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Integral of y dx?

  1. Dec 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve the differential equation: dy/dx = y+3


    2. Relevant equations



    3. The attempt at a solution
    When I try to integrate with respect to x, I get ∫dy/dxdx - ∫ydx=∫3dx→y-∫ydx=3x
    So what does one do about the integral of y with respect to x? There is nothing in the text about this type of thing.
     
  2. jcsd
  3. Dec 22, 2013 #2

    SammyS

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    Start with [itex]\displaystyle \frac{dy}{dx}=y+3[/itex] .

    Multiply both sides by [itex]\displaystyle \frac{1}{y+3}[/itex] .

    Then integrate (both sides) with respect to [itex]x[/itex] .
     
  4. Dec 22, 2013 #3

    D H

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    What text? What class?

    This is a simple problem that can easily be attacked by a number of approaches. Does your text talk about homogeneous and particular solutions? There is an easy-to-find particular solution of y'(x)=y+3. What is the form of the solutions to the homogeneous ODE y'(x)=y?
     
  5. Dec 22, 2013 #4
    Ron Larson
    Bruce Edwards
    Calculus 10E
     
  6. Dec 24, 2013 #5

    NascentOxygen

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    SammyS has mapped out a clear approach which you should closely follow. :approve:

    Start with ...

    Multiply both sides by ....
    and tidy up by performing any 'cancels'

    Multiply both sides by dx
    and tidy up by performing any 'cancels' (maybe you didn't know you can cancel dx/dx )

    Now, write an integral sign in front of the expressions on each side

    Look closely at what you have on each side

    Perform the easy integrations involved on each side :smile:

    Good luck with your study! http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
  7. Dec 24, 2013 #6

    HallsofIvy

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    To answer your question: "So what does one do about the integral of y with respect to x? "
    one does NOTHING. y is an unknown function of x so we can't integrate.

    We can, as SammyS said, write it as [itex]\frac{dy}{y+ 3}= dx[/itex] and integrate.
     
  8. Dec 24, 2013 #7
    I see you're talking about a well-known and widely-used single-variable calculus textbook here, OP. I presume it's for Calc II or a high school AP Calculus course?

    In any case, perhaps your teacher/professor did not cover it yet, but the best way (and the easiest way) to solve such things is something known as the method of separation of variables (forgive me if you already know this and I sound idiotic right now).

    Try to put all the same variables on one side:

    [itex]\frac{dy}{dx} = y + 3[/itex]

    [itex]\frac{dy}{y + 3} = dx[/itex]

    Let [itex]u = y + 3[/itex]. Then, [itex]\frac{du}{dy} = 1[/itex] and [itex]du = dy[/itex]. Make these substitutions:

    [itex]\frac{du}{u} = dx[/itex]

    Integrate: [itex]\int \frac{du}{u} = \int dx[/itex]

    [itex]ln\left|u\right| = x + c[/itex], where [itex]c[/itex] is a constant dependent on initial conditions. Now just plug back in to have an equation for [itex]y[/itex] in terms of [itex]x[/itex].

    [itex]ln\left|y + 3\right| = x + c[/itex]

    [itex]\left|y + 3\right| = e^{x + c} = e^{x}e^{c}[/itex]

    Solving gives: [itex]y + 3 = \pm e^{c}e^{x}[/itex]

    Since [itex]\pm e^{c}[/itex] is just another constant (a constant raised to a constant's power), we'll rename it as "[itex]k[/itex]". Then we can solve and arrive at our final answer:

    [itex]\boxed{y = -3 + ke^{x}}[/itex], where [itex]k[/itex] is a constant dependent on initial conditions for [itex]y[/itex]. Note: You do NOT lose out on solutions by ignoring the [itex]\pm[/itex]. This is accounted for in the fact that [itex]k[/itex] can be negative or positive.
     
  9. Dec 24, 2013 #8

    NascentOxygen

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    Antiderivative, why are you doing the student's homework? It is for him to work through, it's his homework. He asked how to proceed and has been given ample guidance. It is now up to him to apply that and to do the actual computations himself. That is the whole point of the homework forums.

    Presenting the OP with a fully worked solution is verboten.
     
  10. Dec 24, 2013 #9
    Thanks for the help guys. I didn't even think about doing what SammyS did, but I got it now and I've moved on. BTW, I'm self studying Calc 2.
     
  11. Dec 24, 2013 #10
    My fault, I wasn't aware of that in the PF guidelines. I'll refrain from posting fully-worked-out solutions in the future. Sorry about that guys.
     
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