Integral Equations
Convolution-type integral equation
MJAT_TS1_P2
Grade 12
Question:
Let $f(x)$ be a differentiable function such that
$$f(x) = x^2 + \int_0^x e^{-t}\,f(x-t)\,dt$$
Then $6f(1) =$
Step-by-Step Solution
Key Concept: Use the substitution $u = x-t$ to rewrite the convolution: $\int_0^x e^{-t}f(x-t)\,dt = e^{-x}\int_0^x e^u f(u)\,du$. Differentiate both sides with respect to $x$ to convert to an ODE.
$f'(x) = 2x + e^{-x}\cdot e^x f(x) - e^{-x}(e^x f(x) - e^0 f(0)) + \text{chain} = 2x + x^2$ (after simplification using the original equation). Integrating: $f(x) = \frac{x^3}{3} + x^2 + K$. $f(0)=0 \Rightarrow K=0$. So $f(1) = \frac{1}{3}+1 = \frac{4}{3}$, giving $6f(1) = \mathbf{8}$.
Correct Answer: 8