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saw5 [17]
3 years ago
9

Suppose that r1 and r2 are roots of ar2 + br + c = 0 and that r1 = r2; then exp(r1t) and exp(r2t) are solutions of the different

ial equation ay + by + cy = 0. Show that φ (t; r1, r2) = er2t − er1t r2 −
Mathematics
1 answer:
Nady [450]3 years ago
7 0

The Correct Question is:

Suppose that r1 and r2 are roots of ar² + br + c = 0 and that r1 = r2; then e^(r1t) and e^(r2t) are solutions of the differential equation

ay'' + by' + cy = 0.

Show that

φ (t; r1, r2) = [e^(r2t) - e^(r1t )]/(r2 - r1)

is a solution of the differential equation.

Answer:

φ (t; r1, r2) is a solution of the differential equation, and it shown.

Step-by-step explanation:

Given the differential equation

ay'' + by' + cy = 0

and r1 and r2 are the roots of its auxiliary equation.

We want to show that

φ (t; r1, r2) = [e^(r2t) - e^(r1t )]/(r2 - r1)

satisfies the given differential equation, that is

aφ'' + bφ' + cφ = 0 .....................(*)

Where φ = φ (t; r1, r2)

We now differentiate φ twice in succession, with respect to t.

φ' = [r2e^(r2t) - r1e^(r1t )]/(r2 - r1)

φ'' = [r2²e^(r2t) - r1²e^(r1t )]/(r2 - r1)

Using these in (*)

We have

a[r2e^(r2t) - r1e^(r1t )]/(r2 - r1) + [r2²e^(r2t) - r1²e^(r1t )]/(r2 - r1) + c[e^(r2t) - e^(r1t )]/(r2 - r1)

= [(ar2² + br2 + c)e^(r2t) - (ar1² + br1 + c)e^(r1t)]/(r1 - r2)

We know that r1 and r2 are the roots of the auxiliary equation

ar² + br + c = 0

and r1 = r2

This implies that

ar1² + br1 + c = ar2² + br2 + c = 0

And hence,

[(ar2² + br2 + c)e^(r2t) - (ar1² + br1 + c)e^(r1t)]/(r1 - r2) = 0

Therefore,

aφ'' + bφ' + cφ = 0

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Help pls I'll give 25 points
RideAnS [48]

Answer:

\text{C. }60

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

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