Gregory Emery

2021-12-27

Identify the characteristic equation, solve for the characteristic roots, and solve the 2nd order differential equations.

$(4{D}^{4}-4{D}^{3}-23{D}^{2}+12D+36)y=0$

Jack Maxson

Beginner2021-12-28Added 25 answers

To find: The characteristic equation and the solution.

Solution:

Let$y=A{e}^{mx}(A\ne 0)$ be the solution of the given differential equation is

$(4{D}^{4}-4{D}^{3}-23{D}^{2}+12D+36)y=0$

Then the characteristic equation is

$4{m}^{4}-4{m}^{3}-23{m}^{2}+12m+36=0$

$\Rightarrow ({x}^{2}-4x+4)(4{x}^{2}+12x+9)=0$

$\Rightarrow {(x-2)}^{2}{(2x+3)}^{2}=0$

$\Rightarrow x=2,2,x=-\frac{3}{2},-\frac{3}{2}$

Then the solution is$y\left(x\right)={e}^{2x}(4+{c}_{2}x)+{e}^{-\frac{3}{2}x}({c}_{3}+{c}_{3}x)$

Where$c}_{1},{c}_{2},{c}_{3},{c}_{4$ are arbitrary constant.

Solution:

Let

Then the characteristic equation is

Then the solution is

Where

Natalie Yamamoto

Beginner2021-12-29Added 22 answers

A.E. is $4{m}^{4}-4{m}^{3}-23{m}^{2}+12m+36=0$

If$m=2,64-32-92+24+36=0$

$\Rightarrow m=2$ 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.

$4{m}^{2}+12m+9=0$

${(2m+3)}^{2}=0$

$m=\frac{-3}{2},\frac{-3}{2}$

The roots of the A.E.

are$2,2,\frac{-3}{2},\frac{-3}{2}$

Thus,$y=({C}_{1}+{C}_{2}x){e}^{2x}+({C}_{3}+{C}_{4}x){e}^{-3\frac{x}{2}}$ .

If

If

Again

By synthetic division.

The roots of the A.E.

are

Thus,

Vasquez

Expert2022-01-09Added 669 answers

Solution:

the auxilary equation,

Use long division we get,

The general solution,

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