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Reduction of Order

  1. Oct 24, 2014 #1

    QuantumCurt

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    1. The problem statement, all variables and given/known data

    Given - $$y_1(x)=sin(2x)$$, find a second linearly independent solution to $$y''+4y=0$$


    2. Relevant equations

    I'm using reduction of order to write a second solution as a multiple of the first solution. I think I've gotten to the right answer, my main question is in how I should write the general solution of the equation.


    3. The attempt at a solution

    Using the fact that ##y_1(x)=sin(2x)##, and that ##y_2(x)=vy_1(x)## I'm writing a second solution as $$y_2(x)=vsin(2x)$$
    Now taking the first and second derivatives I get $$y_2'(x)=v'sin(2x)+2vcos(2x)$$
    and $$y_2''(x)=v''sin(2x)+4v'cos(2x)-4vsin(2x)$$

    Now I substitute back into the original equation and get, $$v''sin(2x)+4v'cos(2x)=0$$

    Now to reduce order; ##z=v' and z'=v''##

    Then substituting - $$z'sin(2x)+4zcos(2x)=0$$

    Which leads to $$\frac{dz}{z}=-4cot(2x)dx$$

    Integrating this to logarithms and then exponentiating etc. leads me to - $$z=csc^2(2x)$$

    Since ##z=v'## I can substitute again
    $$v=\int csc^2(2x)dx$$
    $$v=-\frac{1}{2}cot(2x)$$

    Now since ##y_2(x)=vy_1(x)##, I can say that,
    $$y_2(x)=[-\frac{1}{2}cot(2x)][sin(2x)]$$
    which simplifies to
    $$y_2(x)=-\frac{1}{2}cos(2x)$$

    The formula for the general solution is ##y(x)=c_1y_1(x)+c_2y_2(x)##. Given this fact, my general solution is written as - $$y(x)=c_1sin(2x)-\frac{1}{2}c_2cos(2x)$$

    Now...my question -

    Since ##-\frac{1}{2}c_1## is really just an arbitrary constant, can I write my general solution in terms of just a constant rather than a fraction times a constant? This would give me the solution of -
    $$y(x)=c_1sin(2x)+c_2cos(2x)$$

    Is this essentially the same solution, or would it be better to write it in terms of the negative fractional constant?

    Any help would be much appreciated. :)
     
    Last edited: Oct 24, 2014
  2. jcsd
  3. Oct 24, 2014 #2

    vela

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    Just absorb the -1/2 into c2.
     
  4. Oct 24, 2014 #3

    QuantumCurt

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    Thank you. I figured that was the case, but I wanted to be sure.
     
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