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Linear Differential Equations behaviour as t->∞

  1. Sep 13, 2009 #1
    1. Consider IVT problem: y'-1.5y=3t+2et, y(0)= y0 Find y0 value that seperates solutions that grow positively as t->∞ from those that grow negatively. How does the soln that corresponds to this critical value or y0 behave as t->∞?

    2. Basically i'm drawing a direction field first, but how am i supposed to see the graph of the functions y(not)=-3, -2 ,-1 ,0 etc..... to see the behaviour as t->∞ if the function is so complicated. i've solved the differential equation it's: -(24/37)cos(3t)-(4/37)sin(3t)+(y0+24/37)et/2 if we could use graphing calculators i could just plug it in and see the behaviour for values of y(not), but there's no calculators allowed on the test

    3. I solved the differential equations but i have to idea how to graph this to view the behaviours. Is this what i'm supposed to do? Or is there an easier way to see this that i don't know about i'm stuck i have the solution but i don't know what y0 has to do with the behaviour of the DE. And how i determine what y0 values do the DE.
  2. jcsd
  3. Sep 13, 2009 #2
    help? please im begging you
  4. Sep 13, 2009 #3
    The solution is all you need to analyze the behavior. You have the sum of two periodic bounded functions added to a non-periodic unbounded function. The latter is the one that must determine whether the function diverges positively or negatively without bound. Consider its coefficient.
  5. Sep 13, 2009 #4

    sorry can you be more presice and provide a procedure on how to approach this?
  6. Sep 13, 2009 #5
    Which of the three transcendental functions in your solution has the potential to increase or decrease without bound? Which of the three cannot do this? Which part of the function modifies whether the one(s) that can be unbounded tend to positive infinity or tend to negative infinity?
  7. Sep 13, 2009 #6
    do i need to make a direction field to be able to tell? and plug in arbitrary values of y(not)?
  8. Sep 13, 2009 #7
    C>0 causes divergence and positive growth C=0 causes finite number as t->∞ and C<0 causes divergent negative growth?
    therefore i have to make y(not) so that C = 0 for it to have a finite value? is this correct?
    what was the purpose of the direction field to this problem?
  9. Sep 13, 2009 #8
    There is no need. This is just algebra and limits. Cosine and sine are bounded functions, and so is their sum, so they have no effect on whether the solution goes to positive or negative infinity. e^t, on the other hand, increases without bound as t increases without bound. What is the coefficient of this function and how does it affect its behavior?
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