Consider the following equation:
$$ y^{\prime\prime}(x) +x = \int _0 ^x (x-u)y(u)du \qquad y(0)=0 \quad y^{\prime}(0)=1$$
I solved it by Laplace transform and got $-\sinh x$ as a solution. It is easily to check that it satisfy the equation, but $y^{\prime}(0)=-1$. Why?
Let us first solve it without the use of Laplace transforms. Differentiating twice, using the product rule and the fundamental theorem of calculus, we find that $y$ should satisfy $$ y''''(x)=y(x),\quad y(0)=0,\quad y'(0)=1,\quad y''(0)=0,\quad y'''(0)=-1. $$ This differential equation has the solution $y(x)=\sin x$ only.
With $Y(s)=(\mathcal{L}y)(s)$, I get, using the Laplace transform (and the general rules for derivatives and convolutions), that $$ s^2 Y-1+\frac{1}{s^2}=\frac{1}{s^2}Y. $$ Solving for $Y$ gives $$ Y(s)=\frac{1}{1+s^2}. $$ Thus $y(x)=\sin x$.