fiora
Aug 12, 2016

fiora
Jul 4, 2015

fiora
Jun 13, 2015

#1**0 **

ODEs stand for Ordinary differential equations?You can find most notes about differential equation in the following site.

http://faculty.atu.edu/mfinan/nnotes.html

I am studying basic introduction to differential equation,but I don't like the way of my book's authors introduce sicence and enginerring at the same time.If you study science and enginerring first,then study math,then you may understand the derivations easier.

fiora
Nov 17, 2016

#2**+15 **

Now,I want to show you guys some physics and mathematics.

acceleration is the change in velocity in a given amount of time.\(a=\frac{V-Vo}{t}\)

where V is the final veocity and Vo is the initial velocity.

rearrange,we have \(V=at+Vo\)If the object's accelration is constant,then the graph of the velocity is a linear function.

But I would like to express it in y=mx+b form in my graph.And here is my graph

The area of the triangle is the sum of the two base times the hight divide by two.or A=(b1+b2)*h/2

In here ,we could express it as \(A=(V+Vo)*\Delta t/2\)

the area under the curve(line) of function of velocity is the displacement.

Therefore,\(\Delta S=(V+Vo)*\Delta t/2\)\(a=\frac{V-Vo}{t}\Rightarrow t=\frac{V-Vo}{a}\)

If the object start moving at time 0 seconds

\(\Delta t=T-To=t\)

the displacement of the object from time 0 seonds is \(\Delta S=\frac{V^2-Vo^2}{2a}\)

substitute V=at+Vo into \(\Delta S=(V+Vo)*\Delta t/2\)

we have \(\Delta S=(2Vo+at)*t/2\Rightarrow \Delta S=1/2at^2+Vo*t\)

displacement equal final position subract inital position \(\Delta S=S-So\)

\(S=\Delta S+So\)\(S=1/2at^2+Vo*t+So\)

Confirm my previous equation by using integration

\(V=\frac{ds}{dt}=a*t+Vo \Rightarrow \int \frac{ds}{dt}dt=\int at+Vo dt\Rightarrow S=1/2at^2+Vo*t+C\) for some costant C

In here C is the initial position,so \(S=1/2at^2+Vo*t+So\)

http://www.desmos.com/calculator/ibbxgibq5n (my original graph from demos)

fiora
Jan 31, 2016