Differential Equations Exercises

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The only equilibrium solution is y = 0:

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There are no places on this slope field where a horizontal line fits in, so there are no equilibrium solutions.

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There are no places on this slope field where a horizontal line fits in, so there are no equilibrium solutions.

Given a differential equation, we can find equilibrium solutions by setting the derivative equal to zero and solving.

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We factor:

The equilibrium solutions are y = 0, y = 1, and y = -1.

When y > 1 the quantity (1 – y) is negative, and (1 + y) is positive, so the slopes are negative.

When 0 < y < 1 all quantities are positive, so the slopes are positive.

When -1 < y < 0 the quantity y is negative, but (1 – y) and (1 + y) are positive so the slopes are negative.

When y < -1 the quantities y and (1 + y) are negative, but (1 – y) is positive so the slopes are positive.

We end up with y = 1 and y = -1 being stable equilibrium solutions, while y = 0 is unstable:

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These don't require any calculus except to understand what the question is asking. To answer the question, set each derivative equal to zero and solve for the dependent variable.

0 = y

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These don't require any calculus except to understand what the question is asking. To answer the question, set each derivative equal to zero and solve for the dependent variable.

0 = P(1 - P)P = 0, 1

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To answer the question, set the derivative equal to zero and solve for the dependent variable.

0 = Q² + 5Q + 6 = (Q + 2)(Q + 3)Q = -2, -3

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Set each derivative equal to zero and solve for the dependent variable.

0 = y² – y = y(1 – y)y = 0, 1

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This d.e. has no equilibrium solutions because the equation

0 = R² + 4

has no solutions.

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First we factor:

The equilibrium solutions are y = -2 and y = 3.

For y > 3, both (y – 3) and (y + 2) are positive. The product (y – 3) (y + 2) is positive, as are the slopes.

For -2 < y < 3 the quantity (y – 3) is negative and (y + 2) is positive, so the slopes are negative.

For y < -2, both (y – 3) and (y + 2) are negative, so the slopes are positive.

We conclude that y = -2 is a stable equilibrium and y = 3 is an unstable equilibrium:

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Let's factor:

The equilibrium solutions are y = 0, y = -1, and y = 1.

When y > 1 all quantities are positive, so the slopes are positive also.

When 0 < y < 1 the quantity (y – 1) is negative and the others are positive, so the slopes are negative.

When -1 < y < 0 the quantities y and (y – 1) are negative, so the slopes are positive.

Finally, when y < -1 all quantities are negative, so the slopes are negative.

We conclude that 0 is the only stable equilibrium solution:

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First we factor:

The equilibrium solutions are y = 0 and y = 1.

Since the quantity (y – 1)² is always positive it won't affect the signs of the slopes.

When y > 0, the slopes are positive.

When y < 0, the slopes are negative:

Both equilibrium solutions are unstable.

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First we factor:

The equilibrium solutions are y = 0 and y = -1:

Since (y + 1)² is always positive, this factor won't affect the signs of the slopes.

When y > 0 the slopes are positive, and when y < 0 the slopes are negative:

Both equilibrium solutions are unstable.