Frank Guyton

2021-12-28

What is the general solution for the high-order differential equations ${y}^{5}+6{y}^{4}+12y+8y=0$

zurilomk4

The derivatives of this trail solution are: ${y}^{\prime }=m{e}^{mx}$
$y{m}^{2}{e}^{mx}$
Substitute the trail solution and its derivatives in the given D.E.
${y}^{5}+6{y}^{4}+12{y}^{3}+8y=0$
${m}^{5}{e}^{mx}+6{m}^{4}{e}^{mx}+12{m}^{3}{e}^{mx}+8m{e}^{mx}=0$
${e}^{mx}\left({m}^{5}+6{m}^{4}+12{m}^{3}+8m\right)=0$
${m}^{5}+6{m}^{4}+12{m}^{3}+8m=0$
$m\left({m}^{4}+6{m}^{3}+12{m}^{2}+8\right)=0$
$m=0,{m}^{4}+6{m}^{3}+12{m}^{2}+8=0$
$y=0$, no solution for $4\in R$

soanooooo40

${y}^{5}+6{y}^{4}+12{y}^{3}+8y=0$
${m}^{5}{e}^{mx}+6{m}^{4}{e}^{mx}+12{m}^{3}{e}^{mx}+8m{e}^{mx}=0$
${e}^{mx}\left({m}^{5}+6{m}^{4}+12{m}^{3}+8m\right)=0$
${m}^{5}+6{m}^{4}+12{m}^{3}+8m=0$
$m\left({m}^{4}+6{m}^{3}+12{m}^{2}+8\right)=0$
$m=0,{m}^{4}+6{m}^{3}+12{m}^{2}+8=0$
$y=0$, no solution for $4\in R$

karton

$⇒\left({D}^{5}+6{D}^{4}+12{D}^{3}+8{D}^{2}\right)y=0$
where $⇒\left({D}^{5}+6{D}^{4}+12{D}^{3}+8{D}^{2}\right)y=0$
The auxilary equation is given by
$f\left(D\right)=0$

Here we see that the roots y the auxiliary equation are repeated.
Hence complementary function is given by $C.F=\left({C}_{1}+x{c}_{2}\right){e}^{ox}+\left({c}_{3}+x{c}_{4}+{x}^{2}{c}_{5}{\right)}^{{e}^{-2x}}$
Hence general solution is
$y=\left({c}_{1}+x{c}_{2}\right)+\left({c}_{3}+x{c}_{4}+{x}^{2}{c}_{5}\right){e}^{-2x}$

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