Solution of y” + 3y’ + 2y = 4sin(x) – 6cos(x).

by

(1)   \begin{equation*} y'' + 3y' + 2y = 4sin(x) - 6cos(x) \end{equation*}

By solving left hand side,

(2)   \begin{equation*} y'' + 3y' + 2y = 0 \end{equation*}

(3)   \begin{equation*} r^2 + 3r + 2 = 0 \end{equation*}

Solving for r, we get:

(4)   \begin{equation*} r_1 = -1, \quad r_2 = -2 \end{equation*}

Thus, the complementary function is:

(5)   \begin{equation*} y_c = c_1 e^{-x} + c_2 e^{-2x} \end{equation*}

To find the particular solution, we use the method of undetermined coefficients. Since the right-hand side of the given differential equation is of the form A\sin(x) + B\cos(x), we assume a particular solution of the form:

(6)   \begin{equation*} y_p = a\sin(x) + b\cos(x) \end{equation*}

(7)   \begin{equation*} y_p' = a\cos(x) - b\sin(x), \quad y_p'' = -a\sin(x) - b\cos(x) \end{equation*}

(8)   \begin{equation*} (-a\sin(x) - b\cos(x)) + 3(a\cos(x) - b\sin(x)) + 2(a\sin(x) + b\cos(x)) = 4\sin(x) - 6\cos(x) \end{equation*}

Simplifying this equation, we get:

(9)   \begin{equation*} (2a - b)\cos(x) + (-a + 2b)\sin(x) = 4\sin(x) - 6\cos(x) \end{equation*}

Equating the coefficients of \sin(x) and \cos(x) on both sides of the equation, we get:

(10)   \begin{equation*} \begin{aligned} 2a - b &= -6 \ -a + 2b &= 4 \end{aligned} \end{equation*}

Solving for a and b, we get:

(11)   \begin{equation*} a = -2, \quad b = -1 \end{equation*}

Thus, the particular solution is:

(12)   \begin{equation*} y_p = -2\sin(x) - \cos(x) \end{equation*}

The general solution is:

(13)   \begin{equation*} y = y_c + y_p = c_1 e^{-x} + c_2 e^{-2x} - 2\sin(x) - \cos(x) \end{equation*}

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