Solve \(f(D)y=G(x)\) for particular solution

Method 1, Undetermined Coefficients

Apply polynomial theory to \(G(x)=0\) find roots and \(y_p\). And then take derivatives on \(y_p\), substitute back to original differential equation to get fixed values for constant coefficients.

Method 2, Variation of Parameters

Use complementary function to make particular solution.

Method 3, Inverse differential operators

Notation,

\[Dy = g(x)\]

\[y=\frac{1}{D}g(x)=\int g(x)dx\]

Method 4, Superposition

If \(y_1\) is a particular solution of \(f(D)y = G_1(x)\) and \(y_2\) is a particular solution of \(f(D)y = G_2(x)\) then

\[y = y_1 + y_2\]

is a particular solution of \(f(D)y = G_1(x) + G_2(x)\).

Method 5, \(G(x)=G_0\) is a constant

\(a_n\) is the nth coefficient in operator.

Case 1. \(a_n \ne 0\). The solution is \(y_p = G_0/a_n\).

Case 2. nth coefficient is zero, lowest order derivative down to \(k\),

\[y_p = \frac{G_0x^k}{k!a_{n-k}}\]