Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th
1 answer:
Answer:
The solutions are linearly independent because the Wronskian is not equal to 0 for all x.
The value of the Wronskian is
Step-by-step explanation:
We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.
Thus replacing the functions of the exercise we get:
Working with the determinant we get
Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.
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Note:
The slope of a straight line passing through the points (x₁, y₁) and (x₂, y₂) is
The two given points are (-3,4) and (4,-1). Therefore the slope is
Answer:
The slope of a straight line passing through the points (x₁, y₁) and (x₂, y₂) is
The two given points are (-3,4) and (4,-1). Therefore the slope is
Answer:
Answer:
f(x) = + 2x³ - 4x² - 6x + 3
Step-by-step explanation:
Note that radical zeros occur in conjugate pairs, thus
- 1 + is a zero then - 1 -
is also a zero
is a zero then -
is also a zero
Thus the corresponding factors are
(x - (- 1 + ) ), (x - (- 1 -
) ), (x -
), (x - (-
)), that is
(x + 1 - ), (x + 1 +
), (x -
), (x +
)
The polynomial is then the product of the roots
f(x) = (x + 1 - )(x + 1 +
)(x -
)(x +
)
= ((x + 1)² - ()²)((x² - (
)²)
= (x² + 2x + 1 - 2)(x² - 3)
= (x² + 2x - 1)(x² - 3) ← distribute
= - 3x² + 2x³ - 6x - x² + 3
= + 2x³ - 4x² - 6x + 3
So we start by combining like terms;
Now we have;
We then subtract 1 from both sides;
Then we divide both sides by 5;
Leaving the final answer to be h = -2