Part (a) Keep in mind that dy/dx is the SLOPE! We simply need to substitute x and y into the differential equation and represent each answer as a slope.

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Presentation transcript:

Part (a) Keep in mind that dy/dx is the SLOPE! We simply need to substitute x and y into the differential equation and represent each answer as a slope on the graph.

Part (a) dy/dx = x2 (y-1) dy/dx = -12 (3-1) dy/dx = 2

Part (a) dy/dx = x2 (y-1) dy/dx = 02 (1-1) dy/dx = 0

Part (a) dy/dx = x2 (y-1) dy/dx = 12 (2-1) dy/dx = 1

Part (a) Repeating this process on the other nine points gives us the following slope field…

“Where is the slope positive?” Part (b) dy/dx = x2 (y-1) “Where is the slope positive?”

Part (b) dy/dx = x2 (y-1) x2 can’t be negative, but it can be zero. This occurs when x=0 (or on the y-axis). For our slopes to be positive, we must stay off of the y-axis.

(Above this dotted line) Part (b) dy/dx = x2 (y-1) For the entire derivative to be positive, (y-1) will need to be positive. This occurs when y>1. (Above this dotted line)

Part (b) dy/dx = x2 (y-1) Therefore, the derivative will be positive at any point (x,y) where x = 0 and y > 1.

(Separate the variables) (Integrate both sides) Part (c) ln y-1 = 1/3 x3 + C dy/dx = x2 (y-1) (Separate the variables) (y-1)-1 dy = x2 dx (Integrate both sides) ln y-1 = 1/3 x3 + C

(Exponentiate both sides) (Using the laws of exponents from Algebra) Part (c) ln y-1 = 1/3 x3 + C (Exponentiate both sides) y-1 = e1/3 x3 + C (Using the laws of exponents from Algebra) y = 1 + (e1/3 x3) (eC) y-1 = +(e1/3 x3) (eC)

Part (c) eC = +2 2 = + (e0) (eC) 3 = 1 + (e1/3 03) (eC) y = 1 + (e1/3 x3) (eC)

Part (c) eC must be +2, because -2 doesn’t work for the given initial condition. eC = 2 y = 1 + (e1/3 x3) (2) y = 1 + (e1/3 x3) (eC) y = 1 + 2e1/3 x3