Differential Equations Part 3

Concept:The given differential equation is linear in $y$.We solve using integrating factor, find the particular solution, then differentiate to locate the local maximum.Formula:Linear differential equation: $\frac{dy}{dx} + P y = Q$.Integrating factor: $I.F. = e^{\int P \, dx}$.Solution: $y \cdot I.F. = \int Q \cdot I.F. \, dx + C$.Solution:Step 1: Rewrite the equation:$(x^2-4)y' - 2xy = -2x(4-x^2)^2$.Note $(4-x^2)^2 = (x^2-4)^2$.Divide by $x^2-4$: $y' - \frac{2x}{x^2-4} y = -2x(x^2-4)$.Here $P = -\frac{2x}{x^2-4}$, $Q = -2x(x^2-4)$.Step 2: Find I.F.:$I.F. = e^{\int -\frac{2x}{x^2-4} dx} = e^{-\ln(x^2-4)} = \frac{1}{x^2-4}$.Step 3: Solution: $\frac{y}{x^2-4} = \int -2x(x^2-4) \cdot \frac{1}{x^2-4} dx + C = \int -2x \, dx + C = -x^2 + C$.Thus $y = (x^2-4)(C - x^2)$.Step 4: Use point $(3,15)$: $15 = (9-4)(C-9) = 5(C-9) \Rightarrow C-9=3 \Rightarrow C=12$.So $f(x) = (x^2-4)(12-x^2) = -x^4 + 16x^2 - 48$.Step 5: Find critical points: $f'(x) = -4x^3 + 32x = -4x(x^2-8)$.For $x>2$, set $f'(x)=0 \Rightarrow x^2=8 \Rightarrow x=2\sqrt{2}$.Step 6: Check sign of $f'$ around $x=2\sqrt{2}$:For $x=2.5$ (left): $f'(2.5) = -10(6.25-8)=17.5 > 0$.For $x=3$ (right): $f'(3) = -12(9-8) = -12 < 0$.Sign changes from positive to negative, so $x=2\sqrt{2}$ is a local maximum.Step 7: Local maximum value: $f(2\sqrt{2}) = ((8)-4)(12-8) = 4 \times 4 = 16$.Answer:The local maximum value of $f$ is $16$. (Option A)