Slope Fields

Section: 
11.2
Date: 
Monday, October 25, 2010 - 14:00 - Tuesday, October 26, 2010 - 15:00
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fall2010math2300_11-2-diff-eq-slope-fields.pdf20.47 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Slope Fields – October 25–26, 2010

Slope Field Examples

  1. Find an infinite collection of solutions to the differential equation

    y^{{\prime}}=-\frac{x}{y}.

    Solution: It's nearly impossible to ask for even a single solution to this differential equation without knowing more about it. You can graph out the slope field yourself (perhaps not quite as neatly as a computer or calculator, but you can do it). You get a slope field like this.

    Now it appears a bit more obvious that we're dealing with circles centered at the origin. Let's see if that is actually correct by taking a generic circle, centered at the origin, and determining if it satisfies the differential equation. In what follows, we use implicit differentiation to view y as a function of x.

    \displaystyle x^{2}+y^{2} \displaystyle=r^{2}
    \displaystyle\frac{d}{dx}\left[x^{2}+y^{2}\right] \displaystyle=\frac{d}{dx}r^{2}
    \displaystyle 2x+2y\frac{dy}{dx} \displaystyle=0
    \displaystyle 2y\frac{dy}{dx} \displaystyle=-2x
    \displaystyle\frac{dy}{dx} \displaystyle=-\frac{2x}{2y}
    \displaystyle\frac{dy}{dx} \displaystyle=-\frac{x}{y}
    \displaystyle y^{{\prime}} \displaystyle=-\frac{x}{y}.

    So, each circle centered at the origin satisfies the given differential equation.

  2. The differential equation y^{{\prime}}=y=f(x) is graphed in the following slope field along with a particular solution (the solid line). An initial condition representing the given solution is (x,y)=(2,1). Find the equation representing the given solution.

    Solution: We know that y^{{\prime}}=y has the solution y=e^{x}. But, here's where knowing some algebra can help us. Consider changing y=e^{x} to

    y=e^{{x+C_{1}}}

    where C_{1} is some constant. Then we also have y^{{\prime}}=y, and so we have made this solution a bit more general. But, now, consider

    y=C_{2}e^{{x+C_{1}}}

    for another constant C_{2}. This also satisfies y^{{\prime}}=y. At first, if you properly understand graph transformations, this seems to contradict the notion of a slope field. (We can shift left and right using the constant C_{1}, and we can ALSO expand/shrink vertically using the constant C_{2}.) It seems like these two constants could conflict and allow us to produce a function not fitting in our slope field. Let's assume C_{2}>0 for now. Then you can find some C_{3} such that

    C_{2}=e^{{C_{3}}}\quad\text{and so}\quad y=C_{2}e^{{x+C_{1}}}=e^{{C_{3}}}e^{{x+C_{1}}}=e^{{x+(C_{1}+C_{3})}}.

    So, this is really equivalent to just selecting the C_{1} appropriately. Likewise, if C_{2}<0 then we write C_{2}=-|C_{2}| and use the same reasoning to write

    y=C_{2}e^{{x+C_{1}}}=-|C_{2}|e^{{x+C_{1}}}=-e^{{C_{3}}}e^{{x+C_{1}}}=-e^{{x+(C_{1}+C_{3})}}.

    Thus, if we have some nonzero y-coordinate in an initial condition, we need to only consider a single constant. Since we know that our initial condition has a positive y-coordinate, (x,y)=(2,1) we must then find a constant C_{1} such that

    1=e^{{2+C_{1}}}.

    Taking the natural log of both sides tells us that 0=2+C_{1} and so C_{1}=-2. Thus, the equation satisfying this differential equation plus our initial condition is

    y=e^{{x-2}}.
  3. Use the following slope field for the differential equation y^{{\prime}}=\frac{1}{x} to approximate the value of \ln(2).

    Solution: Since y=\ln(x) is a solution to the differential equation y^{{\prime}}=\frac{1}{x}, we may use the initial condition (x,y)=(1,0) from \ln(1)=0 to follow a path through the slope field until we are above x=2.

    It appears as though the curve representing \ln(x) will reach a vertical height of around 0.65 to 0.7 when x=2. As it turns out, the precise value of about \ln(2)\approx 0.69315.

© 2011 Jason B. Hill. All Rights Reserved.