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Thread: Singular Points

  1. #1
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    Singular Points

    Need help with solving this question i have no idea how to even start so any help would be appreciated thanks

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  2. #2
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    Why not just start by solving the DE:

    $\displaystyle y'\pm\frac{2}{x}y=3$

    via integrating factor and obtain the two solutions. What can you say about either solution at x=0?
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  3. #3
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    Quote Originally Posted by shawsend View Post
    Why not just start by solving the DE:

    $\displaystyle y'\pm\frac{2}{x}y=3$

    via integrating factor and obtain the two solutions. What can you say about either solution at x=0?
    y do i need to solve that DE ? thats just the example in question i think i need to solve the DE in part a) which i dont knw how to
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  4. #4
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    Ok, then here's how I'd do (A):

    $\displaystyle xy'+2y=3x,\quad y(0)=y_0$ and dividing by $\displaystyle x$:

    $\displaystyle y'+2/x y=3$ for which the integrating factor is $\displaystyle x^2$ so we have:

    $\displaystyle x^2(y'+2/x y)=3x^2$ which the left side is the differential of $\displaystyle x^2y$ so:

    $\displaystyle \int d(x^2y)=\int 3x^2$ or:

    $\displaystyle y(x)=x+\frac{c}{x^2}$ and if $\displaystyle y_0=0$ then I can let $\displaystyle c=0$ and obtain the solution $\displaystyle y=x$. However, if $\displaystyle y_0\neq 0$ say $\displaystyle 3$ for example, then I'd obtain $\displaystyle 3=0+\frac{c}{0}$ which has no solution.

    Ok, so I'm a little uneasy about dividing through by $\displaystyle x$ in the first step since this is valid if $\displaystyle x\ne 0$ then obtain the solution and solve it for $\displaystyle x=0$.
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