Question

If the solution curve $y = f(x)$ of the differential equation $(x^2 - 4) y' - 2xy + 2x(4 - x^2)^2 = 0, x>2$, passes through the point $(3, 15)$, then the local maximum value of $f$ is ___

Hint:

Check for Linear Differential Equation form $dy/dx + Py = Q$.

Answer 1

Correct Answer: 20

Rewrite: $y' - \frac{2x}{x^2-4}y = -2x(x^2-4)$.
Integrating Factor $I.F. = e^{\int \frac{-2x}{x^2-4} dx} = e^{-\ln(x^2-4)} = \frac{1}{x^2-4}$.
Solution: $y \cdot \frac{1}{x^2-4} = \int -2x dx = -x^2 + C$.
$y = (x^2-4)(C-x^2)$.
Use point $(3, 15)$: $15 = (9-4)(C-9) \implies 3 = C-9 \implies C=12$.
$y = (x^2-4)(12-x^2) = -x^4 + 16x^2 - 48$.
For local maximum, $y' = -4x^3 + 32x = 0 \implies x^2 = 8$ (since $x>2$).
Max value $y = (8-4)(12-8) = 4 \times 4 = 16$.