# Integrate second order differential equation

• Aug 31st 2010, 04:44 PM
cheme
Integrate second order differential equation
I need to integrate the function d^2 y/d x^2=0

I know I need to begin by...

d^2 y/d x^2 =

d/dx * dy/dx =

d/dy * dy/dx * dy/dx

.... after that I am lost.
• Aug 31st 2010, 04:58 PM
pickslides
How did you get from one step to the other?

Quote:

Originally Posted by cheme

d/dx * dy/dx =

d/dy * dy/dx * dy/dx

Are you trying to find y?
• Aug 31st 2010, 06:26 PM
cheme
Yes I am trying to find y. However, I just tried to factor out differentials. I am confused because there is nothing else to go along with the equation, thus preventing me from using any DiffEq technique.
• Aug 31st 2010, 06:38 PM
Khonics89
Do this...

d^2y/dx^2=0

dy/dx=A where A= constant

y=Ax+b where B = constant..

hence y=Ax+b

This satisfies the given differential :)
• Aug 31st 2010, 06:39 PM
pickslides
You are trying to get?

$\frac{d^2y}{dx^2} = 0$

$\frac{d}{dx} \frac{dy}{dx} = 0$

$\frac{d}{dx} \frac{d}{dx} (y) = 0$

Maybe then intgrate twice on both sides?

$y = \int \int 0 ~dx~dx$
• Aug 31st 2010, 06:54 PM
cheme
The final answer should relate y to x.

I know the final answer is y=(y2-y1)(x-x1)/(x2-x1) +y1
• Aug 31st 2010, 08:24 PM
Khonics89
Cheme: It's effectively the same thing

y=(y2-y1)(x-x1)/(x2-x1) +y1 = Ax+b

where A= y2-y1/x2-x1 and B=((y2-y1)*x1/x2-x1) +y1

They gave you the solution in general form with gradient and everything from y-y1=m(x-x1) form..

All I did is say

y''(x)=0 --> y'(x)=A -->y(x)=Ax+B where A,B are constant..