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    How would I go about this?
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    What is "reduction of order"? I've never heard of it.

    With a bit of thought, you should be able to recognize the first 2 terms as the derivative of a product, while the last term is the derivative of (y')^2.
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    (Original post by DFranklin)
    What is "reduction of order"? I've never heard of it.

    With a bit of thought, you should be able to recognize the first 2 terms as the derivative of a product, while the last term is the derivative of (y')^2.
    Reduction of order is where you 'reduce' the order of the 2nd order ODE into a first order ODE which is solvable.

    Define v:=\frac{dy}{dx}

    yv\frac{dv}{dy}+v^2-2yv=0

    \frac{dv}{dx}=2-\frac{v}{y}

    \int\frac{1}{2-v}dv=\int\frac{1}{y}dy

    Would this be a valid start (not sure about dividing by v and y) and if so how would I integrate the LHS?
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    Looks OK (other than you mean dv/dy in the 2nd to last line).

    Do you really not know how to integrate 1/(2-v)?
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    (Original post by DFranklin)
    Looks OK (other than you mean dv/dy in the 2nd to last line).

    Do you really not know how to integrate 1/(2-v)?
    Lack of confidence on these things I'm afraid

    Would it be \frac{1}{2}\int\frac{1}{1-\frac{1}{2}v}dv which integrates to -ln(1-\frac{1}{2}v) or is there a \frac{1}{4} in front of that?
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    The first. (Don't forget, you can always check by differentiating. With something like this I think "well, it's going to be *like* ln(2-v). But derivative of ln(2-v) is -1/(2-v), so in fact we want -ln(2-v).")

    N.B. ln(2-v) = ln(1-1/2v)+ln(2), so this is the same as your answer up to the arbitrary constant.
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    (Original post by DFranklin)
    The first. (Don't forget, you can always check by differentiating. With something like this I think "well, it's going to be *like* ln(2-v). But derivative of ln(2-v) is -1/(2-v), so in fact we want -ln(2-v).")

    N.B. ln(2-v) = ln(1-1/2v)+ln(2), so this is the same as your answer up to the arbitrary constant.
    So from here

    ln(y)=-ln(1-\frac{1}{2}v)+c

    ln(y)+ln(1-\frac{1}{2}v)=c

    ln(\frac{y}{1-\frac{1}{2}v})=c

    \frac{y}{1-\frac{1}{2}v}=A, (A=e^c)

    This can be manipulated down to

    \int\frac{A}{2A-2y}dy=\int dx

    \frac{A}{2}\int\frac{1}{A-y}dy=x+D

    \frac{1}{2}\int\frac{1}{1-\frac{y}{A}}dy=x+D

    -\frac{A}{2}ln(1-\frac{y}{A})=x+D

    This can be manipulated down to

    ln(1-\frac{y}{A})=-\frac{2x-2D}{A}

    y=2x+2D+A

    This doesn't quite seem right with ending up with 2 constant terms. Can you please check?
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    I don't see how you get from the 2nd to last line to the last line. (It looks wrong to me).
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    when adding logs dont u multiply?

    2nd to 3rd lines???

    i might be being dumn tho...
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    (Original post by lee_91)
    when adding logs dont u multiply?

    2nd to 3rd lines???

    i might be being dumn tho...
    No you're right stupid mistake :yep:
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    (Original post by DFranklin)
    I don't see how you get from the 2nd to last line to the last line. (It looks wrong to me).
    After correctly manipulating the log and collecting like variables I've now arrived at

    \frac{1}{2}\int\frac{y}{y-A}dy=\int dx

    Does this seem right and if so how to integrate LHS?
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    y/(y-A) = (y-A+A)/(y-A) = 1 + A / (y-A)
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    (Original post by DFranklin)
    y/(y-A) = (y-A+A)/(y-A) = 1 + A / (y-A)
    and this =\int dy+\int\frac{1}{\frac{y}{A}-1}dy

    but I still don't see how to integrate?
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    log...
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    (Original post by TheEd)
    Reduction of order is where you 'reduce' the order of the 2nd order ODE into a first order ODE which is solvable.

    Define v:=\frac{dy}{dx}

    yv\frac{dv}{dy}+v^2-2yv=0

    \frac{dv}{dx}=2-\frac{v}{y}.....(1)

    \int\frac{1}{2-v}dv=\int\frac{1}{y}dy.....(2)


    Would this be a valid start (not sure about dividing by v and y) and if so how would I integrate the LHS?

    Perhaps I am being particularly thick, but equation (2) doesn't seem to follow from equation (1), even allowing for the dv/dx in (1) should be dv/dy as DFranklin pointed out.
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    I wouldn't normally suggest a change of tack this far in, but assuming my previous posting is valid, then:

    As DFranklin pointed out in his first post the first two terms are the derivative of a product, and the last term is actually a constant multiple of that product, which suggest z=yy' might be a useful substitution to use....
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    Re: Ghostwalkers post above the last one: Good point. I'm not batting 100% on this Q.
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    (Original post by ghostwalker)
    Perhaps I am being particularly thick, but equation (2) doesn't seem to follow from equation (1), even allowing for the dv/dx in (1) should be dv/dy as DFranklin pointed out.
    I think you're right. Is this equation even separable as I can't separate like variables onto either side!
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    (Original post by TheEd)
    I think you're right. Is this equation even separable as I can't separate like variables onto either side!
    I was thinking the same thing as Ghostwalkers and wondered if I was being completey thick.

    Equation 1 as Ghostwalkers has labeled it isn't separable. Take the the v/y over to the other side.

    Then use the integrating factor method I showed you in the thread the radioactive decay thread. A quicker way would be to multiply the rearagned equation by y and note that LHS= d(vy)/dy.
    These are both actually the same method. You will notice that the integrating factor w, as I called it in the other thread, will be w=y. In this case the integrating factor is quite easy to find just by looking but the method in the other thread is the general method to finding integrating factors to solve the equation

    dy/dx + P(x)y = Q(x)

    when the integrating factor is not easy to spot.
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    (Original post by thebadgeroverlord)
    I was thinking the same thing as Ghostwalkers and wondered if I was being completey thick.

    Equation 1 as Ghostwalkers has labeled it isn't separable. Take the the v/y over to the other side.

    Then use the integrating factor method I showed you in the thread the radioactive decay thread. A quicker way would be to multiply the rearagned equation by y and note that LHS= d(vy)/dy.
    These are both actually the same method. You will notice that the integrating factor w, as I called it in the other thread, will be w=y. In this case the integrating factor is quite easy to find just by looking but the method in the other thread is the general method to finding integrating factors to solve the equation

    dy/dx + P(x)y = Q(x)

    when the integrating factor is not easy to spot.
    Thanks; so would the following be correct?

    \frac{dv}{dy}+\frac{v}{y}=2

    y\frac{dv}{dy}+v=2y

    ( y\frac{dv}{dy}+v=\frac{d}{dy}(vy  ) )

    vy=\frac{2y^2}{2}+c

    vy=y^2+c

    y\frac{dy}{dx}=y^2+c

    Now how to solve for y?
 
 
 
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