Elementary differential equations question - constant We have the following 2 differential equations:


*

*$ \dfrac{dy}{dt} = 2t $

*$ \dfrac{dy}{dt} = 2y $
So the solution to the first one is quite easy, because it's really the same as just $\displaystyle \int 2x dx$, so you get $y = t^2 + c_1$.
However, the second one is a bit different. Here the solution is $e^{2t} \times c_1$.


*

*Why is it plus a constant in the first one and times a constant in the second one? 

*What would the easiest way be to solve the second one? Wolfram alpha divides both sides by $y$, but I don't intuitively understand $\dfrac{dy}{dt}/y$ (I actually don't intuitively understand $\dfrac{dy}{dx}$ at all, but no one at a high school level does, because it is never really explained).
 A: Any first order differential equation will have one free parameter in the solution which has to be determined by an initial condition.
Here's a standard way of how we solve those equations.(You probably know this already)
$$
\frac{\mathrm{d}y}{\mathrm{d}t} = 2t \\
\mathrm{d}y = 2t \mathrm{d}t \\
\int\mathrm{d}y = \int2t \mathrm{d}t \\
y  = t^2 + c_1
$$
$$
\frac{\mathrm{d}y}{\mathrm{d}t} = 2y \\
\frac{\mathrm{d}y}{y} = 2\mathrm{d}t \\
\int\frac{\mathrm{d}y}{y} = \int2 \mathrm{d}t \\
\ln\left|y\right|  = 2t + c_1\\
y = e^{2t+c_1} \\
y = e^{c_1}e^{2t} \\
y = c_2 e^{2t}
$$
Observe that we are indeed getting a $+c_1$ initially in both the derivations as an integration constant. But because the second DE has $\ln\left|y\right|$ on RHS the constant which was initially additive becomes multiplicative on exponentiation.
I hope that clarified some of the issues you had. I suggest you read the formal definitions of limits, differentiation and Riemann Integration though it is not in your syllabus. It's not at all difficult to understand and also gives you an intuitive picture of what's going on.
