Suppose \(f\) is a continuous function satisfying the equation \(f(x) = 5 + \int_0^x 3f(t)\,dt\) for all real \(x\). Argue that \(f\) must be differentiable and then find all such function(s) explicitly.
Solution
1. The function \(f\) is differentiable and satisfies\[ f'(x)=3f(x) \]for every \(x\in\mathbb R\).
Because \(f\) is continuous, the function \(3f\) is continuous. By the Fundamental Theorem of Calculus, \[ F(x)=\int_0^x 3f(t)\,dt \] is differentiable and \(F'(x)=3f(x)\). Since the given equation is \(f(x)=5+F(x)\), the function \(f\) is differentiable and \(f'(x)=3f(x)\).
2. The initial value is\[ f(0)=5. \]
Substituting \(x=0\) into the integral equation gives \[ f(0)=5+\int_0^0 3f(t)\,dt=5. \]
3. The only differentiable function satisfying 1 and 2 is\[ f(x)=5e^{3x}. \]
Define \(h(x)=e^{-3x}f(x)\). Then \[ h'(x)=e^{-3x}(f'(x)-3f(x))=0, \] so \(h\) is constant. Since \(h(0)=f(0)=5\), we have \(h(x)=5\) for all \(x\), hence \(f(x)=5e^{3x}\).
4. This function indeed satisfies the original integral equation.
For \(f(x)=5e^{3x}\), \[ 5+\int_0^x 3f(t)\,dt =5+15\int_0^x e^{3t}\,dt =5+5(e^{3x}-1) =5e^{3x} =f(x). \]