Use the Method of Variation of Parameters to determine the general solution for the following problems. a. \(y^{\prime \prime}+y=\tan x\). b. \(y^{\prime \prime}-4 y^{\prime}+4 y=6 x e^{2 x}\).

The complementary function for the equation \(y^{\prime \prime}+y=\tan x\) is given by the solution of the homogeneous differential equation \(y^{\prime \prime}+y=0\), which is \(y = c_1 \cos x + c_2 \sin x\). Then, calculate the Wronskian, \(W = y_1y_2^{\prime} - y_2y_1^{\prime} = \sin x \cos x - \cos x \sin x = 0 \), which is not suitable for Variation of Parameters formula, thus, this equation can't be solved through Variation of Parameters.

For the equation \(y^{\prime \prime}-4 y^{\prime}+4 y=6 x e^{2 x}\), the complementary function is given by the solution of the homogeneous differential equation \(y^{\prime \prime}-4 y^{\prime}+4 y=0\), which is \(y = e^{2x}(c_1 + c_2 x)\). Calculate the Wronskian, \(W=y_1y_2'-y_2y_1'\), which results in \(W = e^{4x}\). The method of variation of parameters requires us to compute the functions \(v_1\) and \(v_2\), which are given as \(v_1 = -\int\frac{y_2f}{W}dx\) and \(v_2 = \int\frac{y_1f}{W}dx\). For this problem, we then have \(v_1 = -\int\frac{x e^{4x} 6x e^{2x}}{e^{4x}} dx = -3 \int x^2 e^{2x} dx\) and \(v_2 = \int\frac{e^{4x} 6x e^{2x}}{e^{4x}} dx= 3\int x e^{2x} dx\). The result after integration by parts on \(v_1\) and \(v_2\) is \(v_1 = -3(x^2 e^{2x} - 2x e^{2x} + 2e^{2x})\), and \(v_2 = 3(x e^{2x} - e^{2x})\). The general solution would be \(y = y_1v_1 + y_2v_2 = e^{2x}(c_1 + c_2 x) - 3e^{2x}(x^2 e^{2x} - 2x e^{2x} + 2e^{2x}) + 3xe^{2x}(x e^{2x} - e^{2x})\).