If $$y = y(x)$$ is the solution curve of the differential equation $$(x^2 - 4)dy - (y^2 - 3y)dx = 0$$, $$x > 2$$, $$y(4) = \frac{3}{2}$$ and the slope of the curve is never zero, then the value of $$y(10)$$ equals :

We rewrite the differential equation in the standard $$\dfrac{dy}{dx}$$ form:

$$(x^{2}-4)\,dy-(y^{2}-3y)\,dx = 0 \;\;\Longrightarrow\;\; \dfrac{dy}{dx} = \dfrac{y^{2}-3y}{x^{2}-4}$$

This is a separable equation, so we separate the variables:

$$\dfrac{dy}{y^{2}-3y} = \dfrac{dx}{x^{2}-4}$$

Factor each denominator:

$$y^{2}-3y = y(y-3), \qquad x^{2}-4 = (x-2)(x+2)$$

Hence

$$\dfrac{dy}{y(y-3)} = \dfrac{dx}{(x-2)(x+2)}$$

Partial fractions on the left:

$$\dfrac{1}{y(y-3)} = \dfrac{A}{y} + \dfrac{B}{y-3}$$  Set $$1 = A(y-3)+By$$.  Putting $$y = 0$$ gives $$1 = -3A \Longrightarrow A = -\dfrac{1}{3}$$.  Putting $$y = 3$$ gives $$1 = 3B \Longrightarrow B = \dfrac{1}{3}$$.

So $$\dfrac{1}{y(y-3)} = -\dfrac{1}{3y} + \dfrac{1}{3(y-3)}$$

Partial fractions on the right:

$$\dfrac{1}{(x-2)(x+2)} = \dfrac{C}{x-2} + \dfrac{D}{x+2}$$  Set $$1 = C(x+2)+D(x-2)$$.  Putting $$x = 2$$ gives $$1 = 4C \Longrightarrow C = \dfrac{1}{4}$$.  Putting $$x = -2$$ gives $$1 = -4D \Longrightarrow D = -\dfrac{1}{4}$$.

Thus $$\dfrac{1}{(x-2)(x+2)} = \dfrac{1}{4(x-2)} - \dfrac{1}{4(x+2)}$$

Integrate both sides:

$$\int\!\Bigl(-\dfrac{1}{3y} + \dfrac{1}{3(y-3)}\Bigr)\,dy = \int\!\Bigl(\dfrac{1}{4(x-2)} - \dfrac{1}{4(x+2)}\Bigr)\,dx + C$$

Left integral: $$-\dfrac{1}{3}\ln|y| + \dfrac{1}{3}\ln|y-3| = \dfrac{1}{3}\ln\Bigl|\dfrac{y-3}{y}\Bigr|$$

Right integral: $$\dfrac{1}{4}\ln|x-2| - \dfrac{1}{4}\ln|x+2| = \dfrac{1}{4}\ln\Bigl|\dfrac{x-2}{x+2}\Bigr|$$

Therefore

$$\dfrac{1}{3}\ln\Bigl|\dfrac{y-3}{y}\Bigr| = \dfrac{1}{4}\ln\Bigl|\dfrac{x-2}{x+2}\Bigr| + C$$

Multiply by $$12$$ to clear denominators:

$$4\ln\Bigl|\dfrac{y-3}{y}\Bigr| = 3\ln\Bigl|\dfrac{x-2}{x+2}\Bigr| + C'$$

Exponentiating gives

$$\Bigl|\dfrac{y-3}{y}\Bigr|^{4} = K\,\Bigl|\dfrac{x-2}{x+2}\Bigr|^{3}, \qquad K = e^{C'} \gt 0$$

Use the initial condition $$y(4)=\dfrac{3}{2}$$:

At $$x = 4$$, $$\Bigl|\dfrac{y-3}{y}\Bigr| = \Bigl|\dfrac{\tfrac{3}{2}-3}{\tfrac{3}{2}}\Bigr| = \Bigl|\dfrac{-\tfrac{3}{2}}{\tfrac{3}{2}}\Bigr| = 1$$

So left side $$= 1^{4} = 1$$. For $$x = 4,$$ $$\Bigl|\dfrac{x-2}{x+2}\Bigr|^{3} = \Bigl|\dfrac{2}{6}\Bigr|^{3} = \Bigl(\dfrac{1}{3}\Bigr)^{3} = \dfrac{1}{27}$$

Thus $$1 = K \cdot\dfrac{1}{27} \;\Longrightarrow\; K = 27$$

The relation between $$x$$ and $$y$$ is now

$$\Bigl|\dfrac{y-3}{y}\Bigr|^{4} = 27\Bigl|\dfrac{x-2}{x+2}\Bigr|^{3}$$

Find $$y(10)$$:

For $$x = 10$$ ($$x \gt 2$$), $$\Bigl|\dfrac{x-2}{x+2}\Bigr| = \dfrac{10-2}{10+2} = \dfrac{8}{12} = \dfrac{2}{3}$$

Hence $$\Bigl|\dfrac{y-3}{y}\Bigr|^{4} = 27\Bigl(\dfrac{2}{3}\Bigr)^{3} = 27\cdot\dfrac{8}{27} = 8$$

Since the solution started with $$0 \lt y \lt 3$$ and the slope $$\dfrac{dy}{dx}$$ is negative in this region, $$y$$ remains between $$0$$ and $$3$$. Thus $$y-3 \lt 0$$ and $$y \gt 0$$, making $$\Bigl|\dfrac{y-3}{y}\Bigr| = \dfrac{3-y}{y}$$.

Therefore $$\Bigl(\dfrac{3-y}{y}\Bigr)^{4} = 8 \;\;\Longrightarrow\;\; \dfrac{3-y}{y} = 8^{1/4}$$

Let $$8^{1/4}=t$$. Then $$3 - y = ty \;\;\Longrightarrow\;\; 3 = y(1+t) \;\;\Longrightarrow\;\; y = \dfrac{3}{1+t}$$

Since $$t = 8^{1/4}$$, we obtain

$$y(10) = \dfrac{3}{1 + 8^{1/4}}$$

This matches Option A.

Answer: Option A