General Solution to Differential EquationDate: 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 Adam, 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 zero. 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 METHOD. 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 integration!): \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 http://mathforum.org/dr.math/ |
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