General and Particular Solutions
Date: 01/08/99 at 05:57:52 From: Abbas Subject: Differential Equations Can you please explain me in simple terms the difference between a general solution and a particular solution of a differential equation? When does this occur, and what is its physical significance?
Date: 01/08/99 at 10:29:49 From: Doctor Rob Subject: Re: Differential Equations This is analogous to talking about a general triangle, or a particular one with sides 7, 8, and 10. The general solution encompasses all solutions, and a particular solution is just one of those. It always occurs when you are solving differential equations. Its physical significance is that the particular solution solves the equation, but may not satisfy the initial conditions you are interested in. The solution that does that will be (usually) a different particular solution of the equation, which you can discover from the general solution (which depends on some parameters, such as constants of integration) by applying the initial conditions and solving for the parameter values which make those conditions true. Example: Solve the differential equation: dy/dx = y + 1, y(0) = 1 The general solution is: y = c*e^x - 1 for any real constant c. The particular solution you are interested in is the one where y(0) = 1, that is, c*e^0 - 1 = 1, that is, c = 2, that is: y = 2*e^x - 1 - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Date: 01/09/99 at 00:28:14 From: Abbas Bookwala Subject: Re: Differential Equations Dear Sir, I am sorry for not forwarding the question with a bit of extra information. I was in fact interested in knowing those general and particular solutions occurring in certain equations which are added and the sum is called a solution. If my memory serves me right (back to college days), it was a solution of a non-homogeneous equation. I want to know the physical significance of general and particular solutions in this context. Regards, Abbas S Bookwala
Date: 01/11/99 at 11:01:10 From: Doctor Rob Subject: Re: Differential Equations This new situation you describe is not unrelated to the situation I gave in my previous answer. When you have a linear differential equation of the form: a(0,x)*y + a(1,x)*y' + a(2,x)*y" + ... = f(x) often it is not terribly hard to find the general solution of the corresponding homogeneous diffential equation: a(0,x)*y + a(1,x)*y' + a(2,x)*y" + ... = 0 Call that y_0(x,p1,p2,...). It depends on some parameters p1, p2, etc. Often you can also find a particular solution (depending on no parameters) y_1(x). Since the equation is linear, then y = y_0 + y_1 is also a solution of the original equation, for any values of the parameters. This is pretty easy to see by substituting for y its equal y_0 + y_1, using the fact that the derivative of a sum is the sum of the derivatives of the summands, and the fact that y_0 satisfies the homogeneous equation and that f_1 satisfies the original equation. As for the physical significance, I suppose you can think of the homogeneous equation as the one which drives the behavior of the solutions under some kind of ideal or simplified conditions, and the presence of the function f(x) is forcing some kind of perturbation to that behavior. Once you find how that perturbation affects any single solution (such as y_1), that tells you how it affects all the rest, because the lefthand side is linear. The same example works here. Write the equation in the form: y' - y = 1 The corresponding homgeneous equation is: y' - y = 0 whose general solution is y_0(x,c) = c*e^x. We can look for a particular solution which is a polynomial in x, say y = a*x + b, so y' = a: a - (a*x + b) = 1 which must be an identity. This forces a = 0, b = -1, and, sure enough, y_1(x) = -1 is a solution to the inhomogeneous equation. Putting these together, the general solution to the original equation is: y(x) = y_0(x,c) + y_1(x) = c*e^x - 1 Perhaps a more complicated example would be more enlightening: x*y' - y = x^2 The homogeneous equation is: x*y' - y = 0 y'/y = 1/x log|y| = log|x| + C y = e^C*x y = D*x for any value of D. This gives us y_0(x,c) = D*x. Now we want the particular solution. Again, we try a polynomial of low degree: y = a*x^2 + b*x + c y' = 2*a*x + b x*(2*a*x + b) - (a*x^2 + b*x + c) = x^2 (2*a-a-1)*x^2 + (b-b)*x - c = 0 c = 0, a = 1, b = anything y_1(x) = x^2 Then the general solution is: y(x) = D*x + x^2 - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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