Solving Integral Equation by Converting to Differential Equations

Question

Consider the problem

$$\phi(x) = x - \int_0^x(x-s)\phi(s)\,ds$$

How can we solve this by converting to a differential equation?

Answer 1

Differentiating both sides using Leibniz rule :

$${\phi }'(x)=1-\int_{0}^{x}{\phi (s)ds}$$

Differentiate again:

$${\phi }''(x)=-\phi (x)$$

Answer 2

We have that $$\phi(x)=x-x\int_0^x \phi(s) \mathrm{d} s + \int_0^x s \phi(s)\mathrm{d}s$$ From this, we can see that $\phi(0)=0$. We can differentiate both sides and use the product rule and the FTC1 to get: $$\phi'(x)=1-\int_0^x \phi(s) \mathrm{d}s -x \phi(x)+x\phi(x)$$ $$\phi'(x)=1-\int_0^x \phi(s) \mathrm{d} s$$ From this, we can see that $\phi'(0)=1$. We can differentiate it again: $$\phi''(x)=-\phi(x)$$ Which is an alternative definition of the $\sin$ function.