I'm currently studying the method of solving integral and differential equations via Laplace transform and something is leaving me confused. Let's say i have the following equation:
$$y'(t) + \int_{0}^{t} y(\tau)d\tau = 0$$
Now, all i need is one initial condition, because, if apply the Laplace transform on the equation:
$$sY(s) – y(0) + \frac{1}{s} Y(s) = 0$$
Where $Y(s) = \mathcal{L}\{f(t)\}$. So here i only need the value of $y(0)$ to completely describe the system. Now, if i decide to apply the operator $\frac{d}{dt}$ on the first equation:
$$y''(t) + y(t) = 0$$
Which requires both $y(0) $ and $y'(0)$ to be solved. How is that possible?
What i am thinking is that, in the first equation, one should define $f(t)$ such that
$$\frac{df(t)}{dt} = y(t)$$
And know the value of $f(0)$. I've choosen to define $f$ this way (as the antiderivative of $y$) because, if i choose to define it the "usual" way:
$$f(t) = \int_{0}^{t} y(\tau) d\tau$$
Then one would always obtain $f(0) = 0$. So, is this correct? There are indeed two degrees of freedom in the first equation, and the second one you should now is $f(0)$? Or is there some other condition (like a constant from an integral) or it is the case that integral equations indeed require less information to descibre a system?
Thanks in advance.
Best Answer
As explained in comments, the integral form of the equation implies $y'(0)=0$, which is why only one initial value $y(0)=0$ is needed to determine the solution.
It is possible to write down a first-order integral equation of the same type which does not prescribe the value of $y'$ anywhere. For example $$ y'(t) - \int_{-\infty}^t y(s)\,ds = 0 $$ which has the family of solutions $y(t)=Ce^t$. The differential form $y''-y=0$ has a two-dimensional family of solutions $C_1e^t+C_2e^{-t}$, but now the integral form excludes some of them because of the divergence of the integral.
With the $+$ sign we don't get to start integration at infinity, because the integral wouldn't converge for nontrivial solutions of $y''+y=0$.