Checking whether the function $u(x)=e^x$ solves the integral equation $u(x) + \lambda \int_0^1\sin(xt) u(t) dt=1$

Question

The function being $u(x)=e^x$ and the integral equation is $$u(x) + \lambda \int_0^1\sin(xt) u(t) dt=1$$

I can't do the integration and I'm confused about how to deal with the parameter $\lambda$ here.

Answer 1

In order to see if $u(x)=e^x$ is , or not, solution of the integral equation $$u(x) + \lambda \int_0^1\sin(xt) u(t) dt=1$$ It is sufficient to put $u=e^x$ into the equation and check the result $$e^x + \lambda \int_0^1\sin(xt) e^t dt=e^x + \lambda \frac{x+(\sin x+x\cos x)e}{1+x^2}$$ Obviously, the left term is not equal to the right term $=1$.

Hense, $e^x$ is NOT solution of the equation.

Another way, without need to compute the integral consists in computing the second derivative : $$u''(x) = \lambda \int_0^1\sin(xt) t^2 u(t) dt$$ $$u''_{(x=0)}=0$$ Comparing to the second derivative of $e^x$ at $x=0$, which is equal to $1$, both second derivatives are not equal. So the the two functions are different.

The conclusion is the same : $e^x$ is not solution of the equation.