Step-by-step explanation:
[tex]y' + y = y^2[/tex]
We can rewrite the differential equation above as
[tex]\dfrac{dy}{dx} + y = y^2[/tex]
[tex]dy = (y^2 - y)dx[/tex]
or
[tex]\dfrac{dy}{y^2 -y} = dx[/tex]
We can rewrite the left side of the equation above as
[tex]\dfrac{dy}{y^2-y}=\dfrac{dy}{y(y-1)}= \left(\dfrac{1}{y-1} - \dfrac{1}{y} \right)dy[/tex]
We can the easily integrate this as
[tex]\displaystyle \int \left(\dfrac{1}{y-1} - \dfrac{1}{y} \right)dy = \int dx[/tex]
or
[tex]\displaystyle \int \dfrac{dy}{y-1} - \int \dfrac{dy}{y} = \int dx[/tex]
This will then give us
[tex]\ln |y-1| - \ln |y| + \ln |k| = x[/tex]
where k is the constant of integration. Combining the terms on the left hand side, we get
[tex]\ln \left|\dfrac{k(y-1)}{y} \right| = x[/tex]
or
[tex]\dfrac{y-1}{y} = \frac{1}{k}e^x[/tex]
Solving for y, we get
[tex]y= \dfrac{1}{1- \frac{1}{k} e^x}=\dfrac{k}{k-e^x}[/tex]
We know that [tex]y(0)= \frac{1}{3}[/tex], so when we substitute [tex]x=0[/tex], we find that [tex]k = -\frac{1}{2}[/tex].
Therefore, the final form of the solution to the differential equation above is
[tex]y = \dfrac{1}{1+2e^x}[/tex]
