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Particular Solution of Differential Equation

Date: 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


  (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

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

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