# The solution of the differential equation dy/dx=y^2/1-3xy

$\begin{array}{l}\text{Given that,}\\ \frac{dy}{dx}=\frac{y^2}{1-3xy}\\ \Rightarrow\frac{1-3xy}{y^2}=\frac{dx}{dy}\\ \Rightarrow\frac{dx}{dy}+\frac{3x}{y}=\frac{1}{y^2}\\ \text{It is solve by IF factor.}\\ \frac{dx}{dy}+P\left(y\right)x=Q\left(y\right)\\ \text{so,}\ x\cdot e^{\int_{ }^{ }P\left(y\right)dy}=\int_{ }^{ }Q\left(y\right).e^{\int_{ }^{ }P\left(y\right)dy}dy\ +\ c\\ \text{Here we will apply given equcation,}\\ x\cdot e^{\int_{ }^{\frac{1}{y}}dy}=\int_{ }^{ }\frac{1}{y^2}\cdot e^{\int_{\frac{1}{y}}^{ }dy}dy\ +\ c\\ \Rightarrow x\cdot e^{\ln y}=\int_{ }^{ }\frac{1}{y^2}\cdot e^{\ln y}dy\ +\ c\\ \Rightarrow x\cdot y=\int_{ }^{ }\frac{1}{y^2}\cdot y\ dy\ +c\\ \Rightarrow xy=\ln y+\ c\end{array}$

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