INSTRUCTORS Raffi Hovasapian John Zhu

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### Differential Equations Slope Fields

• Way of showing slope at every point of function
• Draw table of values with desired dimensions
• Fill in with values
• Draw small slope lines at each coordinate
• Calculation times can be reduced by recognizing patterns on graph

### Differential Equations Slope Fields

Given y [dy/dx] + 2x = 0, find the values where [dy/dx] is undefined.
• y [dy/dx] + 2x = 0
• y [dy/dx] = −2x
• [dy/dx] = [(−2x)/y]
[dy/dx] is undefined when y = 0
Solve for y given y [dy/dx] + 2x = 0
• y [dy/dx] + 2x = 0
• y [dy/dx] = −2x
• ∫y  dy = −2 ∫x  dx
• [(y2)/2] = −x2 + c
• y2 = −2x2 + C
y = ±√{C − 2x2}
Given [dy/dx] + 2x  ey = 0, find the values where [dy/dx] is undefined.
• [dy/dx] = −2x ey
• 2x and ey are both continuous
• The product of two continuous functions will also yield a continuous function
[dy/dx] is defined everywhere.
Find the x and y coordinates where [dy/dx] reaches its maximum, given [dy/dx] = [(1 − x2 + x)/(y2 + 1)]
• [dy/dx] will reach a maximum when the numerator is as large as possible and the denominator is as small as possible.
• We know a function will hit a relative maximum or minimum when the slope is equal to zero
• [d/dx] (1 − x2 + x) = 0
• −2x + 1 = 0
• 2x = 1
• x = [1/2]
• Is this a minimum or a maximum? We can try test points around the critical point, or we can use the second derivative test
• f"(x) = −2 always concave down, implies a maximum
• [d/dy] (y2 + 1) = 0
• 2y = 0
• y = 0
• f"(x) = 2 always concave up, implies a minimum
• As a sanity check, let's make sure these values result in a positive number
• [dy/dx] = [(1 − [1/4] + [1/2])/1] = [3/4]
[dy/dx] reaches a maximum at [[1/2], 0]
Find the conditions such that [dy/dx] = 0 if [dy/dx] = x − y
• [dy/dx] = x − y = 0
• x = y
[dy/dx] = 0 when x = y
Sketch the slope field of [dy/dx] = x − y
•  x = -2 x = -1 x = 0 x = 1 x = 2 y = -2 0 1 2 3 4 y = -1 -1 0 1 2 3 y = 0 -2 -1 0 1 2 y = 1 -3 -2 -1 0 1 y = 2 -4 -3 -2 -1 0
Sketch the slope field of [dy/dx] = x  y2
•  x = -2 x = -1 x = 0 x = 1 x = 2 y = -2 -8 -4 0 4 8 y = -1 -2 -1 0 1 2 y = 0 0 0 0 0 0 y = 1 -2 -1 0 1 2 y = 2 -8 -4 0 4 8
Find y given [dy/dx] = x  y2 and y(0) = 2
• [dy/dx] = x  y2
• [1/(y2)] [dy/dx] = x
• [1/(y2)] dy = x  dx
• ∫[1/(y2)] dy = ∫x  dx
• − [1/y] = [(x2)/2] + C
• −1 = ([(x2)/2] + C)y
• y = [1/(−[(x2)/2] + C)]
• y = [2/(c − x2)]
• 2 = [2/(c − 0)]
• c = 1
y = [2/(1 − x2)]
Find the values where [dy/dx] is undefined given [dy/dx] = [(x + 1)/(y − 1)]
• [dy/dx] is undefined when the denominator is equal to zero
• y − 1 = 0
• y = 1
[dy/dx] is undefined when y = 0
Find the values where [dy/dx] is undefined or not real given [dy/dx] = [(x2)/(√{1 − y3})]
• The function is undefined and/or not real when 1 − y3≤ 0. The result will not be real if you take the square root of a negative number.
• 1 − y3≤ 0
• −y3≤ −1
• y3≥ 1
• y ≥ 1
[dy/dx] is undefined and/or not real when y ≥ 1

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.