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Vlada [557]
3 years ago
9

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist

. (Enter your answer using interval notation.) (t − 5)y' + (ln t)y = 6t, y(1) = 6
Mathematics
1 answer:
yuradex [85]3 years ago
3 0

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

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Applying trigonometric ratios, the missing segments lengths in the image attached below are: <em>a = 8√3; b = 12.</em>

<h3>The Trigonometric Ratios</h3>
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  • CAH is cos ∅ = adj/hyp
  • TOA is tan ∅ = opp/adj
  • SOHCAHTOA is used to solve right triangles.

Given the right triangle in the mage attached below, we would find the missing sides as follows:

<em>Find a:</em>

Reference angle (∅) = 30°

opposite = 4√3

hypotenuse = a

  • <em>Apply SOH:</em>

sin 30 = (4√3)/a

a = (4√3)/sin 30

a = (4√3)/(1/2) (sin 30 = 1/2)

a = (4√3)×2

a = 8√3

<em>Find b:</em>

Reference angle (∅) = 60°

opposite = b

adjacent = 4√3

  • <em>Apply TOA:</em>

tan 60 = b/(4√3)

b = tan 60 × 4√3

b = √3 × 4√3

b = 12

Therefore, applying trigonometric ratios, the missing segments lengths in the image attached below are: <em>a = 8√3; b = 12.</em>

Learn more about trigonometric ratios on:

brainly.com/question/4326804

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Step-by-step explanation:

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"" =3.14×7^2

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