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

The Correct Option is A

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