$$\begin{aligned} & \left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0 \\ & \Rightarrow \int \frac{d y}{y^2-3 y}=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \\ & \Rightarrow \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \end{aligned}$$

$$\begin{aligned} & \text { At } \mathrm{x}=4, \mathrm{y}=\frac{3}{2} \\ & \therefore \mathrm{C}=\frac{1}{4} \ln 3 \\ & \therefore \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\frac{1}{4} \ln (3) \\ & \text { At } \mathrm{x}=10 \\ & \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3) \\ & \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\ln 2^{3 / 4}, \forall \mathrm{x}>2, \frac{\mathrm{dy}}{\mathrm{dx}}<0 \\ & \text { as } \mathrm{y}(4)=\frac{3}{2} \Rightarrow \mathrm{y} \in(0,3) \\ & -\mathrm{y}+3=8^{1 / 4} \cdot \mathrm{y} \\ & \mathrm{y}=\frac{3}{1+8^{1 / 4}} \end{aligned}$$