How to solve this equation y'''-4y"+2y'-16y=4x+1 using Method of Undetermined Coefficient, Variation of Parameters and Laplace Transformation

Solve:
The equation is $y^{‴}-4y^{″}+2y^{\prime }-16y=4x+1$
Variation of Parameters:
The homogeneous equation is $y^{‴}-4y^{″}+2y^{\prime }-16y=0$
The homogeneous solution is ${\displaystyle y_h=c_1{e}^{4.37796x}+e^{-0.18898x}(c_2\cos \left(1.90236x\right)+c_3\sin \left(1.90236x\right))}$
Obtain the Wro
ians:
${\displaystyle W=\left|\begin{array}{ccc}{e}^{ax}& e^{bx}\cos cx& e^{bx}\sin cx\\ a{e}^{ax}& b{e}^{bx}\cos cx-c{e}^{bx}\sin cx& b{e}^{bx}\sin cx-c{e}^{bx}\cos cx\\ a^2{e}^{ax}& (b^2-c^2)e^{bx}\cos cx-2bc{e}^{bx}\sin cx& (b^2-c^2)e^{bx}\sin cx-2bc{e}^{bx}\cos cx\end{array}\right|}$
${\displaystyle W_1=\left|\begin{array}{ccc}0& e^{bx}\cos cx& e^{bx}\sin cx\\ 0& b{e}^{bx}\cos cx-c{e}^{bx}\sin cx& b{e}^{bx}\sin cx-c{e}^{bx}\cos cx\\ 1& (b^2-c^2)e^{bx}\cos cx-2bc{e}^{bx}\sin cx& (b^2-c^2)e^{bx}\sin cx-2bc{e}^{bx}\cos cx\end{array}\right|}$
The values of ${\displaystyle a=4.37796,b=-0.18898\text{ and }c=1.90236}$
Further we have
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