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Questions on basic differential equations

  1. Aug 26, 2009 #1
    Well this is my first diff eqs homework and im totally lost, i have no idea what to do here are teh questions that i have...

    1. The problem statement, all variables and given/known data
    1)An object released from a height h meters above the gorund with a veritcal velocity of Vo m/s htis teh ground after To seconds. Neglecting fricitonal forces set up and solve the inital value problem governing the motion and use your solution to show that....
    Vo=(2h-gto^2)/(2To)

    2) Determine a solution to the differential equation
    (1-x^2)y'' - xy' +4y = 0
    of the form Y(x) = a0 + a1x + a2x^2 satisfying the normalization condition y(1) = 1

    3) Determine the differntial equaiton giving the slope of teh tangent line at teh point (x,y) for the given curve
    x^2 + y^2 = 2cx


    3. The attempt at a solution
    1)on that one i got till the point where i have
    1/2gt^2 + C1t+ c2 = y(t)


    2) On this one i have no idea how to start maybe help on where to start will be enough....
    3) on this one i also have no idea where to start...

    im still working on other ones ill see if im able to work them out


    any help is appreciated
    thanks
     
  2. jcsd
  3. Aug 26, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Okay, you know that y(0)= (1/2)g(0^2)+ C1(0)+ C2= h (An object released from a height h meters above the ground) and that y'(0)= g(0)+ C1= V0 (with a vertical velocity of Vo) so you can determine C1 and C2 from those.

    Do it! You are asked for a solution of the form [itex]y= a_0+ a_1x+ a_2x^2[/itex] so [itex]y'= a_1+ 2a_2x[/itex] and [itex]y"= 2a_2[/itex]. Put those into the differential equation and solve for [itex]a_0[/itex], [itex]a_1[/itex], and [itex]a_2[/itex] by, for example, choosing three values for x to get three equations.

    You have a "family" of curves given by [itex]x^2+ y^2= 2cx[/itex] and you want an equation involving y' and/or y" but not c. Now, you could just differentiate with respect to x twice and that would eliminate c: [itex]2x+ 2y y'= 2c[/itex] and then [itex]2+ 2(y')^2+ 2y y"= 0[/itex]. But since there is only one "c" you should be able to do this with just one differentiation. Differentiating once gives [itex]2x+ 2y y'= 2c[/itex] so [itex]c= x+ y y'[/itex]. Replace c in [itex]x^2+ y^2= 2cx[/itex] with that!

     
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