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Integrating differential equations that have ln

  1. Oct 16, 2015 #1
    Hey guys, I have a question concerning the rewriting of a differential equation solution.

    seperation of variables.png

    In the example above, they rewrite [y=(plus/minus)e^c*sqrt(x^2+4)] as [y=C*sqrt(x^2+4)]. I understand that the general solution we get as a result represents all the possible functions, but if we were to attempt to find a particular solution given an initial condition (a point on the graph of the equation), I would think that we would plug in the coordinate into not the general solution but the one above it with the (plus/minus). The problem is that in some of the problems that is not the case - you plug in the coordinate into the general solution to find your equation.

    The reason I would plug in the coordinate into the (plus/minus) equation is because that is the original equation contains above the x-axis values and below the x-axis values (when you graph the resulting equation). But if you plug a coordinate into the general solution, you only get the top portion of bottom portion of the graph depending on if y were less than or greater than 0 (because of the absolute value symbol). So plugging into the general solution kind of makes you lose a portion of the equation's graph.

    Hopefully you guys can help clear up this confusion for me. If you need any clarification, please ask.

  2. jcsd
  3. Oct 16, 2015 #2


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    One obtains the lower portion of the graph from the general solution (last line of your post) by choosing C<0.
  4. Oct 17, 2015 #3
    Does that mean we would be missing the bottom portion of the graph if we only choose C > 0. To see the whole equation, we'd have to choose C and -C. If we only chose C, shouldn't we mention that we are only seeing the y > 0 portion of the graph?

    I guess my confusion comes from problems like this, where they don't put the y in absolute value (I don't know why), which you are supposed to do when integrating a ln derivative. By not putting the y in derivatives, aren't they losing the bottom portion of the graph? And if they did put the y in absolute values, when solving for the particular solution given a coordinate, wouldn't you have to use the (plus/minus) equation to get the top and bottom?

    problem answer 1.png
  5. Oct 17, 2015 #4


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    You have to choose one branch or the other, or else y is not a function. A function of x must have a unique value for each value of x.
  6. Oct 17, 2015 #5
    Thank you! Succinct and insightful :)
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