)/y\]
\[e^(x) e^c = (y-1) /y\]
\[e^x C=(y-1)/y\]

\[y e^x C = y-1\]
\[Cy e^x - y =-1\]
\[y(Ce^x -1)=-1\]
\[y= 1/ (C e^x+1)\]