To find the $\phi(x)$ for particular value of x.

Question

Let $\phi(x)$ be solution of $$x=\int_{0}^{x}\exp(x-t)\phi(t)dt, \quad x\gt0.$$ Then $\phi(1)$ is given by

1. -1
2. 0
3. 1
4. 2

Now I know how to verify whether a function is solution of integral equation or not but here I guess $\phi(x)=x$ gives $\phi(1)=1$. Am I right? Please help me to find the correct option

Answer 1

Use the fundamental theorem of calculus

$$\dfrac{d}{dx}\int_{0}^{x}\exp(x-t)\phi(t)dt=\exp(x-x)\phi(x)+\int_{0}^{x}\dfrac{d\exp(x-t)}{dx}\phi(t)dt$$

and differentiate the given equation with respect to $x$ to obtain:

$$1=\exp(x-x)\phi(x)+\int_{0}^{x}\exp(x-t)\phi(t)dt$$ $$1=\phi(x)+x\implies \phi(x)=1-x$$

Answer 2

Use Laplace method and get $\dfrac{1}{s^2}=\dfrac{1}{s-1}{\cal L}(\phi)$. Then find $\phi$.