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Slav-nsk [51]
4 years ago
11

A line segment is partitioned with a ratio 2:5, how many total parts has the segment been split into

Mathematics
1 answer:
schepotkina [342]4 years ago
7 0

Answer:

7 parts

Step-by-step explanation:

To get the answer to this, we only simply to know the number of times in which the division was made.

This means that the total ratio of the lines is 2 +5 = 7. Meaning we have 7 different divisions.

The first division carries 2/7 while the second division carrues 5/7

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First check the characteristic solution:

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y_p = (At+B)e^{-2t}

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y_p = (At^3+Bt^2)e^{-2t}

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{y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}

{y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}

Substituting these into the left side of the ODE gives

(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}

so that 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 0.

For the second solution corresponding to -8t-12, we use

y_p = Ct + D

with derivative

{y_p}' = C

{y_p}'' = 0

Substituting these gives

4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12

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Then the general solution to the ODE is

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-2 = C_1 - 1 \implies C_1 = -1

1 = -2C_1 + C_2 - 2 \implies C_2 = 1

and so the particular solution satisfying these conditions is

y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1

or

\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}

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