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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.

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/
```
Associated Topics:
College Calculus

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