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Chris L T521 · #7 May 25th, 2008, 01:23 AM

Technique #2 : The "Annihilator" Method:

An alternative to the first technique would be the Annihilator method. As it's name foretells, we annihilate the non-homogeneous term and make the equation homogeneous.

To use the Annihilator technique, you must rewrite the DE using Differential Operator Notation:

$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x) \implies aD^2y+bDy+cy=f(x) \implies \left(aD^2+bD+c\right)(y)=f(x)$

Note that we "factor" out a y (I use this term very loosely; you really can't factor out the y, but as you will see, it will work out to our advantage

)

You can apply the annihilator to any of the following families of functions that $f(x)$ can be:

1. $1,x,x^2,x^3,\cdots ,x^{n-1}$

2. $e^{\alpha x}, xe^{\alpha x}, x^2e^{\alpha x},\cdots ,x^{n-1}e^{\alpha x}$

3. $e^{\alpha x}\cos(\beta x), xe^{\alpha x}\cos(\beta x), x^2e^{\alpha x}\cos(\beta x),\cdots,x^{n-1}e^{\alpha x}\cos(\beta x)$

4. $e^{\alpha x}\sin(\beta x), xe^{\alpha x}\sin(\beta x), x^2e^{\alpha x}\sin(\beta x),\cdots,x^{n-1}e^{\alpha x}\sin(\beta x)$

The Annihilators:

$\color{red}\boxed{D^{n}}$ will annihilate $1,x,x^2,x^3,\cdots ,x^{n-1}$ .

I will leave it for you to prove the other two:

$\color{red}\boxed{\left(D-\alpha\right)^n}$ will annihilate $e^{\alpha x}, xe^{\alpha x}, x^2e^{\alpha x},\cdots ,x^{n-1}e^{\alpha x}$ .

$\color{red}\boxed{\left(D^2-2\alpha D+(\alpha^2+\beta^2)\right)^n}$ will annihilate $e^{\alpha x}\cos(\beta x), xe^{\alpha x}\cos(\beta x), x^2e^{\alpha x}\cos(\beta x),\cdots,x^{n-1}e^{\alpha x}\cos(\beta x)$ and
$e^{\alpha x}\sin(\beta x), xe^{\alpha x}\sin(\beta x), x^2e^{\alpha x}\sin(\beta x),\cdots,x^{n-1}e^{\alpha x}\sin(\beta x)$ .

Example 17:

Solve $y''-4y=\cosh (2x)$ .

Solving the homogeneous equation $y'' -4y=0$ , we see that $r^2-4=0 \implies r=\pm 2$ . Thus the complimentary solution is $y_c=c_1e^{-2x}+c_2e^{2x}$ .

Now, to apply the annihilator to the DE, we need to rewrite it in differential form:

$y''-4y=\cosh (2x) \implies \left(D^2-4\right)(y)=\cosh (2x)$

noting that $\cosh u =\frac{1}{2}\left( e^{u}+e^{-u}\right)$ , we have the DE:

$\left(D^2-4\right)(y)=\frac{1}{2}\left(e^{2x}+e^{-2x}\right)$

Let us determine the proper annihilator:

$D-2 \ \text{annihilates} \ e^{2x} \ \text{and} \ D+2 \ \text{annihilates} \ e^{-2x}$ . This may pose a problem: we have two different annihilators! so which one do we apply to the DE? the answer is both. There is a theorem that states something like the following:

If there are two functions $f(x) \ \text{and} \ g(x)$ and their annihilators are $L_1 \ \text{and} \ L_2$ respectively,
then the product of the two annihilators $\left(L_1L_2\right)$ will annihilate $f(x)+g(x)$ .

Thus, $\left(D-2\right)\left(D+2\right)$ will annihilate $\frac{1}{2}\left(e^{2x}+e^{-2x}\right)$ .

Applying the newly found annihilator to both sides we get:

$\left(D+2\right)\left(D-2\right)\left(D^2-4\right)(y)=0$

Now rewrite the equation so we get the characteristic equation:

$\left(r+2\right)\left(r-2\right)\left(r^2-4\right)=0$

Solving for r, we get $r=2$ with multiplicity 2 and $r=-2$ with multiplicity 2.

Note that 2 of the r values were values used to determine the complimentary solution! Thus, the general solution will be:

$y=\underbrace{c_1e^{-2x}+c_2e^{2x}}_{y_c}+\underbrace{c_3xe^{-2x}+c_4xe^{2x}}_{y_p}$

Now find the coefficients $c_3 \ \text{and} \ c_4$ (which I will denote by A and B, respectively).

$y_p=Axe^{-2x}+Bxe^{2x}$
$y_{p}^{/}=\left(A-2Ax\right)e^{-2x}+\left(2Bx+B\right)e^{2x}$
$y_{p}^{//}=\left(4Ax-4A\right)e^{-2x}+\left(4Bx+4B\right)e^{2x}$

Substituting these values into the DE, we get:

$\left(4Ax-4A\right)e^{-2x}+\left(4Bx+4B\right)e^{2x}-4\left[Axe^{-2x}+Bxe^{2x}\right]=\frac{1}{2}\left(e^{2x}+e^{-2x}\right)$

$\implies -4Ae^{-2x}+4Be^{2x}=\frac{1}{2}\left(e^{2x}+e^{-2x}\right)$

Comparing the coefficients, we get:

$\begin{aligned}-4A &= \frac{1}{2}\\4B &= \frac{1}{2}\end{aligned}$

This gives us $A=-\frac{1}{8}$ and $B=\frac{1}{8}$

Therefore the general solution is:

$\color{red}\boxed{y=c_1e^{-2x}+c_2e^{2x}-\frac{1}{8}\left(xe^{-2x}-xe^{2x}\right)}$

Example 18:

Solve $y''-6y'+13y=xe^{3x}\sin(2x)$ (WARNING!! THIS IS A VERY TEDIOUS PROBLEM TO SOLVE!!!

)

Here's the easy part (solve the homogeneous equation):

$y''-6y+13=0 \implies r^2-6r+13=0\implies r=\frac{6\pm \sqrt{36-52}}{2}$
$\implies r=\frac{6\pm \sqrt{-16}}{2}\implies r=3\pm 2i$

Thus the complimentary solution will be $y_c=e^{3x}\left(c_1\cos(2x)+c_2\sin(2x)\right)$

The next part isn't that bad (solving the non-homogeneous equation):

$y''-6y+13=xe^{3x}\sin(2x) \implies \left(D^2-6D+13\right)(y)=xe^{3x}\sin(2x)$

Now we can find the annihilator:

$\left(D^2-2(3)D+(3^2+2^2)\right)^2 \implies \left(D^2-6D+13\right)^2$ will annihilate $xe^{3x}\sin(2x)$ .

Applying the annihilator to both sides, we get:

$\left(D^2-6D+13\right)^3(y)=0$

Converting it to the characteristic equation, we have:

$\left(r^2-6r+13\right)^3=0 \implies r=3\pm 2i$ with multiplicity three (note that one of these roots form the complementary solution)

Now here comes the nasty part: find $y_p$

$y_p=xe^{3x}\left(A\cos(2x)+B\sin(2x)\right)+x^2e^{3x}\left(C\cos(2x)+D\sin(2x)\right)$
$\implies y_p=\left(Ax+Cx^2\right)e^{3x}\cos(2x)+\left(Bx+Dx^2\right)e^{3x}\sin(2x)$

I'll leave it for you to show that

$\begin{aligned}y_{p}^{/}=&\left[\left(3A+2B+2C\right)x+\left(3C+2D\right)x^2+A\right]e^{3x}cos(2x)\\&+\left[\left(3B-2A+2D\right)x+\left(3D-2C\right)x^2+B\right]e^{3x}\sin(2x)\end{aligned}$

$\begin{aligned}y_{p}^{//}=&\left[\left(5A+12B+12C+8D\right)x+\left(12D+5C\right)x^2+3A+2B\right]e^{3x}\cos(2x)\\&+\left[\left(5B-12A-8C+12D\right)x+\left(5D-12C\right)x^2+3B-2A\right]e^{3x}\sin(2x)\end{aligned}$

Substituting this into the DE (

), we have (...get ready...)

$\begin{aligned}&\left(\left[\left(5A+12B+12C+8D\right)x+\left(12D+5C\right)x^2+3A+2B\right]e^{3x}\cos(2x)\right.\\&+\left.\left[\left(5B-12A-8C+12D\right)x+\left(5D-12C\right)x^2+3B-2A\right]e^{3x}\sin(2x)\right)\\&-6\left[\left[\left(3A+2B+2C\right)x+\left(3C+2D\right)x^2+A\right]e^{3x}cos(2x)\right.\end{aligned}$
$\begin{aligned}&+\left.\left[\left(3B-2A+2D\right)x+\left(3D-2C\right)x^2+B\right]e^{3x}\sin(2x)\right]\\&+13\left[\left(Ax+Cx^2\right)e^{3x}\cos(2x)+\left(Bx+Dx^2\right)e^{3x}\sin(2x)\right]=xe^{3x}\sin(2x)\end{aligned}$

...well, after a decent amount of cancellations, we get:

$\left[2D-3A+8Dx\right]e^{3x}\cos(2x)+\left[-3B-2A-8Cx\right]e^{3x}\sin(2x)=xe^{3x}\sin(2x)$

Comparing the coefficients, we get:

$\begin{aligned}2D-3A&=0 \\8D&=0 \\-8C&=1 \\-3B-2A&=0\end{aligned}$

Solving this system, we get $A=0 \text{,} \ B=0 \text{,} \ C=-\frac{1}{8} \text{,} \ \text{and} \ D=0$

Therefore, $y_p=-\frac{1}{8}x^2e^{3x}\cos(2x)$

Therefore, the general solution is:

$\color{red}\boxed{y=e^{3x}\left(c_1\cos(2x)+c_2\sin(2x)\right)-\frac{1}{8}x^2e^{3x}\cos(2x)}$

I will discuss Variation of Parameters tomorrow (hopefully)

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