19. Method of Undetermined Coefficients (Annihilator Operator Approach 2)

Dated: 01-12-2024

This method is limited to non homogeneous linear differential equations with following factors

Constant coefficients
\(g(x)\) has a specific form

Form of \(g(x)\)

Any of the following are valid

A constant function \(k\).
A polynomial function
An exponential function
The trigonometric functions \(\sin (\beta x), \cos(\beta x)\)
Finite sums and products of these functions

The Method

Write the non homogeneous linear differential equation in the form

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

Find the complementary solution \(y_c\) by finding general solution of the associated homogeneous differential equation

\[L(y) = 0\]

Operate on both sides of the non homogeneous equation with a differential operator \(L_1\) that annihilates the function \(g(x)\).
Find the general solution of the higher order homogeneous differential equation

\[L_1L(y) = 0\]

Delete all those terms from step 4 that are duplicated in \(y_c\) found in step 2
Form a linear combination \(y_p\) of the terms that remain. This is the form of a particular solution of the non homogeneous differential equation

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

Substitute \(y_p\) found in step 6 into the given non-homogeneous linear differential equation \(L(y) = g(x)\). Match coefficients of various functions on each side of the equality and solve the resulting system of equations for the unknown coefficients in \(y_p\).
With the particular integral found in step 7, form the general solution of the given differential equation as

\[y = y_c + y_p\]

Example

Solve

\[\frac{d^{2}y}{dx^{2}}+3\frac{dy}{dx}+2y=4x^{2}.\]

Solution

Step 1

since

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

Therefore, the given differential equation can be written as

\[(D^{2}+3D+2)y=4x^{2}\]

Step 2

To find the complementary function \(y_{c}\), we consider the associated homogeneous differential equation

\[(D^{2}+3D+2)y=0\]

The auxiliary equation is

\[m^{2}+3m+2=(m+1)(m+2)=0\]

\[\implies m=-1,-2\]

Therefore, the auxiliary equation has two distinct real roots.

\[m_{1}=-1,\quad m_{2}=-2\]

Thus, the complementary function is given by

\[y_{c}=c_{1}e^{-x}+c_{2}e^{-2x}\]

Step 3

In this case the input function is

\[g(x)=4x^{2}\]

Further

\[D^{3}g(x)=4D^{3}x^{2}=0\]

Therefore, the differential operator \(D^3\) annihilates the function \(g\). Operating on both sides of the equation in step 1, we have

\[D^3(D^2+3D+2)y=4D^3x^2\]

\[D^3(D^2+3D+2)y=0\]

This is the homogeneous equation of order 5. Next we solve this higher order equation.

Step 4

The auxiliary equation of the differential equation in step 3 is

\[m^3(m^2+3m+2)=0\]

\[m^3(m+1)(m+2)=0\]

\[m=0,0,0,-1,-2\]

Thus its general solution of the differential equation must be

\[y=c_1+c_2x+c_3x^2+c_4e^{-x}+c_5e^{-2x}\]

Step 5

The following terms constitute \(y_c\)

\[c_4e^{-x}+c_5e^{-2x}\]

Therefore, we remove these terms and the remaining terms are

\[y=c_1+c_2x+c_3x^2\]

Step 6

This means that the basic structure of the particular solution \(y_p\) is

\[y_p=A+Bx+Cx^2\]

Where the constants \(c_1\), \(c_2\), and \(c_3\) have been replaced, with \(A\), \(B\), and \(C\), respectively.

Step 7

Since

\[y_p=A+Bx+Cx^2\]

\[y_p'=B+2Cx\]

\[y_p''=2C\]

Therefore

\[y_p''+3y_p'+2y_p=2C+3B+6Cx+2A+2Bx+2Cx^2\]

\[y_p''+3y_p'+2y_p=(2C)x^2+(2B+6C)x+(2A+3B+2C)\]

Substituting into the given differential equation, we have

\[(2C)x^2+(2B+6C)x+(2A+3B+2C)=4x^2+0x+0\]

Equating the coefficients of \(x^2\), \(x\), and the constant terms, we have

\[2C=4\]

\[2B+6C=0\]

\[2A+3B+2C=0\]

Solving these equations, we obtain

\[A=7,\quad B=-6,\quad C=2\]

Hence

\[y_p=7-6x+2x^2\]

Step 8

The general solution of the given non-homogeneous differential equation is

\[y=y_c+y_p\]

\[y=c_1e^{-x}+c_2e^{-2x}+7-6x+2x^2\]