Thread: [Simple] Differential Equations

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  1. #1 [Simple] Differential Equations 
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    1) Give a Differential Equation with the solution y=Ae^4x + Be^5x (Solved)
    2) Give a Differential Equation with the solution y=Ax^5 + Bx^5 (Solved)
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    Ray
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    instead of finding a solution of d^2y/dx^2 + b dy/dx + cy = 0 which is the generic 2nd order ode problem, you're given the general solution instead.

    substitute y=e^(kx), finding values of k so that y is a solution of a d^2y/dx^2 + b dy/dx + cy = 0, as the auxillary equation of this is ak^(2) + bk + c = 0 where y=e^(kx). it's literally like solving a quadratic


    so for 1)

    the general solution is: y_(cf)(x) = Ae^(4x) + Be^(5x), the two solutions are y=e^(4x) and y=^(5x) and the auxillary equation can be factorised as (k-4)(k-5) = 0 and so the required values of k are 4 and 5. now finding the auxilary equation from these two 'roots': k^2 - 9k + 20 = 0 and hence d^2y/dx^2 - 9* dy/dx + 20y = 0

    another method is by differentiating the general solution but i prefer this method but it's all essentially the same shit

    try do the same for 2), it's a little easier than the first
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    Quote Originally Posted by Ray View Post
    instead of finding a solution of d^2y/dx^2 + b dy/dx + cy = 0 which is the generic 2nd order ode problem, you're given the general solution instead.

    substitute y=e^(kx), finding values of k so that y is a solution of a d^2y/dx^2 + b dy/dx + cy = 0, as the auxillary equation of this is ak^(2) + bk + c = 0 where y=e^(kx). it's literally like solving a quadratic


    so for 1)

    the general solution is: y_(cf)(x) = Ae^(4x) + Be^(5x), the two solutions are y=e^(4x) and y=^(5x) and the auxillary equation can be factorised as (k-4)(k-5) = 0 and so the required values of k are 4 and 5. now finding the auxilary equation from these two 'roots': k^2 - 9k + 20 = 0 and hence d^2y/dx^2 - 9* dy/dx + 20y = 0

    another method is by differentiating the general solution but i prefer this method but it's all essentially the same shit

    try do the same for 2), it's a little easier than the first
    I realized it for the first one, so I wrote that (y-4)(y-5) is one solution since it gives e^4x and e^5x. But I'm not sure if it works the same way for the second question. I usually read the books that's given but I'm out at parents house without a stable wifi atm so hard to do it.
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    Quote Originally Posted by Ray View Post
    instead of finding a solution of d^2y/dx^2 + b dy/dx + cy = 0 which is the generic 2nd order ode problem, you're given the general solution instead.

    substitute y=e^(kx), finding values of k so that y is a solution of a d^2y/dx^2 + b dy/dx + cy = 0, as the auxillary equation of this is ak^(2) + bk + c = 0 where y=e^(kx). it's literally like solving a quadratic


    so for 1)

    the general solution is: y_(cf)(x) = Ae^(4x) + Be^(5x), the two solutions are y=e^(4x) and y=^(5x) and the auxillary equation can be factorised as (k-4)(k-5) = 0 and so the required values of k are 4 and 5. now finding the auxilary equation from these two 'roots': k^2 - 9k + 20 = 0 and hence d^2y/dx^2 - 9* dy/dx + 20y = 0

    another method is by differentiating the general solution but i prefer this method but it's all essentially the same shit

    try do the same for 2), it's a little easier than the first
    If you could assist me with #2 I would be very grateful
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