Find the solution of the differential equation xy' + y = y^2 that satisfies the initial condition y(1) = −2.

Answer 1

First separate equation with y and x on opposite sides

y'/(y²-y) = 1/x

Integrate both sides:

ln|y-1| - ln|y| = ln|x| + C

Simply:

ln|(y-1)/y| = ln |x| + C

y-1/y = x + C

Solve for y:

y = 1/(c-x)

Plug in initial values y = -2, x = 1 to get C

C = 1/2

Final solution:

y = 1/(1/2 - x) or 2/(1-2x)

The hardest part is first separating the equation then integrating. Let me know if you have any questions or need further explanations. I didn't show all of my work here, just broad steps of breaking down the problem.