Find f(x) $\int_0^x f(u)du - f'(x) = x$

Question

Find f(x)

$$\int_0^x f(u)du - f'(x) = x$$

I was not given f(0) which makes it difficult for me to find f(x). This is what I have thus far:

$$\frac{F(p)}{p}-pF(p)+f(0)=\frac{1}{p^2}$$ $$\frac{F(p)}{p}-pF(p)=\frac{1}{p^2}-f(0)$$ $$F(p)(\frac{1}{p}-p)=\frac{1}{p^2}-f(0)$$ $$F(p)=\frac{\frac{1}{p^2}-f(0)}{(\frac{1}{p}-p)}$$

Can I even take the Inverse Laplace to get f(x) or do I have to do something entirely different. Once again I was not given f(0). And I used the formula:

$$L(\int_0^t f(u)du)=\frac{1}{p} F(p)$$

where L=Laplace Transform

Answer 1

$$\int_{0}^{x}f(u)\space\text{d}u-f'(x)=x\Longleftrightarrow$$ $$\mathcal{L}_{x}\left[\int_{0}^{x}f(u)\space\text{d}u-f'(x)\right]_{(s)}=\mathcal{L}_{x}\left[x\right]_{(s)}\Longleftrightarrow$$ $$\mathcal{L}_{x}\left[\int_{0}^{x}f(u)\space\text{d}u\right]_{(s)}-\mathcal{L}_{x}\left[f'(x)\right]_{(s)}=\mathcal{L}_{x}\left[x\right]_{(s)}\Longleftrightarrow$$ $$\frac{f(s)}{s}-sf(s)+f(0)=\frac{1}{s^2}\Longleftrightarrow f(s)\left[\frac{1}{s}-s\right]=\frac{1}{s^2}-f(0)\Longleftrightarrow$$ $$f(s)=\frac{\frac{1}{s^2}-f(0)}{\frac{1}{s}-s}\Longleftrightarrow f(s)=\frac{1-f(0)s^2}{s-s^3}\Longleftrightarrow$$ $$\mathcal{L}_{s}^{-1}\left[f(s)\right]_{(x)}=\mathcal{L}_{s}^{-1}\left[\frac{1-f(0)s^2}{s-s^3}\right]_{(x)}\Longleftrightarrow f(x)=1+(f(0)-1)\text{cosh}(x)$$

Answer 2

A start: Differentiate. By the Fundamental Theorem of Calculus,we have $f(x)-f''(x)=1$. Solve this differential equation.

By substituting in the original equation, we can see that $f'(0)=0$.

Answer 3

Edit: I add done a mistake in my previous answer (taking $f(0)=0$ instead of $f'(0)=0$). It is corrected in this version:

Taking $x=0$, one finds

$$f'(0)=0 \ \ (1)$$

Thus this integro-differential equation is equivalent, by derivation, to

$$f(x)-f''(x)=1 \ \ \text{with condition (1)}$$

which has solution

$$1+Ae^{x}+Be^{-x} \ \ \text{with condition (1)}$$

(particular solution of the diff. equ. + general solution of the homogeneous associated diff. equ. $f(x)-f''(x)=0$.)

Taking (1) into account, and because $f'(x)=Ae^{x}-Be^{-x}$, we have $f'(0)=A-B=0$; thus, the solution to the question is:

$$f(x)=1+Ae^{x}+Ae^{-x}=1+a\cosh{x}$$

$a$ being an arbitrary real constant.

Of course, treating the initial equation by Laplace Transform yields the same results (as confirmed by the result of @Jan Eerland)