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Math Help - A Linear Equation; IVP; Piecewise Defined Funciton

  1. #1
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    A Linear Equation; IVP; Piecewise Defined Funciton

    \frac{dy}{dx}+2xy=f(x) such that y(0)=2

    f(x)=\left\{\begin{array}{cc} x, &\mbox { if } 0\leq x< 1\\ 0,&\mbox { if } x\geq 1\end{array}\right

    The integrating factor is e^{\int 2xdx}=e^{x^2}, so:

    \int\frac{d}{dx}\left[e^{x^2}y\right]dx=\int xe^{x^2}dx

    e^{x^2}y=\frac{1}{2}e^{x^2}+C_1

    The point (0,2) gives:

    C_1=\frac{3}{2}

    y=\frac{1}{2}+\frac{3}{2e^{x^2}}

    For x\geq 1 the equation is separable.

    \frac{dy}{dx}=-2xy

    \int \frac{1}{y}\frac{dy}{dx}dx=-\int 2x dx=-x^2+C_2

    ln\mid y\mid =-x^2+C_3

    \mid y \mid = C_4e^{-x^2}

    y=\pm C_4e^{-x^2}

    Problem: The solution in the book gives \left(\frac{1}{2}e+\frac{3}{2}\right)e^{-x^2} for x \geq 1 and I don't see how to find C_4 because the point  (0,2) isn't on the interval [1,\infty)
    Last edited by adkinsjr; January 23rd 2011 at 11:45 AM.
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  2. #2
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    Your solution looks fine, I don't see how they could have done that either.

    On a side note, you could also have used the Integrating Factor method with \displaystyle x \geq 1 (and it would probably have been easier, since you already had it).
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    Quote Originally Posted by Prove It View Post
    Your solution looks fine, I don't see how they could have done that either.
    Yeah, the family of functions y=Ce^{-x^2} is a solution for any value of C, and basically they have C=\frac{1}{2}e+\frac{3}{2} which is a constant. The problem is that you can't get to this value...

    It turns out that this constant makes the piecewise defined solution continous at x=1 if you change the original intervals to 0\leq x \leq 1 and x>0. The you can say:

    lim_{x->1}Ce^{-x^2}=y(1)=\frac{1}{2}+\frac{3}{2e^{(1)^2}}=\frac{1  }{2}+\frac{3}{2e}

    lim_{x->1}Ce^{-x^2}=\frac{C}{e}

    \frac{C}{e}=\frac{1}{2}+\frac{3}{2e}

    C=\frac{1}{2}e+\frac{3}{2}

    However, this isn't correct because y(1)\neq \frac{1}{2}+\frac{3}{2e} on the original interval. But I'm pretty sure I see their mistake now.
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