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Ludmilka [50]
3 years ago
10

Let y′′′−9y′′+20y′=0. find all values of r such that y=erx satisfies the differential equation. if there is more than one correc

t answer, enter your answers as a comma separated list.
Mathematics
1 answer:
DerKrebs [107]3 years ago
4 0

we are given

differential equation as

y'''-9y''+20y'=0

we are given

y=e^{rx}

Firstly, we will find y' , y'' and y'''

those are first , second and third derivative

First derivative is

y'=re^{rx}

Second derivative is

y''=r*re^{rx}

y''=r^2e^{rx}

Third derivative is

y'''=r^2*re^{rx}

y'''=r^3e^{rx}

now, we can plug these values into differential equation

and we get

r^3 e^{rx}-9r^2 e^{rx}+20re^{rx}=0

now, we can factor out common terms

e^{rx}(r^3 -9r^2 +20r)=0

we can move that term on right side

(r^3 -9r^2 +20r)=0

now, we can factor out

r(r^2 -9r +20)=0

r(r-5)(r-4)=0

now, we can set them equal

r=0

r-5=0

r=5

r-4=0

r=4

so, we will get

r=0,4,5...............Answer

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Answer:

<u>The perimeter of the kite is 29 units.</u>

Step-by-step explanation:

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Step-by-step explanation:

1/4 × 4 = 1

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