Recent content by ballajr

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    Help with 2 Problems in Differential Equations

    There was a typo in #2: 2) Prove or disprove that there are two constants A and B such that: t2D2 - tD - 8I = (tD + AI)(tD + BI)
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    Help with 2 Problems in Differential Equations

    Homework Statement 1) Find the General Solution to: (D3 - D2 + D - I)[y] = t5 + 1 2) Prove or disprove that there are two constants A and B such that: t2D - tD - 8I = (tD + AI)(tD + BI) Homework Equations The Attempt at a Solution 1) I can't figure out how to attempt this...
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    Differential Equations Homework

    My teacher gave the following hint: The theorem states the existence of such two functions, and we don't know what they are. What we know is that there ARE two such functions. That's what the theorem is about, namely, the existence of such two functions. Now, not knowing what those are, we pull...
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    Differential Equations Homework

    Thank you so much! That seems so easy, but I don't think I would have ever known that method, but now it makes so much sense. Thank you! Any input on the other 2 problems?
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    Differential Equations Homework

    For some reason, it didn't subscript the c's and y's and z's. All of those are supposed to be written as subscripts, not superscripts. The only superscripts are just the D's and they are squared.
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    Differential Equations Homework

    Homework Statement 1. Let D and I be the differential operator and the identity operator, respectively. Find two real-valued functions f(t) and g(t) such that: D^{2} + I = (D + f(t)I)(D + g(t)I): 2. Use this theorem to prove the corollary given below. Theorem: There are two nonzero...
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    Differential Equations: Find all solutions of these equations

    Oh okay. There's going to be more than one solution for all three of those right? My teacher likes to trick us like that. The second one i haven't solved yet. The fourth one is a bit confusing. I need some help with that.
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    Differential Equations: Find all solutions of these equations

    Homework Statement 1) y' + ycos(t) = 0 2) y' - 2yt = 0 3) y'tan(t) + ysec2(t) = 0 4) Let p(x) be an arbitrary polynomial function. Note that the following is a polynomial solution to y" = p(x) + y: y = -p - p" - p(4) - p(6) - ... - p(n) where n is the smallest positive even...