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B Second derivative differential equations in terms of y?

  1. Jul 21, 2017 #1
    Firstly I know how to do this with first derivatives in differential equations - for example say we had ##\frac{dy}{dx}=4y^2-y##, and we're also told that ##y=1## when ##x=0##.

    ##\frac{dy}{dx}=4y^2-y##
    ##\frac{dx}{dy}=\frac{1}{4y^2-y}=\frac{1}{y\left(4y-1\right)}=\frac{4}{4y-1}-\frac{1}{y}##
    ##\int _{ }^{ }\frac{dx}{dy}dy=\int _{ }^{ }\frac{4}{4y-1}dy-\int _{ }^{ }\frac{1}{y}dy##
    ##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)+c##

    Substituting in ##y=1## and ##x=0##, we can find ##c##:
    ##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)+c##
    ##0=\ln \left(3\right)+c,\ c=-\ln \left(3\right)##
    ##x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)-\ln \left(3\right)##
    ##x=\ln \left(\frac{\left|4y-1\right|}{3\left|y\right|}\right)##
    ##3e^x=\left|4-\frac{1}{y}\right|##
    ##4-\frac{1}{y}=-3e^x,4-\frac{1}{y}=3e^x##

    Because we know that ##y## has to equal postive ##1## when ##x## equals ##0##;
    ##4-1=-3e^0,\ 4-1=3e^0##
    ##3=-3,\ 3=3##
    As the first statement is false, the second one must be true.
    ##∴4-\frac{1}{y}=3e^x##
    ##\frac{1}{y}=4-3e^x##
    ##y=\frac{1}{4-3e^x}##.

    So as you can see it's relatively simple to solve a differential equation of the form ##\frac{dy}{dx}=f\left(y\right)##, but I'm wondering if it would be possible to solve one in the form of ##\frac{d^2y}{dx^2}=f\left(y\right)##. I've tried looking it up online but I haven't been able to find anything useful.
     
  2. jcsd
  3. Jul 21, 2017 #2

    fresh_42

    Staff: Mentor

    One can introduce ##z:=\frac{dy}{dx}=f\left(x,y\right)## and get a system of two differential equations of order one. The other equation is ##\frac{dz}{dx}=f\left(x,y,z\right)##. This process can be iterated to higher orders. You can look for ODE (ordinary differential equations) to solve such systems.
     
  4. Jul 21, 2017 #3

    Mark44

    Staff: Mentor

    It's not necessarily that simple -- you have to be able to integrate ##\int \frac{dy}{f(y)}##.
     
  5. Jul 22, 2017 #4

    pasmith

    User Avatar
    Homework Helper

    Multiply both sides by [itex]y'[/itex] and integrate once with respect to [itex]x[/itex] to obtain [tex]
    \left(\frac{dy}{dx}\right)^2 = 2\int f(y)\,dy.[/tex] Now just determine the constant of integration from the initial conditions, and work out whether you need the positive root or the negative root, and the problem is reduced to a first-order separable ODE.
     
  6. Jul 24, 2017 #5
    Thank you for you're help but I still find myself a little confused about how you arrived at [itex]\left(\frac{dy}{dx}\right)^2 = 2\int f(y)\,dy.[/itex]

    I tried multiplying both sides by ##\frac{dy}{dx}## and then integrating both sides with respect to ##x## and this is what I got;
    [tex]\frac{d^2y}{dx^2}=f\left(y\right)[/tex]
    [tex]\frac{dy}{dx}\cdot \frac{d}{dx}\cdot \frac{dy}{dx}=f\left(y\right)\cdot \frac{dy}{dx}[/tex]
    [tex]\left(\frac{dy}{dx}\right)^2\cdot \frac{d}{dx}=f\left(y\right)\cdot \frac{dy}{dx}[/tex]
    [tex]\int _{ }^{ }\left(\frac{dy}{dx}\right)^2\cdot \frac{d}{dx}\cdot dx=\int _{ }^{ }f\left(y\right)\cdot \frac{dy}{dx}\cdot dx[/tex]
    [tex]\int _{ }^{ }\left(\frac{dy}{dx}\right)^2\cdot d=\int _{ }^{ }f\left(y\right)\cdot dy[/tex]

    I am unsure of what to do on the left hand side - how am I supposed to integrate without a variable for which to integrate with respect to. I'm also not sure why you have a two out the front of the integral on the right hand side, but I'm guessing that the answer to my first question will probably explain that as well.
     
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