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Homework Help: First order linear ode, initial condition problem

  1. Apr 26, 2012 #1

    i'm having trouble with this question,

    x y' - y = x2cosx

    the solution is

    y= xc + xsinx

    and we are asked to solve the equation in the following two cases,

    1, y(0)=0

    and 2, y(0) = 1

    but applying these conditions to the general solution gives no information,

    in both cases its 0 = 0, so doing this you cant find the constant.

    is there another way to solve the equation for these conditions?

    even mathematica can't solve it, but it was a question in a test and i was marked wrong for saying neither cases can be solved (as in the constant cant be found)

  2. jcsd
  3. Apr 26, 2012 #2


    User Avatar
    Science Advisor

    Has it occured to you that the whole point of this exercise is that there is NO unique solution?

    I suspect that you either have just had or will soon have the "Fundamental existance and uniqueness theorem" for first order equations:

    If f(x,y) is continuous and "Lipschitz in y" (a condition that is partway between "continuous" and "differentiable"- "differentiable" is sufficient and many texts just use "differentiable in y") in some region around the point [itex](x_0, y_0)[/itex] then there exist a unique function, y(x), satisfying dy/dx= f(x,y) and [itex]y(x_0)= y_0[/itex].

    In this example, solving for y' gives [itex]y'= (x^2 cos(x)- y)/x[/itex] so that [itex]f(x,y)= (x^2 cos(x)- y)/x[/itex] is not continuous in any neighborhood containing x= 0. Thus, the "existance and uniqueness theorem" does not apply. Your first example is one in which a solution exists but is not unique. Your second is one in which a solution does not exist.
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