 # Solve the following differential Equation gorovogpg 2021-12-26 Answered
Solve the following differential Equation
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${D}^{4}y-8{D}^{3}y+24{D}^{2}y-32Dy+16y=0$
$\left({D}^{4}-8{D}^{3}+24{D}^{2}-32D+16\right)y=0$
Auxiliary Equation
${m}^{4}-8{m}^{3}+24{m}^{2}-32m+16=0$
Let $m=2,f\left(2\right)={2}^{4}-8{\left(2\right)}^{3}+24{\left(2\right)}^{2}-32\left(2\right)+16$
$=16-64+96-64+16$
$=128-128=0$
$\therefore {m}^{2}-6{m}^{2}+12m-8=0$
${\left(m-2\right)}^{3}=0$
$\therefore {\left(m-2\right)}^{3}=0⇒\left(m-2\right)\left(m-2\right)\left(m-2\right)$
$\therefore m=2,2,2,2$ are the four repeated roots
The solution is of the form
${y}_{c}\left(t\right)=\left({C}_{1}+{C}_{2}t+{C}_{3}{t}^{2}+{C}_{4}{t}^{3}\right){e}^{mt}$
$=\left({C}_{1}+{C}_{2}t+{C}_{3}{t}^{2}+{C}_{4}{t}^{3}\right){e}^{2t}$
Result: we obtain real and equal roots with multiplicity 4
$y\left(t\right)=\left({C}_{1}+{C}_{2}t+{C}_{3}{t}^{2}+{C}_{4}{t}^{3}\right){e}^{2t}$
###### Not exactly what you’re looking for? karton

$\begin{array}{}{D}^{4}y-8{D}^{3}y+24{D}^{2}y-32Dy+16y=0\\ \left({D}^{4}-8{D}^{3}+24{D}^{2}-32D+16\right)y=0\\ {m}^{4}-8{m}^{3}+24{m}^{2}-32m+16=0\\ m=2,f\left(2\right)={2}^{4}-8\left(2{\right)}^{3}+24\left(2{\right)}^{2}-32\left(2\right)+16\\ =16-64+96-64+16\\ =128-128=0\\ {m}^{2}-6{m}^{2}+12m-8=0\\ \left(m-2{\right)}^{3}=0\\ \left(m-2{\right)}^{3}=0⇒\left(m-2\right)\left(m-2\right)\left(m-2\right)\\ m=2,2,2,2\\ {y}_{c}\left(t\right)=\left({C}_{1}+{C}_{2}t+{C}_{3}{t}^{2}+{C}_{4}{t}^{3}\right){e}^{mt}\\ =\left({C}_{1}+{C}_{2}t+{C}_{3}{t}^{2}+{C}_{4}{t}^{3}\right){e}^{2t}\\ y\left(t\right)=\left({C}_{1}+{C}_{2}t+{C}_{3}{t}^{2}+{C}_{4}{t}^{3}\right){e}^{2t}\end{array}$