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Pavel [41]
2 years ago
6

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

ting it into the differential equation. (Enter the terms in the order given. Enter c1c1 as c1 and c2c2 as c2.)
Mathematics
1 answer:
Harrizon [31]2 years ago
4 0

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.

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3 years ago
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Answer:

Prime

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Answer:

The completed proof is presented as follows;

The two column proof is presented as follows;

Statements    {}                                               Reason

1. \overline {HI} ║ \overline {KL}, J is the midpoint of \overline {HL} {}         1. Given

2. ∠IHJ ≅ ∠JLK{}                                            2. Alternate angles are congruent

3. ∠IJH ≅ ∠KJL   {}                                         3. Vertically opposite angles

4.  \overline {HJ} ≅ \overline {JL}   {}                                              4. Definition of midpoint

5. ΔHIJ ≅ ΔLKJ  {}                                         5. By ASA rule of congruency

Step-by-step explanation:

Alternate angles formed by the crossing of the two parallel lines \overline {HI} and \overline {KL}, by the transversal \overline {HL} are equal

Vertically opposite angles formed by the crossing of two straight lines \overline {IK} and \overline {HL} are always equal

A midpoint divides a line into two equal halves

Angle-Side-Angle, ASA rule of congruency states that two triangles ΔHIJ and ΔLKJ, that have two congruent angles, ∠IHJ in ΔHIJ ≅ ∠JLK{} in ΔLKJ and ∠IJH in ΔHIJ ≅ ∠KJL in ΔLKJ, and that the included sides between the two congruent angles is also congruent \overline {HJ} ≅ \overline {JL}, then the two triangles are congruent, ΔHIJ ≅ ΔLKJ.

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