Differential Equations
Linear ODE — Local Maximum via Bernoulli/Linear Form
nta_pyq_2026_jan
Grade 12
Question:
If the solution curve $y=f(x)$ of the differential equation $\left(x^2-4\right)y'-2xy+2x\left(4-x^2\right)^2=0$, $x>2$, passes through the point $(3,15)$, then the local maximum value of $f$ is _____.
Step-by-Step Solution
Key Concept: Rewrite: $\dfrac{dy}{dx}-\dfrac{2x}{x^2-4}y=-2x(x^2-4)$. IF $=e^{-\int\frac{2x}{x^2-4}dx}=\dfrac{1}{x^2-4}$. $\dfrac{d}{dx}\!\left(\tfrac{y}{x^2-4}\right)=-2x\Rightarrow y/(x^2-4)=-x^2+C$.
$f(x)=(x^2-4)(12-x^2)$. Local max $=16$ at $x=2\sqrt{2}$.
Correct Answer: 16