\begin {align*} y^{\prime } & =ay-by^{2}\\ y\left ( 0\right ) & =y_{0} \end {align*}
A solution exist an is unique. Integrating gives\begin {align*} \int \frac {dy}{ay-by^{2}} & =\int dx\qquad ay-by^{2}\neq 0\\ \frac {1}{a}\ln y-\frac {1}{a}\ln \left ( by-a\right ) & =x+c_{1}\\ \ln y-\ln \left ( by-a\right ) & =ax+ac_{1}\\ \frac {y}{by-a} & =e^{ax+ac_{1}}\\ \frac {y}{by-a} & =c_{2}e^{ax}\\ y & =c_{2}bye^{ax}-ac_{2}e^{ax}\\ y\left ( 1-c_{2}be^{ax}\right ) & =-ac_{2}e^{ax}\\ y & =\frac {-ac_{2}e^{ax}}{1-c_{2}be^{ax}}\\ & =\frac {ac_{2}e^{ax}}{c_{2}be^{ax}-1}\\ & =\frac {ac_{2}}{c_{2}b-e^{-ax}}\\ & =\frac {a}{b-\frac {1}{c_{2}}e^{-ax}}\\ & =\frac {a}{b+c_{3}e^{-ax}} \end {align*}
Applying IC\begin {align*} y_{0} & =\frac {a}{b+c_{3}}\\ \left ( b+c_{3}\right ) y_{0} & =a\\ by_{0}+c_{3}y_{0} & =a\\ c_{3} & =\frac {a-by_{0}}{y_{0}} \end {align*}
Hence the solution becomes\begin {align*} y & =\frac {a}{b+\left ( \frac {a-by_{0}}{y_{0}}\right ) e^{-ax}}\\ & =\frac {ay_{0}}{by_{0}+\left ( a-by_{0}\right ) e^{-ax}} \end {align*}