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General Solution to Differential Equation

Date: 07/13/2004 at 07:52:05
From: Adam
Subject: First order differential equations

How do I find the general solution of d^3x/dy^3 = 0?

I just don't know where to begin.  Surely multiplying both sides by
dy^3 will leave zero on the right hand side.  So, how do I end up with
an integration that allows me to find a general solution?

Date: 07/13/2004 at 18:39:02
From: Doctor Willae
Subject: Re: First order differential equations


There are a couple of ways to go about solving your problem.  You 
could just make an educated guess and check it.  Since it's pretty
straightforward to check a solution to a differential equation, this
is often a good strategy.  The other option is to work through all the
integrals until you get an answer.  Let's try both.

Here's a question for you.  What happens when you differentiate a
constant function (a.k.a. a 0-degree polynomial)?  You get zero.  What
happens when you differentiate a linear function twice?  Well, you get
a constant function when you differentiate once.  So you must get zero
when you differentiate twice.  Let's take this one step further: What
happens when you differentiate a quadratic function 3 times?  You get

And that's precisely what your differential equation is asking you. 
What function, when differentiated 3 times, yields zero?  The answer,
as we just reasoned, is any quadratic function.  Let's check it:

  f(x)    = ax^2 + bx + c
  f'(x)   = 2ax + b
  f''(x)  = 2a
  f'''(x) = 0

If you continue to study differential equations, I think you'll find
that it's very useful to build up your intuition.  It's very handy to
be able to solve simple differential equations simply by inspection. 
However, if that doesn't work for you, there's always the BRUTE FORCE

Because this is plain email, I'm going to have to use some interesting
notation.  I'll use \int to mean the integral sign.  And I'll use y'
instead of dy/dx, y'' instead of d^2y/dx^2, etc..  Now, let's start
with your differential equation:

  y''' = 0

Integrate both sides (and don't forget about the constants of

  \int y''' = \int 0

   y'' + k1 = 0 + k2

        y'' = k2 - k1

            = 2a

Since the difference of two constants is just another constant, I've
elected to replace k2 - k1 with 2a.  It'll make more sense after you
work out the rest of the problem.  If you just continue in this style,
I'm sure you'll arrive at the same answer we guessed and verified above.

Let me know if this ODE gives you any more trouble.

- Doctor Willae, The Math Forum 
Associated Topics:
College Calculus

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