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--- title: "Simple Differential Equations" source: https://www.jemoka.com/posts/kbhsimple_differential_equations/ --- Here is the most simple Differential Equation one could imagine: \begin{equation} \dv{x}{t} = f(t,x) \end{equation} Or, perhaps, we have a second order differential equation which is the same thing but in the second degree: \begin{equation} \dv[2]{x}{t} = f\qty(t,x,\dv{x}{t}) \end{equation} Then in which case, we have that the first most simple type of differential equation to be as follows: \begin{equation} \dv{x}{t} = x(t) \end{equation} If we can solve this, we can generalize this to most of other First-Order Differential Equations. where, the function \(f(t,x)=x(t)\). \begin{align} & \dv{x}{t} = x(t) \\ \Rightarrow\ & \frac{1}{x(t)}\dd{x} = \dd{t} \end{align} At this point, you may ask yourself, why not construct it such that we have \(\dd{x} = x(t)\dd{t}\)? Well, its because our \(x\) is a variable in \(t\), so if we constructed it that way we’d have to integrate a function \(\dd{t}\) with usub and the reverse chain rule, etc. etc. If we are instead integrating it on \(\dd{x}\), it becomes much easier because our variable of interest no longer considers the \(t\). Continuing on, then: \begin{align} &\frac{1}{x(t)}\dd{x} = \dd{t} \\ \Rightarrow\ &\int \frac{1}{x(t)}\dd{x} = \int \dd{t} \\ \Rightarrow\ & \ln (x(t)) = t \\ \Rightarrow\ & x(t) = e^{t} \end{align} Awesome. It should’t be hard also to see that, generally: \begin{equation} x(t) = e^{ct} \end{equation} is the solution to all equations \(\dv{x}{t} = cx\). Turns out (not proven in the book), this holds for complex valued equations as well. So, we have some: \begin{align} &x(t) = e^{it} \\ \Rightarrow\ & \dv{x}{t} = ix \end{align} Of course, from elementary calculus we also learned the fact that \(e^{x}\) can be represented as a power series; so check that out for now we connect it. This equation leads us to solve: \begin{equation} \dv{x}{t} + ax = b(t) \end{equation} In order to do this, we neeed to find a replacement of the property that: \begin{equation} \dv t\qty(e^{at}x) = e^{at}\qty(\dv{x}{t} +at) \end{equation} A more general result of the above form is \begin{equation} \dv{x}{t} + a(t)x = b(t) \end{equation} This is fine, but now we need to leverage to chain rule to have \(\dv t a(t)\) would be simply changing the above result to \(a’(t)\). But anyways through this we will end up with the same solution we get from solving differential equations.