Question

(D^5-13D^3+26D^2+82D+104)y=0​

Answer 1

Answer:

Please mark it as the brainliest

Step-by-step explanation:

We need to get the auxillary equation of the given differential equation above...

The auxillary equation would be:

r5+r4−7r3−11r2−8r−12=0

We need to get the roots of the polynomial r5+r4−7r3−11r2−8r−12=0 using the rational zero theorem...

The rational zero theorem states that if P(x) is a polynomial with integer coefficients and if is a zero of P(x)(P(pq)=0), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x).

With that in mind....

The factors of p are:

−1,12,1,−12

−2,6,2,−6

−3,4,3,−4

The factors of q are:

1,1,−1,−1

Then the values of pq would be:

−1,1,2,−3,4,−4,12,−12,6,−6,−2,3

It was found out that only r=−2 and r=3 could make the polynomial r5+r4−7r3−11r2−8r−12=0 true, instead of 5

Now....if I only got 2 real and distinct roots out of possible 5, then the solution of the differential equation (D5+D4−7D3−11D2−8D−12)y=0 would be:

y=c1e−2x+c2e3x+(unknown term)+(unknown term)+(unknown term)

I'm stuck....How do you get the solution of the given differential equation above?

UPDATE

After some feedback from intrepid answerers....I was able to factor out the polynomial r5+r4−7r3−11r2−8r−12=0 into (r+2)(r+2)(r−3)(r2+1)=0, so I think the solution to the differential equation above would be:

y=c1e−2x+c2xe−2x+c3e3x+ex(c4cos(−x)+c5sin(−x))

or

y=c1e−2x+c2xe−2x+c3e3x+ex(c4cosx−c5sinx)