 # How to solve this equation y'''-4y"+2y'-16y=4x+1 using Method of Undetermined Coefficient, Variation of Parameters and Laplace Transformation Line 2020-11-08 Answered

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

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Solve:
The equation is ${y}^{‴}-4{y}^{″}+2{y}^{\prime }-16y=4x+1$
Variation of Parameters:
The homogeneous equation is ${y}^{‴}-4{y}^{″}+2{y}^{\prime }-16y=0$
The homogeneous solution is ${y}_{h}={c}_{1}{e}^{4.37796x}+{e}^{-0.18898x}\left({c}_{2}\mathrm{cos}\left(1.90236x\right)+{c}_{3}\mathrm{sin}\left(1.90236x\right)\right)$
Obtain the Wro
ians:
$W=|\begin{array}{ccc}{e}^{ax}& {e}^{bx}\mathrm{cos}cx& {e}^{bx}\mathrm{sin}cx\\ a{e}^{ax}& b{e}^{bx}\mathrm{cos}cx-c{e}^{bx}\mathrm{sin}cx& b{e}^{bx}\mathrm{sin}cx-c{e}^{bx}\mathrm{cos}cx\\ {a}^{2}{e}^{ax}& \left({b}^{2}-{c}^{2}\right){e}^{bx}\mathrm{cos}cx-2bc{e}^{bx}\mathrm{sin}cx& \left({b}^{2}-{c}^{2}\right){e}^{bx}\mathrm{sin}cx-2bc{e}^{bx}\mathrm{cos}cx\end{array}|$
${W}_{1}=|\begin{array}{ccc}0& {e}^{bx}\mathrm{cos}cx& {e}^{bx}\mathrm{sin}cx\\ 0& b{e}^{bx}\mathrm{cos}cx-c{e}^{bx}\mathrm{sin}cx& b{e}^{bx}\mathrm{sin}cx-c{e}^{bx}\mathrm{cos}cx\\ 1& \left({b}^{2}-{c}^{2}\right){e}^{bx}\mathrm{cos}cx-2bc{e}^{bx}\mathrm{sin}cx& \left({b}^{2}-{c}^{2}\right){e}^{bx}\mathrm{sin}cx-2bc{e}^{bx}\mathrm{cos}cx\end{array}|$
The values of
Further we have