Question

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

• A. $\frac{3}{1 + (8)^{1/4}}$
• B. $\frac{3}{1 + 2\sqrt{2}}$
• C. $\frac{3}{1 - 2\sqrt{2}}$
• D. $\frac{3}{1 - (8)^{1/4}}$

Answer 1

Answer: (A)

$\frac{3}{1 + (8)^{1/4}}$

Answer 2

Given:
$(x^2 - 4)dy/dx = (y^2 - 3y)dx = 0.$

Rearranging:
$\frac{dy}{y(y - 3)} = \frac{dx}{x^2 - 4}.$

Using partial fractions:
$\frac{1}{y(y - 3)} = \frac{1}{3} \left(\frac{1}{y - 3} - \frac{1}{y}\right).$

So:
$\frac{1}{3} \left(\frac{1}{y - 3} - \frac{1}{y}\right) dy = \frac{dx}{x^2 - 4}.$

Integrating both sides:
$\frac{1}{3} (\ln|y - 3| - \ln|y|) = \frac{1}{4} \ln\left|\frac{x - 2}{x + 2}\right| + C.$

Simplifying:
$\frac{1}{3} \ln \frac{y - 3}{y} = \frac{1}{4} \ln\left|\frac{x - 2}{x + 2}\right| + C.$

Given $x = 4$ and $y = \frac{3}{2}$, substituting these values:
$\frac{1}{3} \ln \frac{\frac{3}{2} - 3}{\frac{3}{2}} = \frac{1}{4} \ln \left|\frac{4 - 2}{4 + 2}\right| + C.$

$\frac{1}{3} \ln \frac{-\frac{3}{2}}{\frac{3}{2}} = \frac{1}{4} \ln \frac{2}{6} + C.$

Calculating $C$:
$C = \frac{1}{4} \ln 3.$

At $x = 10$:
$\frac{1}{3} \ln \frac{y - 3}{y} = \frac{1}{4} \ln \left|\frac{10 - 2}{10 + 2}\right| + \frac{1}{4} \ln 3.$

Simplifying:
$\ln \frac{y - 3}{y} = \ln 2^{3/4}.$

Thus:
$\ln \frac{y - 3}{y} = \ln 2^{3/4}.$

Given that $y(4) = \frac{3}{2}$ and $y \in (0, 3): \frac{dy}{dx} < 0.\

The Correct answer is: ( \frac{3}{1 + (8)^{1/4}} $