Differential Equations
Differential Equations
nta_pyq_2025_apr
Grade 12
Question:
Let $f:(0,\infty)\to\mathbb{R}$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2 f'(x) = 2xf(x)+3$, with $f(1) = 4$. Then $2f(2)$ is equal to:
Step-by-Step Solution
Key Concept: Rearrange to $\dfrac{x^2 f'(x)-2xf(x)}{x^4} = \dfrac{3}{x^4}$, i.e., $\dfrac{d}{dx}\!\left(\dfrac{f(x)}{x^2}\right) = \dfrac{3}{x^4}$; integrate directly.
$\dfrac{d}{dx}\!\left(\dfrac{f(x)}{x^2}\right) = \dfrac{3}{x^4}$.
Integrating: $\dfrac{f(x)}{x^2} = -\dfrac{1}{x^3}+C \Rightarrow f(x) = -\dfrac{1}{x}+Cx^2$.
At $x=1$: $4 = -1+C \Rightarrow C=5$. So $f(x) = -\dfrac{1}{x}+5x^2$.
$$2f(2) = 2\!\left(-\frac{1}{2}+20\right) = 2\cdot\frac{39}{2} = 39.$$
Correct Answer: 1