【答案】\(3x^{2}-4xy + 5y^{2}=4\) 

【解析】 

解： 

\[

\begin{aligned}

&(2y - 3x)dx + (2x - 5y)dy = 0 \\

\Rightarrow \quad &2ydx + 2xdy - 3xdx - 5ydy = 0 \\

\Rightarrow \quad &d(2xy) - d\left(\frac{3}{2}x^{2}\right) - d\left(\frac{5}{2}y^{2}\right) = 0

\end{aligned}

\]

即 

\[ \begin{aligned} &d\left(2xy-\frac{3}{2}x^{2}-\frac{5}{2}y^{2}\right)= 0 \\ \Rightarrow &2xy-\frac{3}{2}x^{2}-\frac{5}{2}y^{2}= c \end{aligned} \]

又因为

\[y(1) = 1\]

则

\[2 - \frac{3}{2} - \frac{5}{2} = c,\Rightarrow c = -2\]

即

\[2xy - \frac{3}{2}x^2 - \frac{5}{2}y^2 = -2\]

\[3x^2 - 4xy + 5y^2 = 4\]

则所求方程为\(y = \frac{2x + \sqrt{20 - 11x^2}}{5}\)  。