5.5: Annihilation
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section we consider the constant coefficient equation
\[\label{eq:5.4.1} ay''+by'+cy=f(x)\]
From Theorem 5.4.2, the general solution of Equation \ref{eq:5.4.1} is \(y=y_p+c_1y_1+c_2y_2\), where \(y_p\) is a particular solution of Equation \ref{eq:5.4.1} and \(\{y_1,y_2\}\) is a fundamental set of solutions of the homogeneous equation
\[ay''+by'+cy=0. \nonumber \]
In Section 5.3 we showed how to find \(\{y_1,y_2\}\). In this section we’ll show how to find \(y_p\). The procedure that we’ll use is called the method of annihilation.
The annihilation method is exactly what the name implies; we will attempt to annihilate \(f(x)\) in \ref{eq:5.4.1} (make it \(0\)) and create a new homogeneous equation that we now know how to solve.
Note: again, we are transforming a current equation that we don't know how to solve into one that we do.
So, the next obvious questions are: how, exactly, do we annihilate a function and can we annihilate all functions that we will come across. We'll discuss the first question here and the second in the following section.
An operator \(L\) is an annihilator of \(f(x)\) if
\[ L(f(x)) = 0.\nonumber\]
Annihilators in this class involve derivatives and we will use \(D\) to stand for derivative; sometimes called a differential operator in linear algebra. In other words, \[D={d\over dx},\quad D^2={d^2\over dx^2},\quad etc.\nonumber\]
Annihilating Polynomials
We will first consider simple polynomials. As noted above, annihilators are derivatives and as you'll recall from calculus, derivatives lower the power on each term of the polynomial, eventually leading to \(0\) if you differentiate enough.
If \(f(x)=x^2\), find an annihilator \(L\) that annihilates \(f(x)\).
Solution
Annihilators involve derivatives, so let's see what happens if we start repeatedly taking derivatives:
\[D(x^2)=2x; \quad D(2x)=2; \quad D(2)=0\nonumber\]
So, it takes three derivatives to annihilate \(f(x)=x^2\), so in this example \(L=D^3\).
Or, \[D^3(x^2)=0\nonumber\]
If \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), then \(D^{n+1}\) annihilates \(f(x)\).
Find an annihilator \(L\) for \(f(x)=3x^5-7x^2+2x-9\)
Solution
Since the highest power of \(x\) is 5, we will need one more derivative, so \(L=D^6\), to annihilate \(f(x)\). It is left to the reader to show this.
Although powers of \(D\) higher than \(n+1\) will annihilate \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), it is best to choose the smallest annihilator, as we will see later.
Annihilating \(e^{\alpha x}\)
In the case of functions like \(e^{\alpha x}\), derivatives alone do not annihilate it. However, exponential functions are what we call cyclic functions; that is, they come back to themselves eventually, and in the case of the exponential it happens immediately. We can use this to formulate an annihilator for \(e^{\alpha x}\).
If \(f(x)=e^{5x}\), find an annihilator \(L\) that annihilates \(f(x)\).
Solution
If we take a derivative we get \(D(e^{5x})=5e^{5x}\), and you can see that we once again have \(e^{5x}\) but multiplied by 5, so to annihilate \(f(x)\) we need to address that as well. The following will do just that:
\[(D-5)e^{5x}=D(e^{5x})-5(e^{5x})=5e^{5x}-5e^{5x}=0\nonumber\]
So, in this example \(L=D-5\).
Note: it may seem odd that we can "distribute" a derivative like a number, but that is because a derivative is what in linear algebra is called a linear transformation. If you have had linear algebra you will recall this; if not...just trust me.
If \(f(x) = a_nx^ne^{\alpha x} + a_{n-1}x^{n-1}e^{\alpha x} + \ldots + a_1xe^{\alpha x} + a_0e^{\alpha x}\), then \((D-\alpha)^{n+1}\) annihilates \(f(x)\).
Note: we are adding on to what we did earlier. \(D-\alpha\) is for the exponential part and the power \(n+1\) is for the polynomial.
Find an annihilator \(L\) for \(f(x)=x^2e^{5x}+2xe^{5x}\)
Solution
Since the highest power of \(x\) is 2 and \(\alpha\) is 5, the annihilator is \(L=(D-5)^3\). It is left to the reader to show this.
Annihilating \(\sin \beta x\) and \(\cos \beta x\)
Similar to \(e^{\alpha x}\), \(\sin \beta x\) and \(\cos \beta x\) are cyclic functions; that is, they come back to themselves eventually, and in this case it takes two derivatives to come back to itself. We can use this to formulate an annihilator for \(\sin \beta x\) and \(\cos \beta x\).
If \(f(x)=\sin{3x}\), find an annihilator \(L\) that annihilates \(f(x)\).
Solution
As we did in Example 1, let's see what happens if we start repeatedly taking derivatives:
\[D(\sin{3x})=3\cos 3x; \quad D(3\cos 3x)=-9\sin 3x\nonumber\]
Similar to the exponential function, we got back to sine again but multiplied by -9 this time, so to annihilate \(f(x)\) we need to address that as well. The following will do just that:
\[(D^2+9)\sin{3x}=-9\sin{3x}+9\sin{3x}=0.\nonumber\]
So, in this example \(L=D^2+9\).
If \(f(x) = a_nx^ne^{\alpha x} \cos{\beta x} + \ldots + a_0e^{\alpha x} \cos{\beta x} + b_nx^ne^{\alpha x} \sin{\beta x} + \ldots + b_0e^{\alpha x} \sin{\beta x}\) then \(((D-\alpha)^2+\beta^2)^{n+1}\) annihilates \(f(x)\).
Note: we are again adding on to what we did earlier. \(D^2+\beta^2\) is for sine and cosine, \(D-\alpha\) is for the exponential part, and the power \(n+1\) is for the polynomial.
Find an annihilator for \(f(x)=x^3e^{2x}\cos 5x+3xe^{2x}\cos 5x+9\sin 5x\)
Solution
Since the highest power of \(x\) is 3, \(\alpha\) is 2, and \(\beta\) is 5, the annihilator is \(((D-2)^2+25)^4\). It is left to the reader to show this...but...you probably don't want to do that.
Annihilating Linear Combinations of Functions
Our goal again is to take \ref{eq:5.4.1} and annihilate \(f(x)\) using the smallest annihilator possible. so, the next question we're going to consider is what happens when we need to annihilate several different types of functions all at once. The next theorem addresses this - the proof is beyond the scope of this class.
If \(L_1\) annihilates \(f_1(x)\) and \(L_2\) annihilates \(f_2(x)\), then \(L_1\)\(L_2\) annihilates any linear combination of \(f_1(x)\) and \(f_2(x)\). In other words,
\[L_1L_2(c_1f_1(x)+c_2f_2(x))=0.\nonumber\]
Note: When dealing with constant coefficient differential equations, as we are in this section, annihilators are commutative. In other words, \(L_1L_2=L_2L_1\).
Find an annihilator for \(f(x)=7x^4+2x^3-5+2xe^{2x}+9e^{2x}-3x^5e^{3x}\cos 4x\)
Solution
\(D^5\) annihilates \(7x^4+2x^3-5\), \((D-2)^2\) annihilates \(2xe^{2x}+9e^{2x}\), and \(((D-3)^2+16)^6\) annihilates \(-3x^5e^{3x}\cos 4x\), so
\(D^5(D-2)^2((D-3)^2+16)^6\) annihilates \(f(x)\). In other words,
\[D^5(D-2)^2((D-3)^2+16)^6(7x^4+2x^3-5+2xe^{2x}+9e^{2x}-3x^5e^{3x}\cos 4x)=0\nonumber\]
Again, it would not be fun to show this, but there is no need to do this.
Solving Nonhomogeneous Differential Equations Using Annihilation
We can write differential equations using the differential operator \(D={d\over dx}\) as well. For example:
\[y''-y'-6y=D^2y-Dy-6y=(D-3)(D+2)y\nonumber\]
Not that we know how to annihilate several types of common functions, let's see how we can use this idea to solve nonhomogeneous differential equations of the form \ref{eq:5.4.1}. We will first solve the homogeneous version of \ref{eq:5.4.1} before we solve the nonhomogeneous version. The reason will quickly become apparent in Example 5.5.9.
Find the general solution of
\[\label{eq:5.4.2} y''-7y'+12y=4e^{2x}.\]
Solution
We first solve \[\label{eq:5.4.3} y''-7y'+12y=0\]
by finding the characteristic equation \(r^2-7r+12=(r-3)(r-4)=0\), which gives us a fundamental set of solutions \(\{e^{3x},e^{4x}\}\).
Therefore, \[y_h=c_1e^{3x}+c_2e^{4x}\]
We now go back to Equation \ref{eq:5.4.2} and write it using differential operator notation:
\[D^2y-7Dy+12y=4e^{2x}\nonumber\]
or \[(D-4)(D-3)y=4e^{2x}\nonumber\]
Note: The factorization of the differential operators is exactly the same as the characteristic equation - upon inspection the reason why should be obvious.
Now, our goal here is to take this nonhomogeneous equation and make it into a new homogeneous equation. We do this by annihilating \(4e^{2x}\) by applying \(D-2\) to both sides we get
\[(D-2)(D-4)(D-3)y=(D-2)4e^{2x}\nonumber\] giving us
\[(D-2)(D-4)(D-3)y=0\nonumber\]
This new homogeneous equation has the characteristic equation \((r-2)(r-3)(r-4)=0\) giving us roots \(r=2,3,4\).
So, we would get solutions \(\{e^{2x},e^{3x},e^{4x}\}\). However, we know that \(\{e^{3x},e^{4x}\}\) are homogeneous solutions and will give us \(0\) when we use them in \ref{eq:5.4.2}. So, we can disregard those two solutions and focus on \(e^{2x}\). So, we are going to look at solutions of the form
\[y_p=Ae^{2x}\nonumber\]
Note: we are using \(A\) instead of \(c_3\) to remind us that we must actually find \(A\). In other words, we need to find the value of \(A\) so that when we substitute it for \(y\) in our nonhomogeneous equation we get \(4e^{2x}\).
So, if \(y_p=Ae^{2x}\), then \(y_p'=2Ae^{2x}\), and \(y_p''=4Ae^{2x}\). Substituting these into \ref{eq:5.4.2} gives us
\[4Ae^{2x}-7(2Ae^{2x})+12(Ae^{2x})=4e^{2x}\nonumber\]
which gives us
\[2Ae^{2x}=4e^{2x}\nonumber\] which implies \(2A=4\), or \(A=2\). Therefore, we get \[y_p=2e^{2x}\nonumber\] and the general solution of \ref{eq:5.4.2} is
\[y=c_1e^{3x}+c_2e^{4x}+2e^{2x}\nonumber\]
Find the general solution of
\[\label{eq:5.4.4} y''-7y'+12y=5e^{4x}.\]
Solution
We first solve \(y''-7y'+12y=0\), which is the same homogeneous equation that we had in Example 5.5.8; therefore, we have \(y_h=c_1e^{3x}+c_2e^{4x}\).
Now writing \ref{eq:5.4.4} using differential operator notation we get
\[(D-3)(D-4)y=5e^{4x}\nonumber\]
We know \(D-4\) annihilates \(5e^{4x}\), so applying it to both sides we get
\[(D-3)(D-4)^2y=0\nonumber\]
and the roots \(r=3,4,4\). Now we have used 3 and 4 in the homogenous solution, so the new root is the second 4. Now recall from repeated roots that \[y_p=Axe^{4x}.\nonumber\]Note: This is why it's important that we solve the homogeneous equation before the nonhomogeneous.
So, if \(y_p=Axe^{4x}\), then \(y_p'=Ae^{4x}+4Axe^{4x}\), and \(y_p''=8Ae^{4x}+16Axe^{4x}\). Substituting these into \ref{eq:5.4.4} gives us
\[8Ae^{2x}+16Axe^{4x}-7(Ae^{4x}+4Axe^{4x})+12(Axe^{4x})=5e^{4x}\nonumber\] which gives us
\[Ae^{4x}=5e^{4x}\nonumber\] which implies \(A=5\). Therefore, we get \[y_p=5xe^{4x}\nonumber\] and the general solution of \ref{eq:5.4.4} is
\[y=c_1e^{3x}+c_2e^{4x}+5xe^{4x}\nonumber\]
Find the general solution of
\[\label{eq:5.4.6} y''-8y'+16y=2e^{4x}.\]
Solution
Since the characteristic equation of the homogeneous equation
\[\label{eq:5.4.7} y''-8y'+16y=0\]
is \(r^2-8r+16=(r-4)^2=0\), both \(y_1=e^{4x}\) and \(y_2=xe^{4x}\) are solutions of Equation \ref{eq:5.4.7}.
Turning to \ref{eq:5.4.6} and annihilating \(2e^{4x}\) we get
\[(D-4)^3y=0\nonumber\]
giving us a new characteristic equation \((r-4)^3=0\).
Since the new root is the third time 4 is a root, \[y_p=Ax^2e^{4x}\nonumber\]
So, if \(y_p=Ax^2e^{4x}\), then \(y_p'=2Axe^{4x}+4Ax^2e^{4x}\), and \(y_p''=2Ae^{4x}+16Axe^{4x}+16Ax^2e^{4x}\). Substituting these into \ref{eq:5.4.6} gives us
\[2Ae^{4x}+16Axe^{4x}+16Ax^2e^{4x}-8(2Axe^{4x}+4Ax^2e^{4x})+16(Ax^2e^{4x})=2e^{4x}\nonumber\] which gives us
\[2Ae^{4x}=5e^{4x}\nonumber\] which implies \(A=5/2\). Therefore, we get \[y_p={5\over 2}x^2e^{4x}\nonumber\] and the general solution of \ref{eq:5.4.4} is
\[y=c_1e^{4x}+c_2xe^{4x}+{5\over 2}x^2e^{4x}\nonumber\]
Solve the initial value problem \[\label{eq:5.4.18} y''+y=4x+10\sin x; \quad y(\pi)=0, y'(\pi)=2.\]
Solution
We'll leave the homogeneous part to the reader and just start there:
\[y_h=c_1\cos x+c_2\sin x\nonumber\]
Turning to \ref{eq:5.4.18} and annihilating \(4x+10\sin x\) we get
\[D^2(D^2+1)^2y=0\nonumber\]
giving us a new characteristic equation \(r^2(r^2+1)^2=0\).
So, we now have roots \(r=0,0,\pm i,\pm i\). Since the new roots are \(0,0,\pm i\) and since it's the second time \(\pm i\) are roots, we have \[y_p=A+Bx+Cx\cos x+Dx\sin x.\nonumber\]
Following the same process as above we end up with \(A=0, B=4, C=-5, D=0\). Therefore, we get \[y_p=4x-5x\cos x\nonumber\] and the general solution of \ref{eq:5.4.18} is
\[y=c_1\cos x+c_2\sin x+4x-5x\cos x\nonumber\]
Turning to our initial conditions we have \(y(\pi)=-c_1+9\pi=0\), so \(c_1=9\pi\).
Substituting \(c_1\) and taking the derivative we get \(y'=-9\pi\sin x+c_2\cos x+4-5\cos x+5x\sin x\nonumber\) and using
\(y'(\pi)=-c_2+9=2\), so \(c_2=7\) and we get the solution to our initial value problem:
\[y=9\pi\cos x+7\sin x+4x-5x\cos x\nonumber\]
Using the Principle of Superposition
The next example shows how to combine the method of annihilation and Theorem 5.4.3, the principle of superposition.
Find a particular solution of
\[\label{eq:5.4.17} y''-7y'+12y=4e^{2x}+5e^{4x}.\]
Solution
In Example 5.5.8 we found that \(y_{p_1}=2e^{2x}\) is a particular solution of
\[y''-7y'+12y=4e^{2x}, \nonumber\]
and in Example 5.5.9 we found that \(y_{p_2}=5xe^{4x}\) is a particular solution of\[y''-7y'+12y=5e^{4x}. \nonumber\]
Therefore the principle of superposition implies that \(y_p=2e^{2x}+5xe^{4x}\) is a particular solution of Equation \ref{eq:5.4.17}.