Differential Equations
Integral Equation — Differentiation to ODE
nta_pyq_2026_jan
Grade 12
Question:
Let $f$ be a twice differentiable non-negative function such that $\left(f(x)\right)^2=25+\displaystyle\int_0^x\!\left(\left(f(t)\right)^2+\left(f'(t)\right)^2\right)dt$. Then the mean of $f\!\left(\log_e 1\right),\,f\!\left(\log_e 2\right),\,\ldots,\,f\!\left(\log_e 625\right)$ is equal to _____.
Step-by-Step Solution
Key Concept: Differentiate: $2ff'=f^2+(f')^2\Rightarrow(f-f')^2=0\Rightarrow f'=f$. So $f(x)=Ce^x$. At $x=0$: $f(0)^2=25\Rightarrow f(0)=5$, giving $C=5$, $f(x)=5e^x$.
$f(x)=5e^x$. Mean $=1565$.
Correct Answer: 1565