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GalinKa [24]
3 years ago
6

PLEASE HELP WILL MARK BRAINLIEST

Mathematics
1 answer:
Paha777 [63]3 years ago
3 0

Answer: D 12

Step-by-step explanation:

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Simplify.<br> Remove all perfect squares from inside the square root.<br> √180
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Read 2 more answers
Suppose x=c1e−t+c2e3tx=c1e−t+c2e3t. Verify that x=c1e−t+c2e3tx=c1e−t+c2e3t is a solution to x′′−2x′−3x=0x′′−2x′−3x=0 by substitu
Harrizon [31]

The correct question is:

Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)

Answer:

x = c1e^(-t) + c2e^(3t)

is a solution to the differential equation

x''- 2x' - 3x = 0

Step-by-step explanation:

We need to verify that

x = c1e^(-t) + c2e^(3t)

is a solution to the differential equation

x''- 2x' - 3x = 0

We differentiate

x = c1e^(-t) + c2e^(3t)

twice in succession, and substitute the values of x, x', and x'' into the differential equation

x''- 2x' - 3x = 0

and see if it is satisfied.

Let us do that.

x = c1e^(-t) + c2e^(3t)

x' = -c1e^(-t) + 3c2e^(3t)

x'' = c1e^(-t) + 9c2e^(3t)

Now,

x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]

= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)

= 0

Therefore, the differential equation is satisfied, and hence, x is a solution.

4 0
2 years ago
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