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Math Help - First order differential equations

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    First order differential equations

    Consider the first order differential equation:

    d^2y/dx^2 = -((lambda + 1)^2)y

    Write down the general solution.
    Find the particular solution satisfying dy/dx = 0, y = 1 when x = 0.

    I didnt think this was first order, but maybe if you integrate first. I couldnt get an answer that looked right.
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    Quote Originally Posted by BogStandard View Post
    Consider the first order differential equation:

    d^2y/dx^2 = -((lambda + 1)^2)y

    Write down the general solution.
    Find the particular solution satisfying dy/dx = 0, y = 1 when x = 0.

    I didnt think this was first order, but maybe if you integrate first. I couldnt get an answer that looked right.
    this is a second order differential equation with constant coefficients. we proceed by assuming a solution of y = e^{rx} and then forming the characteristic equation. which in this case is:

    r^2 + ( \lambda + 1)^2 = 0 (do you see why?)

    now solve for the two roots of this equation, r_1 and r_2. your general solution is of the form y(x) = Ae^{r_1t} + Be^{r_2t}

    for the particular solution (we want to know what A and B are): you know that y'(0) = 0 and y(0) = 1, we can use these equations to solve for the (constant) unknowns A and B to obtain our particular solution

    can you continue?
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    I would normally be able to solve this for a first order, but for second order im not sure.

    This is an exam question, and the paper says its first order, which has completely thrown me off.

    So im afraid I dont understand your method!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by BogStandard View Post
    I would normally be able to solve this for a first order, but for second order im not sure.

    This is an exam question, and the paper says its first order, which has completely thrown me off.

    So im afraid I dont understand your method!
    there is something amiss here. \frac {d^2y}{dx^2} means it is second order, because that is the second derivative. please type the question exactly as stated
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    Quote Originally Posted by Jhevon View Post
    there is something amiss here. \frac {d^2y}{dx^2} means it is second order, because that is the second derivative. please type the question exactly as stated
    Yeah I know what you're thinking, I thought it too, but that's literally what the question says.
    It's most probably a typo or something, or some really unobvious trick question. Pretty bad for an exam paper.

    So could you explain to me the same method but for second order differential equations?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by BogStandard View Post
    Yeah I know what you're thinking, I thought it too, but that's literally what the question says.
    It's most probably a typo or something, or some really unobvious trick question. Pretty bad for an exam paper.

    So could you explain to me the same method but for second order differential equations?
    my explanation was for second order. first-orders are different. to be first order, you would have to have dy/dx not d^2y/dx^2. if that was what they meant though, they would not have given us y'(0) and y(0). only y(k) is required for first order. so something is diffinately off here. i suspect the only typo is "first order", maybe it should really be a second order DE, in which case you follow the method i gave you.

    for your general edification:

    if you have a first order DE of the form y' = ky, where k is a constant, you proceed as follows


    divide through by y:
    \frac {y'}y = k

    integrate both sides:
    \ln y = kx + C

    solve for y:
    y = e^{kx + C}

    \Rightarrow y = Ae^{kx} .... A is a constant, you can find it if you are given intial data, say y(0) = ...
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