Integrals as Solutions

Integrals as solutions simply means the equation hands you the derivative outright, as a formula with no $y$ in it, so you find $y$ by undoing the derivative. It's the most direct kind of differential equation there is, and it's exactly what you need for motion: know the speed or acceleration over time and you can recover the position. All we're really saying with $y' = f(x) \implies y = \int f(x)\,dx + C$ is: $y'$ (the slope of the unknown function) is given as $f(x)$ (a formula in $x$ alone), so $y$ is an antiderivative $\int f(x)\,dx$ plus $C$ (the constant of integration). It's Calculus 1 wearing a new name, so bear with it.

The simplest differential equations. You've actually been solving these since Calculus 1. If the equation says $y' = f(x)$ and $f$ doesn't contain $y$, then "find $y$" is the same question as "find an antiderivative of $f$":

Every antiderivative works, so the general solution is a whole family of curves, each a vertical shift of the others. That makes sense: shifting a curve up or down doesn't change its slope, so the equation can't tell them apart. An initial condition $y(x_0) = y_0$ picks one.

Second order: integrate twice. For $y'' = f(x)$, integrate once to get $y'$ (constant $C_1$), then again to get $y$ (constant $C_2$). Two constants need two conditions, typically $y(x_0)$ and $y'(x_0)$. Use $y'(x_0)$ right after the first integration, before the second, so you're never solving for two unknowns at once.

A common mistake. It's tempting to integrate twice and write a single $+C$ at the end, because that's what you do for one antiderivative. It isn't enough. Integrating $y'' = \sin x$ once gives $y' = -\cos x + C_1$, and integrating again turns that constant into $C_1x$, a whole term you'd lose. Each integration gets its own constant.

Disguises. Equations rarely arrive tidy. Something like

is still this kind. Multiply through by $q'$ to get $e^{3s} - q' = -\frac2s$, so $q' = e^{3s} + \frac2s$ and $q = \frac13e^{3s} + 2\ln|s| + C$. The unknown function is $q(s)$; the letters are just labels.

Motion. Velocity is the derivative of position, and acceleration is the derivative of velocity. So if you know the acceleration $a(t)$ and where the object starts and how fast it's going, two integrations give its position at every time. That's exactly the $y'' = f(x)$ case.

Engineers recover a robot arm's position from its accelerometer by integrating twice, and a rocket's altitude from its thrust profile the same way. Each integration needs a starting value, which is why sensors are zeroed before a run.