Question

solve -: d^3y/dx^3-2(d^2y/dx^2)+4(dy/dx)-8y=0​

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Answer 1

☯ Explanation,

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$\bigstar \tt \: \bigg(\dfrac{d {}^{3}y }{dx {}^{3} } \bigg) -2 \bigg( \dfrac{d {}^{2}y }{dx {}^{2} } \bigg) - 4 \bigg(\dfrac{dy}{dx} \bigg) + 8y = 0 \\ \\ \\ \tt \: Let , \dfrac{dy}{dx} = R\\ \\ \\ \: : \implies \tt \bigg( R {}^{3} - R {}^{2} - 4R + 8 \bigg)y= 0 \\ \\ \\ : \implies \tt \bigg(R {}^{3} - R {}^{2} - 4R + 8 \bigg) = 0 \\ \\ \\ : \implies \tt \bigg(R {}^{2} (R - 2) - 4(R - 2) \bigg)= 0 \\ \\ \\ : \implies \tt \bigg( (R - 2)(R - 2)(R + 2) \bigg) = 0 \\ \\ \\ : \implies \tt R = \: 2 , \: 2 , \: - 2 \\ \\ \\ \red{ \tt [ \because \: R =\: 2 , \:2 , \:-2 \: are \: the \: roots ]} \\ \\ \\ \therefore \underline{\boxed{ \tt{y = (c_1x + c_2) e {}^{2x} + c_2 e {}^{ - 2x} }}}\bigstar$

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Answer 2

$\large \sf { \red{ Solution :- }}$

$\begin{gathered}\sf{ \color{royalblue} \: {\bigg(\dfrac{d {}^{3}y }{dx {}^{3} } \bigg) -2 \bigg( \dfrac{d {}^{2}y }{dx {}^{2} } \bigg) - 4 \bigg(\dfrac{dy}{dx} \bigg) + 8y = 0 }}\\ \\ \\ \sf \:{ \color{green} {Let , \dfrac{dy}{dx} = R}}\\ \\ \\ \: { \purple{ \implies}} \sf{ \purple{\bigg( R {}^{3} - R {}^{2} - 4R + 8 \bigg)y= 0}} \\ \\ \\ { \purple{ \implies \sf \bigg(R {}^{3} - R {}^{2} - 4R + 8 \bigg) = 0 }}\\ \\ \\ { \purple{ \implies \sf \bigg(R {}^{2} (R - 2) - 4(R - 2) \bigg)= 0}} \\ \\ \\ { \purple{ \implies \sf \bigg( (R - 2)(R - 2)(R + 2) \bigg) = 0}} \\ \\ \\ { \purple{\implies \sf R = \: 2 , \: 2 , \: - 2}} \\ \\ \\ {\pink{ \sf \underline{ [ \because \: R =\: 2 , \:2 , \:-2 \: are \: the \: roots ]}}} \\ \\ \\ \therefore { \color{orange}{\boxed{ \sf{y = (c_1x + c_2) e {}^{2x} + c_2 e {}^{ - 2x} }}}}\end{gathered}$