Before we explain the method of undetermined coefficients we need to make a simple observation about nonhomogeneous or driven equations such as
\begin{equation*}
y'' + ay' + by = f(x).
\end{equation*}
Solving such equations where the right hand side is nonzero will require us to actully find two different solutions \(y_p\) and \(y_h\text{.}\) The \(p\) stands for particular and the \(h\) stands for homogeneous. The following theorem explains why.
Theorem6.3.1
Suppose \(y_p\) is a solution to:
\begin{equation}
y'' + ay' + by = f(x).
\label{men-41}\tag{6.3.1}
\end{equation}
And suppose \(y_1\) and \(y_2\) are solutions to the associated homogeneous equation:
\begin{equation}
y'' + ay' + by = 0.
\label{eqn-associated-homogeneous}\tag{6.3.2}
\end{equation}
Then the function defined by
\begin{align}
y \amp = y_h + y_p \notag\notag\\
y \amp =
\underbrace{c_1 y_1 + c_2 y_2}_{y_h} + y_p\label{eqn-homogeneous-plus-particular}\tag{6.3.3}
\end{align}
is a solution to the original nonhomogeneous system, equation (6.1.3).
Proof
Plug equation (6.3.3) into the governing equation (6.1.3).
\begin{align*}
\amp (c_1 y_1 + c_2 y_2 + y_p)'' + a(c_1 y_1 + c_2
y_2 + y_p)' + b(c_1 y_1 + c_2 y_2 + y_p)\\
= \amp (c_1 y_1'' + c_2 y_2'' + y_p'') + a(c_1
y_1' + c_2 y_2' + y_p') + b(c_1 y_1 + c_2 y_2 +
y_p)\\
= \amp c_1(y_1'' + a y_1' + b y_1) + c_2(y_2'' + a
y_2' + b y_2) + (y_p'' + a y_p' + b y_p)\\
= \amp c_1 \cdot 0 + c_2 \cdot 0 + f(x)\\
= \amp f(x)
\end{align*}
The theorem says that solving nonhomogeneous linear equations is a three step process:
Find a solution, \(y_h\text{,}\) to the associated homogeneous equation.
Find any particular solution, \(y_p\text{,}\) which satisfies the original nonhomogeneous equation.
Form the general solution.
\begin{equation}
y = y_h + y_p.
\label{eqn-general-solution-linear-driven}\tag{6.3.4}
\end{equation}
If \(y_p\) is a solution to the original nonhomogeneous equation, why doesn't that suffice for the general solution? The reason that the particular solution alone is insufficient is because \(y_p\) is exactly one function, but a general solution must be an infinite family of solutions. In chapter one we solved first order differential equations by integration and in so doing introduced a constant of integration to the general solution. If we were given a specific point (initial condition) which the desired solution had to pass through, then that allowed us to determine a particular value for the constant of integration, and we called this a particular solution. In other words, each constant of integration introduces a degree of freedom into the solution set. Second order equations involve two derivatives and thus require two integrations to solve and hence introduce two constants of integration which we usually call \(c_1\) and \(c_2\text{.}\) A second order initial value problem (IVP) will require two initial values or conditions, e.g.
\begin{equation*}
y(a) = b_0, \qquad y'(a) = b_1.
\end{equation*}
An \(n\)th order differential equation requires \(n\) integrations and hence will introduce \(n\) constants of integration or degrees of freedom, which we label: \(c_1,
c_2, \ldots, c_n\text{.}\) These constants afford us enough degrees of freedom to uniquely satisfy all \(n\) initial conditions:
\begin{equation*}
y(a) = b_0, \:\: y'(a) = b_1,\:\: y''(a) = b_2,
\ldots ,\:\: y^{(n-1)} = b_{n-1}.
\end{equation*}