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18. Method of Undetermined Coefficients (Annihilator Operator Approach 1)

Dated: 01-12-2024

Differential Operators

In calculus, we often denote \(\frac d {dx}\) by \(D\).
This is known as differential operator.
This operator possesses linearity, meaning if \(f\) and \(g\) are two differentiable functions then

\[D(a f(x) + b g(x)) = aDf(x) + bDg(x)\]

Therefore, we call it linear differential operator.

\[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=D(Dy)=D^2y\]

The following are also linear differential operators

\[a_nD^n+a_{n-1}D^{n-1}+\cdots+a_1D+a_0\]
\[D+3, D^{2}+3D-4, 5D^{3}-6D^{2}+4D\]

Differential Equation in terms of \(D\)

Any linear differential equation can be expressed in terms of the notation \(D\). Consider a \(2^{nd}\) order equation with constant coefficients

\[ay''+by'+cy=g(x)\]

Since

\[ \frac{dy}{dx}=Dy, \frac{d^2y}{dx^2}=D^2y \]

Therefore the equation can be written as

\[ aD^2y+bDy+cy=g(x) \]

or

\[ (aD^2 + bD + c)y = g(x) \]

Now, we define another differential operator \(L\) as

\[ L = aD^2 + bD + c \]

Then the equation can be compactly written as

\[ L(y) = g(x) \]

The operator \(L\) is a second-order linear differential operator with constant coefficients.

Example

Consider the differential equation

\[ y'' + y' + 2y = 5x - 3 \]

Since

\[ \frac{dy}{dx} = Dy, \quad \frac{d^2y}{dx^2} = D^2y \]

Therefore, the equation can be written as

\[ (D^2 + D + 2)y = 5x - 3 \]

Now, we define the operator \(L\) as

\[ L = D^2 + D + 2 \]

Then the given differential can be compactly written as

\[ L(y) = 5x - 3 \]

Factorization of a Differential Operator

An \(nth\) order linear differential operator

\[L=a_nD^n+a_{n-1}D^{n-1}+\ldots+a_1D+a_0\]

with constant coefficients can be factorized, whenever the characteristics polynomial equation

\[L=a_nm^n+a_{n-1}m^{n-1}+\ldots+a_1m+a_0\]

can be factorized.
The factors of a linear differential operator with constant coefficients commute.

Example

Consider the following \(2^{nd}\) order linear differential operator

\[D^2 + 5D + 6\]

If we treat \(D\) as an algebraic quantity, then the operator can be factorized as

\[D^2 + 5D + 6 = (D + 2)(D + 3)\]

To illustrate the commutative property of the factors, we consider a twice-differentiable function \(y = f(x)\). Then we can write

\[(D^2 + 5D + 6)y = (D + 2)(D + 3)y = (D + 3)(D + 2)y\]

To verify this we let

\[w=(D+3)y=y'+3y\]

Then

\[(D+2)w=Dw+2w\]
\[(D+2)w=(y''+3y')+(2y'+6y)\]
\[(D+2)w=y''+5y'+6y\]

or

\[(D+2)(D+3)y=y''+5y'+6y\]

Similarly if we let

\[w=(D+2)y=(y'+2y)\]

Then

\[(D+3)w=Dw+3w=(y''+2y')+(3y'+6y)\]
\[(D+3)w=y''+5y'+6y\]
\[(D+3)(D+2)y=y''+5y'+6y\]

Therefore, we can write from the two expressions that

\[(D+3)(D+2)y=(D+2)(D+3)y\]

Hence

\[(D+3)(D+2)y=(D+2)(D+3)y\]

Annihilator Operator

Suppose that

  • \(L\) is a linear differential operator with constant coefficients.
  • \(y = f(x)\) defines a sufficiently differentiable function.
  • The function \(f\) is such that \(L(y) = 0\)

Then the differential operator \(L\) is said to be an annihilator operator of function \(f\).

Example

Find a differential operator that annihilates the polynomial function

\[y=1-5x^{2}+8x^{3}.\]

Solution

Since

\[D^{4}x^{3}=0\]

Therefore

\[D^{4}y=D^{4}(1-5x^{2}+8x^{3})=0.\]

Hence, \(D^4\) is the differential operator that annihilates the function \(y\).

Note that the functions that are annihilated by an nth-order linear differential operator \(L\) are simply those functions that can be obtained from the general solution of the homogeneous differential equation

\[L(y) = 0\]

Notes

  • If \(y_1\) and \(y_2\) are functions and \(L(y_1) = 0\) and \(L(y_2) = 0\).
\[L\left(c_1y_1(x) + c_2y_2(x)\right) = 0\]
  • Suppose \(L_1\) and \(L_2\) being linear operators with constant coefficients such that
\[L_1(y_1) = 0, L_2(y_2) = 0\]
\[L_1(y_2) \ne 0, L_2(y_1) \ne 0\]

then

\[L_1L_2(y_1(x) + y_2(x)) = 0\]

To demonstrate this fact we use the linearity property for writing

\[L_1L_2(y_1+y_2) = L_1L_2(y_1) + L_1L_2(y_2)\]

Since

\[L_1L_2 = L_2L_1\]

therefore

\[L_1L_2(y_1+y_2) = L_2L_1(y_1) + L_1L_2(y_2)\]

or

\[L_1L_2(y_1+y_2) = L_2(L_1(y_1)) + L_1(L_2(y_2))\]

But we know that

\[L_1(y_1) = 0, \quad L_2(y_2) = 0\]

Therefore

\[L_1L_2(y_1 + y_2) = L_2(0) + L_1(0) = 0\]

Example

Find a differential operator that annihilates the function

\[f(x)=7-x+6\sin{3x}\]

Solution

Suppose that

\[y_1(x)=7-x,\quad y_2(x)=6\sin{3x}\]

Then

\[D^2y_1(x)=D^2(7-x)=0\]
\[(D^2+9)y_2(x)=(D^2+9)\sin{3x}=0\]

Therefore, \(D^2(D^2+9)\) annihilates the function \(f(x)\).

References

Read more about notations and symbols.