This is modified method of the method from the last lesson (Undetermined coefficients—superposition approach). The DE to be solved has again the same limitations (constant coefficients and restrictions on the right side). Annihilator approach finds $y_c$ and $y_p$ by means of operators explained further. Finally the values of arbitrary constants of particular solution have to be found as was explained.
Prior to explain the method itself we need to introduce some new terms we will use later.
An operator is a mathematical device which converts one function into another. For example the operator $'$ (differential operator) converts $f(x)$ into a new function $f'(x)$. $\intop f(t)\ dt$ converts $f(t)$ into new function $F(x)$.
We also use letter $D$ to denote the operation of differentiation. $D$ is called differential operator.
$$ \begin{eqnarray} D^0y &=& y \nonumber\\ D\color{silver}{^1}y &=& y' \nonumber\\ D^2y &=& y'' \nonumber\\ &\vdots & \nonumber\\ D^ny &=& y^{(n)} \nonumber\\ \end{eqnarray} $$If we use differential operator $D$ we may form a linear combination of differential operators of orders $0$ to $n$:
$$ a_0 + a_1 D + a_2 D + \cdots + a_n D^n $$Thus we a have a handy tool which helps us also to generalize some rules into sample manner. For example if we work with operator in above polynomial form, we may rely also on polynomial behaviour, e.g.:
$$ (D+2)(D+3) = (D+3)(D+2) = (D^2 + 5D + 6),\\ (D^2 + 3) (x^3 + \cos x) = D^2(x^3 + \cos x) + 3(x^3 + \cos x) $$If $L$ is linear differential operator such that
$$ L(f(x)) = 0 $$then $L$ is said to be annihilator. For example
$D^n$ annihilates not only $x^{n-1}$, but all members of polygon
$$ c_0 + c_1 x + c_2 x^2 + \cdots + c_{n-1} x^{n-1} $$A function $e^{\alpha x}$ is annihilated by $(D-\alpha)$:
$$ (D-\alpha)e^{\alpha x} = D e^{\alpha x} - \alpha e^{\alpha x} = \alpha e^{\alpha x} - \alpha e^{\alpha x} = 0 $$$(D-\alpha)^n$ annihilates each of the member
$$ e^{\alpha x}, x e^{\alpha x}, x^2 e^{\alpha x}, \cdots, x^{n-1} e^{\alpha x} $$$(D^2 + \beta^2)$ annihilates $\cos \beta x$ and also $\sin \beta x$:
$$ (D^2 + \beta^2) \cos \beta x = 0, \hskip1em (D^2 + \beta^2) \sin \beta x = 0. $$We are often interested in annihilating a sum of two or more functions $y_1$, $y_2$, ... It can be shown that
Example:
| $7-x$ is being annihilated by | $D^2$ |
| $\sin 4x$ is being annihilated by | $D^2 + 16$ |
| Then $D^2(D^2+16)$ annihilates the linear combination $7-x + 6 \sin 4x$ | |
The differential operator which annihilates given function is not unique. For example $D^2(x) = 0$. But also $D^3(x) = 0$. We want the operator of the lowest possible order.
Homogeneous high order DE can be written also as $L(y) = 0$ and nonhomogeneous as $L(y) = g(x)$ where $L$ is a proper differential operator.
Example:
$$ \begin{eqnarray} y''+3y'+2y &=& 0 \hskip1em \text{can be written as} \nonumber\\ (D^2 + 3D +2)y &=& 0 \nonumber \end{eqnarray} $$The right side containing $g(x)$ can be annihilated by $L_1$:
$$ \begin{eqnarray} L(y) &=& g(x) \nonumber\\ L_1L(y) &=& L_1(g_x)) \hskip1em \text{right side is being annihilated} \nonumber\\ L_1L(y) &=& 0 \nonumber \end{eqnarray} $$If we solve $L_1L(y) = 0$ we get an instance of solution $y=y_c+y_p$.
Then we have to distinguish terms which belong to particular solution $y_p$ and find constants for all these terms.
Solve $y'' + 4y' +4y = 2x +6$
This step is voluntary and rather serves to bring more light into the method. We will find $y_c$ as we are used to:
$$ \begin{eqnarray} &&m^2 + 4m + 4 = 0 \implies m =\{-2, -2\} \nonumber\\ &&\underline{y_c = c_1 e^{-2x} + c_2 x e^{-2x}} \nonumber \end{eqnarray} $$Apply annihilators against $g(x)$.
$$ \begin{eqnarray} (D^2 + 4D+4) y & = & 2x + 6 \nonumber \\ D^2(D^2 + 4D+4) y & = & D^2(2x + 6) \nonumber \\ D^2(D^2 + 4D+4) y & = & 0 \nonumber \\ m^2(m^2 + 4m+4) & = & 0 \implies m = \{0, 0\}, \{-2, -2\} \nonumber \\ \end{eqnarray} $$It can be seen that the solution $m = \{-2, -2\}$ belongs to complementary function $y_c$ and $m=\{0, 0\}$ belongs to particular solution $y_p$.
$$ \begin{align} & y = c_1 e^{-2x} + c_2 x e^{-2x} + c_3 e^0 + c_4 e^0 x \nonumber \\ & \underline{y = c_1 e^{-2x} + c_2 x e^{-2x} + c_3 + c_4 x }\nonumber \end{align} $$The particular solution is not supposed to have its members multiplied by arbitrary constants. We have to find values $c_3$ and $c_4$ in such way, that the solution satisfies DE. To do so, we will use method of undeterminated coefficients as in previous lesson. We know that $y_p$ is a solution of DE. So we can feed $y_p = A + Bx$ and its derivatives into DE and find constants $A$, $B$: $A= 1$, $B=\frac 1 2$.
$$ \underline{\underline{y = 1 + \frac 1 2 x + c_1 e^{_2x} + c_2 e^{-2x}x}} $$Solve $y''' - y'' + y' -y= x e^x - e^{-x} + 7$
Note that the imaginary roots come in conjugate pairs.
Apply annihilators against $g(x)$.
$$ \begin{eqnarray} 7 &\ldots& D \nonumber \\ e^{-x} &\ldots& (D+1) \nonumber \\ x e^{x} &\ldots& (D-1)^2 \nonumber \\ \end{eqnarray} $$ $$ \begin{eqnarray} (D^3 - D^2 + D -1)y &=& x e^x - e^{-x} + 7 \nonumber \\ D(D+1)(D-1)^2(D^3 - D^2 + D -1)y &=& D(D+1)(D-1)^2(x e^x - e^{-x} + 7) \nonumber \\ \nonumber D(D+1)(D-1)^2(D^3 - D^2 + D -1)y &=& 0\\ \nonumber m(m+1)(m-1)^2(m^3 - m^2 + m -1) &=& 0 \implies m = \{1, i, -i\}, \{0, -1, 1, 1\}\\ \nonumber \end{eqnarray} $$ $$ \underline{y_p(x) = A e^0 + B e^{-x} + Ce^x x + D e^x x^2} $$In step 1 the members of complementary function $y_c$ are found from auxiliary equation. However even if step 1 is skipped, it should be obvious which roots belong to $y_c$ and which roots belong to $y_p$ from step 2 itself.
After expressing $y_p'$ and $y_p''$ we can feed them into DE and find constants $A$, $B$, $C$ and $D$ of particular solution. Finally we can form
$$ \underline{\underline{y_c = c_1 e^x + c_2 \cos x + c_3 \sin x - 7 + \frac 1 4 e^{-x} - \frac 1 2 x e^x + \frac 1 4 x^2 e^x}}. $$Solve $y'' - y' + y= x^{2}$
First we rewrite the DE by means of differential operator $D$ and then we have to ask, what is annihilator for $x^2$ on the right side? Is it $D$? It is not: $D$ annihilates only a constant. We have to use $D^3$ to annihilate $x^2$.
$$ \begin{align} D^2y + Dy + y &= x^2 \nonumber \\ D^3(D^2y + Dy + y) &= D^3(x^2) \nonumber \\ D^3(D^2y + Dy + y) &= 0 \nonumber \\ m^3(m^2 + m + 1) &= 0 \nonumber \\ \end{align} $$The member $m^3$ belongs to the particular solution $y_p$ and roots from $m^2 + m + 1$ will form complementary function $y_c$. To find roots we might use calculator able to solve quadratic equation or we might use quadratic formula being taught at high school.
The found roots are $m = \{0,\ 0,\ 0,\ -1/2+i\sqrt{3}/2 ,\ -1/2-i\sqrt{3}/2 \}$. The general solution can be formed as
$$ y = y_c + y_p = c_1e^{-1/2-i\sqrt{3}/2 } + c_2e^{-1/2+i\sqrt{3}/2 } + c_3 + c_4x + c_5x^2 $$The first members involve imaginary numbers and might be also rewritten by means of $\sin()$ and $\cos()$ to avoid complex numbers. These roots comes in conjugate pairs $\alpha + i\beta$ and $\alpha - i\beta$, so they do not repeat. The next three members would repeat based on the value of the root $m=0$, so they are multiplied by $x$ and $x^2$.
The job is not done yet, since we have to find values of constants $c_3$, $c_4$, $c_5$ which are part of particular solution. Note that we have 2nd order DE, so we expect to have two arbitrary constants, not five.
These constants can be obtained by forming particular solution in a more convenient way $y_p=A+Bx +Cx^2$, preparing $y_p',\ y_p''$ ans substituting into DE. It will be found that $A=0,\ B=-2,\ C=1$.
$$ \underline{\underline{ y = c_1e^{-1/2-i\sqrt{3}/2 } + c_2e^{-1/2+i\sqrt{3}/2 } -2x + x^2}} $$