Finding function $f(x)$

Question

How do we find the function(s) $f(x)$ given that $$f(x)=\int_{0}^{x} te^tf(x-t) \ \mathrm{d}t$$

My Try : I first used the property $\int_{0}^{a}g(x) \ \mathrm{d}x=\int_{0}^{a}g(a-x) \ \mathrm{d}x$ and then differentiated to find that $$f'(x)=f(x)-e^x\int_{0}^{x}e^{-t}f(t) \ \mathrm{d}t$$

Then I differentiated again to find that $$f''(x)=2f'(x)-2f(x)$$

I entered this differential equation into wolfram alpha and it returned

$f(x)=c_1e^x\sin x+c_2e^x\cos x$

But I don't think that this satisfies the given initial condition.

Please help me out. Hints or answers appreciated. Thank you.

Answer 1

Hint: $$\int_{0}^{x} te^tf(x-t) dt = \int_{0}^{x} (x-t)e^{x-t}f(t) dt = xe^x\int_{0}^{x} e^{-t}f(t) dt - e^x\int_{0}^{x} te^{-t}f(t) dt $$whose derivative is $$ (x+1)e^x\int_{0}^{x} e^{-t}f(t) dt + xf(x) - e^x\int_{0}^{x} te^{-t}f(t) dt - xf(x) \\= (x+1)e^x\int_{0}^{x} e^{-t}f(t) dt - e^x\int_{0}^{x} te^{-t}f(t) dt $$

Answer 2

my puzzlement arises as follows: $$ f(x)=\int_{0}^{x} te^tf(x-t) \ \mathrm{d}x \\ f(0) = 0 \\ f'(x) = xe^xf(0) + \int_0^x te^tf'(x-t)dt=\int_0^x te^tf'(x-t)dt \\ f'(0) =0 $$ similarly $$ f^{(n)}(0) =0 $$ if $f$ had a Maclaurin expansion about zero then $f \equiv 0$

have i got my arithmetic wrong, or is there something peculiar at $x=0$ which forbids the Maclaurin development?