---
title: "Simple Differential Equations"
type: concept
related: [Power Series, Diffeq Intro]
source: https://www.jemoka.com/posts/kbhsimple_differential_equations/
confidence: high
status: active
---

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} &amp; \dv{x}{t} = x(t) \\ \Rightarrow\ &amp; \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&rsquo;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} &amp;\frac{1}{x(t)}\dd{x} = \dd{t} \\ \Rightarrow\ &amp;\int \frac{1}{x(t)}\dd{x} = \int \dd{t} \\ \Rightarrow\ &amp; \ln (x(t)) = t \\ \Rightarrow\ &amp; x(t) = e^{t} \end{align}
Awesome. It should&rsquo;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} &amp;x(t) = e^{it} \\ \Rightarrow\ &amp; \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 \(a&rsquo;(t)\).
But anyways through this we will end up with the same solution we get from solving differential equations.
