Separation of Variables

Section: 
11.4
Date: 
Wednesday, October 27, 2010 - 14:00 - 15:00
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fall2010math2300_11-4-separation-variables-notes.pdf97.19 KB

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Separation of Variables – October 27, 2010

Definition: We call a differential equation separable if it can be written in the form

\frac{dy}{dx}=g(x)f(y)

for f a function of y and g a function of x.

The Formal Approach to Solving Separable Differntial Equations

If f(y)\neq 0 then we can write

\frac{1}{f(y)}\frac{dy}{dx}=g(x).

Integrating both sides with respect to the variable x gives

\displaystyle\int\frac{1}{f(y)}\frac{dy}{dx}\, dx \displaystyle=\int g(x)\, dx
\displaystyle\int\frac{1}{f(y)}\, dy \displaystyle=\int g(x)\, dx

What will usually happen is that this introduces a natural log in terms of y. Exponentiating and solving for y then provides a solution in terms of x.

Examples:

  1. Using separation of variables, show that solutions to

    \frac{dy}{dx}=-\frac{x}{y}

    are circles centered at the origin.

    Formal Solution:

    \displaystyle\frac{dy}{dx} \displaystyle=-\frac{x}{y}
    \displaystyle y\frac{dy}{dx} \displaystyle=-x
    \displaystyle\int y\frac{dy}{dx}\, dx \displaystyle=-\int x\, dx
    \displaystyle\int y\, dx \displaystyle=-\int x\, dx
    \displaystyle\frac{1}{2}y^{2}+C_{1} \displaystyle=-\frac{1}{2}x^{2}+C_{2}
    \displaystyle x^{2}+y^{2} \displaystyle=C

    Less Formal Solution:

    \displaystyle\frac{dy}{dx} \displaystyle=-\frac{x}{y}
    \displaystyle y\, dy \displaystyle=-x\, dx
    \displaystyle\int y\, dy \displaystyle=-\int x\, dx
    \displaystyle y^{2} \displaystyle=-x^{2}+C
    \displaystyle x^{2}+y^{2} \displaystyle=C
  2. Find the general form of solutions to

    \frac{dy}{dx}=ky

    where k is some constant.

    Solution:

    \displaystyle\frac{dy}{dx} \displaystyle=ky
    \displaystyle\frac{1}{y}\, dy \displaystyle=k\, dx
    \displaystyle\int\frac{1}{y}\, dy \displaystyle=\int k\, dx
    \displaystyle\ln|y| \displaystyle=kx+C
    \displaystyle|y| \displaystyle=e^{{kx+C}}
    \displaystyle|y| \displaystyle=e^{{kx}}e^{C}
    \displaystyle|y| \displaystyle=Ae^{{kx}}
    \displaystyle y \displaystyle=\pm Ae^{{kx}}
    \displaystyle y \displaystyle=Be^{{kx}}

    Here, A satisfies A=e^{C} and B is simply used to rewrite A as either positive or negative.

  3. For k>0 find and graph solutions of

    \frac{dH}{dt}=-k(H-15)

    The graph of solutions for k=1 is as follows.

    Of course, as H becomes closer to 15, we see that the slope becomes closer to zero. A general solution is gound by dividing by H-15.

    \displaystyle\frac{dH}{dt} \displaystyle=-k(H-15)
    \displaystyle\frac{1}{H-15}\frac{dH}{dt} \displaystyle=-k
    \displaystyle\int\frac{1}{H-15}\, dH \displaystyle=-\int kdt
    \displaystyle\ln|H-15| \displaystyle=-kt+C
    \displaystyle|H-15| \displaystyle=e^{{-kt+C}}
    \displaystyle H-15 \displaystyle=\pm e^{{-kt}}e^{C}
    \displaystyle H \displaystyle=Be^{{-kt}}+15

    where B=\pm e^{{C}}.

© 2011 Jason B. Hill. All Rights Reserved.