5.3 Undetermined Coefficients

First find \(x_c\), which is the solution of \[x'' -x' -2x = 0\hspace {0.3cm}:\hspace {0.2cm} m^2 - m - 2 = 0,\hspace {0.4cm} m = -1,2\]

\[x_c(t) = Ae^{-t} + Be^{2t}\]

\begin {align*} \text {Let}\hspace {1cm} x_p & = c_1\cos t + c_2 \sin t \\ x'_p & = c_1 \sin t + c_2 \cos t\\ x''_p & = -c_1 \cos t - c_2 \sin t \end {align*}

\[\big ( - c_1 \cos t - c_2 \sin t \big ) - \big ( -c_1 \sin t + c_2 \cos t\big ) - 2\big ( c_1 \cos t + c_2 \sin t \big ) = 10 \cos t\]

\[\implies \hspace {0.5cm} \big ( -c_1 -c_2 -2c_1 \big ) \cos t + \big ( -c_2 + c_1 -2c_2 \big ) \sin t = 10 \cos t\]

\[\implies -3c_1 - c_2 = 10\hspace {0.1cm}, \hspace {0.2cm} -3c_2 + c_1 = 0 \implies c_1 = 3c_2\]

\[-9c_2 - c_2 = 10 \implies -10c_2 = 10 \implies c_2 = -1\hspace {0.4cm}\text {and}\hspace {0.4cm} c_1 = -3\]

\[x_p = -3\cos t - \sin t\]

\begin {align*} x(t) & = x_c + x_p\\ & = A e^{-t} + B e^{2t} -3\cos t - \sin t \end {align*}

\[\text {But}\hspace {0.5cm} y = x' + 2x\]

\begin {align*} \implies \hspace {1cm} y(t) & = -Ae^{-t} + 2Be^{2t} + 3\sin t - \cos t\\\\ y(t) & = Ae^{-t} + 4Be^{2t} - 7\cos t + \sin t\\\\ \end {align*}

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