Find the solution of the differential equation that satisfies the given initial condition. xy' + y = y^2 , y(1) = -2
I've been attempting to solve it by separable equations, by making the equation xy'=y(y-1), but my answer has not been working correctly.
I've done thiat already, and by using partial fractions, I come up with:
ln(y-1/y) =lnx +c
y-1/y= x + C2
1 - 1/y = x + C2
-1/y = x + C3
y = 1/(-x + C3)
Using my initial condition, I get
-2 = 1/ (-1 + C3)
Therefore, C3 = 1/2
So my solution should be
y= 1/(-x + 1/2)
However, when I tried to check my answer by substituting it back into the original given differential equation, my answer did not work out. Can anyone help me?
No. Taking the exponential of both sides gives (y-1)/y= C2 x where C2=
1 - 1/y = x + C2
-1/y = x + C3
y = 1/(-x + C3)
Using my initial condition, I get
-2 = 1/ (-1 + C3)
Therefore, C3 = 1/2
So my solution should be
y= 1/(-x + 1/2)
However, when I tried to check my answer by substituting it back into the original given differential equation, my answer did not work out. Can anyone help me?