| Line 10: | Line 10: | ||
<font size="3px"> Previously, we learnt to solve equations with numbers as solutions. For example, for linear equations <math>ax+b=0</math> with respect to <math>x</math> and <math>a≠0</math>, the solution is going to be <math>x=-\frac{b}{a}</math>. For quadratic equations <math>ax<sup>2</sup>+bx+c=0</math> with respect to <math>x</math> and <math>a≠0</math>, the solution is going to be <math>x={\frac{-b±\sqrt[2]{b^2-4ac}}{2a}}</math>. <font size="3px"> | <font size="3px"> Previously, we learnt to solve equations with numbers as solutions. For example, for linear equations <math>ax+b=0</math> with respect to <math>x</math> and <math>a≠0</math>, the solution is going to be <math>x=-\frac{b}{a}</math>. For quadratic equations <math>ax<sup>2</sup>+bx+c=0</math> with respect to <math>x</math> and <math>a≠0</math>, the solution is going to be <math>x={\frac{-b±\sqrt[2]{b^2-4ac}}{2a}}</math>. <font size="3px"> | ||
| + | |||
In both examples, the solutions <math>x=-\frac{b}{a}</math> and <math>x={\frac{-b±\sqrt[2]{b^2-4ac}}{2a}}</math> are particular numbers, and mostly we are discussing them within real numbers. When a set of numbers match another set of numbers by a particular definition, a function forms. Therefore, if we apply the function to the solution, (i.e. the equation has a function as a solution) a '''differential equation''' forms. Consider what we have learned in Calculus I. When we are taking derivatives to a function, the result is still going to be a function (either constant functions or not). | In both examples, the solutions <math>x=-\frac{b}{a}</math> and <math>x={\frac{-b±\sqrt[2]{b^2-4ac}}{2a}}</math> are particular numbers, and mostly we are discussing them within real numbers. When a set of numbers match another set of numbers by a particular definition, a function forms. Therefore, if we apply the function to the solution, (i.e. the equation has a function as a solution) a '''differential equation''' forms. Consider what we have learned in Calculus I. When we are taking derivatives to a function, the result is still going to be a function (either constant functions or not). | ||
Revision as of 16:47, 11 October 2017
Introduction to Differential Equations
1.0 Abstract
When I was first learning differential equations myself, I felt hard to understand the textbook and my lecture notes. After winning the battle, now I am trying to build up those concepts again and explain them in an easier and more concise way. It is not that academic, aiming for intuitive understanding.
1.1 Concept
Previously, we learnt to solve equations with numbers as solutions. For example, for linear equations $ ax+b=0 $ with respect to $ x $ and $ a≠0 $, the solution is going to be $ x=-\frac{b}{a} $. For quadratic equations $ ax<sup>2</sup>+bx+c=0 $ with respect to $ x $ and $ a≠0 $, the solution is going to be $ x={\frac{-b±\sqrt[2]{b^2-4ac}}{2a}} $.
In both examples, the solutions $ x=-\frac{b}{a} $ and $ x={\frac{-b±\sqrt[2]{b^2-4ac}}{2a}} $ are particular numbers, and mostly we are discussing them within real numbers. When a set of numbers match another set of numbers by a particular definition, a function forms. Therefore, if we apply the function to the solution, (i.e. the equation has a function as a solution) a differential equation forms. Consider what we have learned in Calculus I. When we are taking derivatives to a function, the result is still going to be a function (either constant functions or not).
