Particular Solution of Differential EquationDate: 11/03/2007 at 16:58:25 From: charles Subject: particular solution of ODE Set up the appropriate form of a particular solution y_p but do not determine the coefficients, of the ODE y" + 4y = 3x cos (2x). Which side of the equation or both effect my answer? Ax^2 cos (2x) + A_2 x cos(2x) + B x^2 sin (2x) + B_2 x sin (2x) Date: 11/04/2007 at 11:51:03 From: Doctor Fenton Subject: Re: particular solution of ODE Hi Charles, Thanks for writing to Dr. Math. In this example, both sides affect the form of the particular solution. If I use the letter D to denote the derivative operator D = d/dx, then y' = Dy and y" = D^2y, so that the left side becomes D^2y + 4y = (D^2+4I)y in operator notation (I is the identity operator). Corresponding to this operator is the polynomial z^2 + 4 (replace D with z), and the roots of z determine the solutions of the homogeneous DE (D^2 + I)y_h = 0 . If r1 and r2 are the roots of the polynomial, then e^(r1*x) and e^(r2*x) are two linearly independent solutions of the homogeneous DE. In your case, the roots of z^2 + 4 are 2i and -2i, giving complex exponential solutions e^(2i*x) = cos(2x) + i*sin(2x) and e^(-2i*x) = cos(2x) - i*sin(2x). If you prefer real solutions, you can use the linear combinations e^(2ix) + e^(-2ix) e^(2ix) + e^(-2ix) cos(2x) = ------------------ and sin(2t) = ------------------ 2 2i which are also linearly independent solutions of the homogeneous DE. This method of solving linear, constant-coefficient, homogeneous DE's will give the general solution if you can determine the roots of the associated polynomial. If the polynomial has repeated roots, i.e. roots of multiplicity higher than 1, then by analogy the case D^ny = 0, whose polynomial z^n has 0 as a root of multiplicity n, and whose solution is a polynomial of degree n-1. If the z-polynomial has a root r of multiplicity k, then e^(kx), te^(kx), x^2*e^*(kx),..., x^(k-1)*e^(kx) are k linearly independent solutions. When solving a linear, constant coefficient DE p(D)y = f(x) where f(x) itself is a solution of a linear, constant-coefficient, homogeneous DE q(D)f = 0, then applying q(D) to the non-homogeneous DE gives q(D)p(D)y = q(D)f(x) = 0 , and any solution of the non-homogeneous DE p(D)y = f(x) must be a solution of the homogeneous DE q(D)p(D)y = 0. Now, you need to list all the roots of the polynomial q(z)p(z), and write the general solution of the DE q(D)p(D)y = 0. From this, you can omit any function which is a solution of the homogeneous DE p(D)y = 0. When you have eliminated these functions, what remains will give you the form of a particular solution of the non-homogeneous DE. In your example, the right side of your DE is 3x cos (2x) which is a solution of a linear constant-coefficient homogeneous DE whose polynomial q(z) has a double root of 2i (the 2i from the fact that cos(2x) is a solution, and the root must be at least a double root to give the factor x multiplying the cosine). The simplest such polynomial q(z) is (z^2+4)^2, so q(D) = (D^2+4I)^2. Any particular solution y_p of the DE y" + 4y = 3x cos (2x) must be a solution of the homogeneous DE (D^2+4I)^2 (D^2+4I)y_p = 0 or (D^2+4I)^3 y_p = 0 . Can you write the general solution of this homogeneous DE and then delete the solutions of (D^2+4I)y = 0? If so, you have the form of y_p, and just need to substitute that form, with unknown constant coefficients, into the non-homogeneous DE, and solve for those undetermined coefficients (which is why this method is called the Method of Undetermined Coefficients). If you have any questions or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/