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    theres this differential equation that i havent seen its kind of type before. its:

    25y" + 4y = 7cosx

    i had a go at it, but i messed it up a bit (checked my answer on wolfram alpha and it was different). i am unsure how to deal with just a y" and y (no y' in there) and unsure how to deal with that cosx (i ended up with a sin instead of a cos in the final answer, which is what it should be). can anyone help me? thanks
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    theres this differential equation that i havent seen its kind of type before. its:

    25y" + 4y = 7cosx

    i had a go at it, but i messed it up a bit (checked my answer on wolfram alpha and it was different). i am unsure how to deal with just a y" and y (no y' in there) and unsure how to deal with that cosx (i ended up with a sin instead of a cos in the final answer, which is what it should be). can anyone help me? thanks
    Start with the normal method for solving 2nd order ODE's, i.e. write out the equation as

     25\frac{d^2y}{dx^2} + 4y =  0

    The auxillary equation is then

     25m^2 + 4 = 0

    Solve this to get the first part of the solution (the complimentary function), which is of the form

     y = Ae^{m_1x} + Be^{m_2x}

    where m1 and m2 are the roots of the quadratic from above.

    Now you need to consider the RHS. To solve this, you need to find the 'particular integral'. This basically means you start with a trial solution that makes sense, differentiate it twice and work out some values of constants.

    for cos x, you need the trial solution

     y = a cos(kx) + b sin(ky)

    differentiate this twice, put in into the LHS and see what values of a,b and k you need to get the result as 7cosx...

    The full solution is the sum of the particular integral and the complimentary function...

    If you're wanting more guidance, check this pdf out

    http://www.ul.ie/~mlc/support/Loughb...hap19/19_6.pdf
 
 
 
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Updated: February 4, 2010
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