5.3 Undetermined Coefficients
Continuing Example 5.2.1, the general solution of (7) is \(x = x_c + x_p\), where \(x_c\) solves the homogeneous equation and \(x_p\) is any particular solution. For the homogeneous equation \(x'' - x' - 2x = 0\), the characteristic equation \(m^2 - m - 2 = 0\) has roots \(m = -1\) and \(m = 2\), so \[x_c(t) = Ae^{-t} + Be^{2t} .\] Since the right-hand side is \(10\cos t\), try \(x_p = c_1\cos t + c_2 \sin t\), with \[x'_p = -c_1 \sin t + c_2 \cos t ,\qquad x''_p = -c_1 \cos t - c_2 \sin t .\] Substituting into (7), \[( - c_1 \cos t - c_2 \sin t ) - ( -c_1 \sin t + c_2 \cos t) - 2( c_1 \cos t + c_2 \sin t ) = 10 \cos t ,\] that is, \[( -3c_1 -c_2 ) \cos t + ( c_1 - 3c_2 ) \sin t = 10 \cos t .\] Comparing coefficients, \(-3c_1 - c_2 = 10\) and \(c_1 - 3c_2 = 0\), so \(c_1 = 3c_2\), \(-10c_2 = 10\), and \(c_2 = -1\), \(c_1 = -3\). Hence \(x_p = -3\cos t - \sin t\) and \[x(t) = A e^{-t} + B e^{2t} -3\cos t - \sin t .\] Finally \(y = x' + 2x\), with \(x' = -Ae^{-t} + 2Be^{2t} + 3\sin t - \cos t\), so \[y(t) = Ae^{-t} + 4Be^{2t} - 7\cos t + \sin t .\] Substituting back into the second equation of the system confirms the solution.
The method of undetermined coefficients applies whenever the right-hand side is a sum of products of polynomials, exponentials, sines and cosines. The trial function has the same form; if a term of it already solves the homogeneous equation, it is multiplied by \(t\) (or \(t^2\)) until it does not.
Solution. The trial \(Ce^{2t}\) fails, since \(e^{2t}\) solves the homogeneous equation. Try \(x_p = Cte^{2t}\): then \(x_p' = C(1 + 2t)e^{2t}\), \(x_p'' = C(4 + 4t)e^{2t}\), and \[x_p'' - x_p' - 2x_p = Ce^{2t}\big [(4 + 4t) - (1 + 2t) - 2t\big ] = 3Ce^{2t} .\] So \(3C = 6\), \(C = 2\), and \(x_p = 2te^{2t}\).
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.