Solution of the following integral Equation $\varphi(x) - \lambda\int\limits_{-1}^1 x e^t\varphi(t) \: dt=x$

Question

Consider that the following equation is solvable then analyze with respect to $\lambda$ $$\varphi(x) - \lambda\int\limits_{-1}^1 x e^t\varphi(t) \: dt=x$$

Can someone tell me how can I solve it ?

Answer 1

If there is no typo in the equation this is very easy. Since $c=\int_{-1}^{1}e^{t}\phi(t)dt$ is just a constant we get $\phi (x)=\lambda cx+x=x(1+c\lambda)$. Now multilply by $e^{t}$ and integrate this. You get $c=\int_{-1}^{1}te^{t}dt (1+c\lambda)$. Solve this for $c$ and you get your solution.

Answer 2

Assuming that $\phi(x)$ is $C^2$, differentiating in $x$ twice gives $\phi''(x)=0$, so $\phi(x)=bx+c$.

Plugging this into the original equation gives:

$bx+c -x\lambda\int_{-1}^1 e^t(bt+c)dt=x$.

This should give you two equations for your two unknowns after you compare powers of $x$.