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Differential Equations y" + y = tanx, solve this DE

  1. May 7, 2015 #1
    1. The problem statement, all variables and given/known data

    1. y" + y = tanx, solve this DE.

    2. dP/dt = P(1 - P) where P = (c1et / 1 + c1et), verify that P is a solution to this DE.

    3. Given the pair of functionsx and y, show they solve this system:
    dx/dt = x + 3y
    dy/dt = 5x+3y

    x = e-2t + 3e6t
    y = -e-2t + 5e6t


    3. The attempt at a solution

    For 1, we want to get x and y on the left and right hand sides of the equation so we can integrate i think.

    so we go

    dy/dx + y = tanx
    dy + ydy = tanxdx
    ∫dy + ∫ydy = ∫tanxdx (is this right?)

    For 2, the equation is not quite in standard form (dy/dx + P(x)y = f(x)), but it doesn't look like we can achieve this form. How do you proceed?

    For 3, i'm not even sure how to solve it. There aren't any examples in this format... :eek:
     
  2. jcsd
  3. May 7, 2015 #2

    Mark44

    Staff: Mentor

    No, this is wrong. Your "ydy" term on the left should be "ydx" since you multiplied both sides by dx. In any case this isn't the right approach. What you have is a first-order nonhomogeneous equation. Your general solution will consist of one part that is a solution to the homogenous equation and another part that is a particular solution to the nonhomogeneous equation. Look in your textbook to see if there are any similar worked examples.
    To proceed, just show that P as defined above is a solution to the equation. You are not asked to solve the equation.

    If you were, the equation is separable, meaning that it isn't too hard to get P and dP on one side and t and dt on the other. Again, though, you are NOT asked to solve the equation.
    You are GIVEN the solution, so there is nothing to solve. Just show that x(t) and y(t) satisfy the system of differential equations.

    My advice to you is to pay closer attention to what is being asked for in a problem. Whenever a problem asks you to verify that such-and-such is a solution of some differential equation, you don't have to solve the equation. All you need to do is to verify that substituting the solutions function(s) into the differential equation(s) results in a statement that is identically true; i.e., for all values of the independent variable.
     
  4. May 7, 2015 #3

    SammyS

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    Staff Emeritus
    Science Advisor
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    Gold Member

    Isn't that the second derivative of y in the given equation ?

    ##\displaystyle\ \frac{d^2y}{dx^2}+y=\tan(x) \ ##
     
  5. May 7, 2015 #4

    Mark44

    Staff: Mentor

    Good eye, Sammy. I missed the double prime (y'') in the equation.
     
  6. May 11, 2015 #5
    Okay so the first question is slightly different as well.

    The solution y = -(cos x) ln(sec x tan x) is given, all that is needed is to verify it is the explicit solution.

    To do this, could i derive y twice then add y" to y and show that the result is tanx?
     
  7. May 11, 2015 #6

    Mark44

    Staff: Mentor

    "Explicit solution" - what is that? If they have given you a solution, then that solution would be explicit, since it was "explained" to you, which is what "explicit" literally means.
    Yes, in a way. If you differentiate y twice (the process to get the first and second derivative - we don't say "derive" in this context, in English), and substitute for y'' and y, the equation y'' + y = tan(x) should be identically true. That's what it means for a particular function to be a solution of a differential equation.
     
  8. May 11, 2015 #7
    From my text, here is what they mean by explicit

    You should be familiar with the terms explicit
    and implicit functions from your study of calculus. A solution in which the dependent variable
    is expressed solely in terms of the independent variable and constants is said to be an
    explicit solution. For our purposes, let us think of an explicit solution as an explicit formula
    y = Φ(x) that we can manipulate, evaluate, and differentiate using the standard rules
     
  9. May 11, 2015 #8

    Mark44

    Staff: Mentor

    Gotcha - explicit function. On the other hand, a solution given as, say, x2 + y3 = 3 would represent a solution given implicitly.
     
  10. May 11, 2015 #9
    Okay so i solved the first 2, all that you gotta do is compute some derivatives, not too bad. The third is a bit trickier though. Do i solve them one at a time, or is there some trick i need to do?
     
  11. May 11, 2015 #10

    Mark44

    Staff: Mentor

    For #3, there's nothing to solve -- they have given you the solutions!
    For the last two equations, find, respectively, dx/dt and dy/dt. Show that dx/dt is identically equal to x + 3y, and that dy/dt is identically equal to 5x + 3y.
     
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