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Math Help - First-order linear ode

  1. #16
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    If you change the K in the denominator to +K, then the answer is y = -2...
    No, because the k in the numerator and the k in the denominator are the same k. They would both have to change.

    I agree with you up to here:

    \displaystyle -2\,\frac{e^{-\frac{1}{2}x^2} + K}{e^{-\frac{1}{2}x^2} - K}.

    Replacing K with -K yields

    \displaystyle -2\,\frac{e^{-\frac{1}{2}x^2} - K}{e^{-\frac{1}{2}x^2} + K}.

    Multiplying the negative sign in yields

    \displaystyle 2\,\frac{K-e^{-\frac{1}{2}x^2}}{K+e^{-\frac{1}{2}x^2}}.

    Now, finally, one more trick: replacing K with 1/K, will, I think, finish off the problem.
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  2. #17
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    Thanks. Where does the other solution y = -2 come from?
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  3. #18
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    Well, you could take the previous step, right before replacing K with -K, and then let K=0 there. Equivalently, let K\to\infty in the final solution.
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