# Identify the characteristic equation, solve for the characteristic roots, and

Identify the characteristic equation, solve for the characteristic roots, and solve the 2nd order differential equations.
$\left(4{D}^{4}-4{D}^{3}-23{D}^{2}+12D+36\right)y=0$
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Jack Maxson
To find: The characteristic equation and the solution.
Solution:
Let $y=A{e}^{mx}\left(A\ne 0\right)$ be the solution of the given differential equation is
$\left(4{D}^{4}-4{D}^{3}-23{D}^{2}+12D+36\right)y=0$
Then the characteristic equation is
$4{m}^{4}-4{m}^{3}-23{m}^{2}+12m+36=0$
$⇒\left({x}^{2}-4x+4\right)\left(4{x}^{2}+12x+9\right)=0$
$⇒{\left(x-2\right)}^{2}{\left(2x+3\right)}^{2}=0$
$⇒x=2,2,x=-\frac{3}{2},-\frac{3}{2}$
Then the solution is $y\left(x\right)={e}^{2x}\left(4+{c}_{2}x\right)+{e}^{-\frac{3}{2}x}\left({c}_{3}+{c}_{3}x\right)$
Where ${c}_{1},{c}_{2},{c}_{3},{c}_{4}$ are arbitrary constant.
###### Not exactly what you’re looking for?
Natalie Yamamoto
A.E. is $4{m}^{4}-4{m}^{3}-23{m}^{2}+12m+36=0$
If $m=2,64-32-92+24+36=0$
$⇒m=2$ is a root of inspection. By synthetic division,
$i.e.,4{m}^{3}+4{m}^{2}-15m-18=0$
If $m=2$
$32+16-30-18=0$
Again $m=2$ is a root.
By synthetic division.
$4{m}^{2}+12m+9=0$
${\left(2m+3\right)}^{2}=0$
$m=\frac{-3}{2},\frac{-3}{2}$
The roots of the A.E.
are $2,2,\frac{-3}{2},\frac{-3}{2}$
Thus, $y=\left({C}_{1}+{C}_{2}x\right){e}^{2x}+\left({C}_{3}+{C}_{4}x\right){e}^{-3\frac{x}{2}}$.
###### Not exactly what you’re looking for?
Vasquez

Solution:
$\left(4{D}^{4}-4{D}^{3}-23{D}^{2}+12D+36\right)y=0$
the auxilary equation, $4{m}^{4}-4{m}^{3}+23{m}^{2}+12m+36=0$
Use long division we get, $m=-\frac{3}{2}$ of order 2 and $m=2$ of order 2
The general solution,
$y\left(x\right)=A{e}^{-\frac{3}{2}x}+Bx{e}^{-\frac{3}{2}x}+C{e}^{2x}+Dx{e}^{2x}$
$y\left(x\right)=\left(A+Bx\right){e}^{-\frac{3}{2}x}+\left(C+Dx\right){e}^{2x}$