Example 2

Find a particular solution of the differential equation

y" - y' - 6y = 5 + 18t².

Since f(t) is a polynomial of degree two, we guess that some particular solution y(t) is a polynomial of degree two,

y(t) = K + Lt + Mt².

First we find the first and second derivatives of y(t).

y' = L + 2Mt, y" = 2M.

Next we substitute these derivatives into the given differential equation.

In the last equation the coefficients for each power of t must be equal. There are three equations, for units, t, and t².

units: 2M - L - 6K = 5

t: -2M - 6L = 0

t²: -6M = 18.

We can now solve the three equations for the three unknowns K, L, and M.

K= -2, L = 1, M = -3.

The required particular solution is then

y(t) = -2 + t - 3t².

It can be shown that whenever the forcing term f(t) is a polynomial of degree n, the differential equation (1) will have a particular solution that is a polynomial of degree n. When f(t) is a polynomial of degree n, the guess y(t) should be a polynomial of degree n with unknown coefficients,

y(t) = A₀ + A₁t + ... + A_ntⁿ.