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Math Help - Homogeneous Differential Equations.

  1. #16
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Here is one way to test and see if its homogeneous:

    We (in a sense) see that N(x,y)=\frac{1-3y}{y} and M(x,y)=\frac{2-3x}{x} after getting our equation in the form of M(x,y)\,dx+N(x,y)\,dy=0
    I see where this is bit is from, so great!

    It is homogeneous if we can observe the following:

    M(tx,ty)=t^{\alpha}M(x,y) AND N(tx,ty)=t^{\alpha}N(x,y).
    For this part, what are tx and ty? It's just that the DE in question is using y's and x's.

    Is t a dummy variable or something?
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  2. #17
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I see where this is bit is from, so great!



    For this part, what are tx and ty? It's just that the DE in question is using y's and x's.

    Is t a dummy variable or something?
    Sorta. The variable t is like a dummy variable, but its something that can help us show [analytically] if the functions M(x,y) and N(x,y) [from the DE] in question are homogeneous.

    If we introduce this dummy variable t and show that if we have t^{\alpha}M(x,y) after manipulating M(tx,ty) we're almost done showing that its homogeneous. The last part is to show that we get t^{\alpha}N(x,y) after manipulating N(tx,ty) AND that the we have the same \alpha exponents. If the exponents are different, OR we can't show that M(tx,ty)=t^{\alpha}M(x,y) or N(tx,ty)=t^{\alpha}N(x,y), then its not homogeneous.

    Does this help?

    --Chris
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  3. #18
    Super Member Showcase_22's Avatar
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    I think i'm understanding it, but there's just something else that's puzzling.

    We have:

    -\frac{2-3x}{x}dx+\frac{3y-1}{y}dy=0

    and we're comparing this with:

    M(x,y)dx+N(x,y)dy=0

    Comparing coefficients we get:

    M(x,y)=-\frac{(2-3x)}{x} and N(x,y)=\frac{3y-1}{y}

    Is this equal to:

    M(x)=-\frac{(2-3x)}{x} and N(y)=\frac{3y-1}{y}?

    For the next part:

    <br />
M(tx,ty)=t^{\alpha}M(x,y)<br />

    \frac{-(2-3tx)}{tx}=t^\alpha \frac{-(2-3x)}{x}

    This gives:

    \frac{-(2-3tx)}{t}=t^\alpha (3x-2)

    3tx-2=t^{\alpha +1} (3x-2)

    and since the -2 on the LHS does not contain a t then there is no value of \alpha for which this is true.

    I haven't tried it with the y's yet since it has to satisfy both these conditions to be homogeneous. Since this one doesn't work the DE is non-homogeneous.

    Is that about right?
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  4. #19
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    I think i'm understanding it, but there's just something else that's puzzling.

    We have:

    -\frac{2-3x}{x}dx+\frac{3y-1}{y}dy=0

    and we're comparing this with:

    M(x,y)dx+N(x,y)dy=0

    Comparing coefficients we get:

    M(x,y)=-\frac{(2-3x)}{x} and N(x,y)=\frac{3y-1}{y}

    Is this equal to:

    M(x)=-\frac{(2-3x)}{x} and N(y)=\frac{3y-1}{y}?

    For the next part:

    <br />
M(tx,ty)=t^{\alpha}M(x,y)<br />

    \frac{-(2-3tx)}{tx}=t^\alpha \frac{-(2-3x)}{x}

    This gives:

    \frac{-(2-3tx)}{t}=t^\alpha (3x-2)

    3tx-2=t^{\alpha +1} (3x-2)

    and since the -2 on the LHS does not contain a t then there is no value of \alpha for which this is true. Correct

    I haven't tried it with the y's yet since it has to satisfy both these conditions to be homogeneous. Since this one doesn't work the DE is non-homogeneous. Correct. You will end up with the same situation with the y's, where there is no \alpha value where this would be true.

    Is that about right?
    It seems like you have the hang of this.

    --Chris
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  5. #20
    Super Member Showcase_22's Avatar
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    Sweet!

    Thanks for your help Chris!

    It was also nice to know that my 200th post was such a success! 8-)
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