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Solving Differential Equations by substitution Watch

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    (FP2 AQA)

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    Have I messed up yet? I'm repeating this question where I got it wrong (there is also no answer for this on the solutions)
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    What you've done doesn't look right, although I'm not entirely clear where one line starts and another begins at points.

    I think you would make life easier for yourself if you wrote dt/dx in terms of 'x', not t; then when you differentiate the 2nd time (w.r.t. x), you just have \dfrac{-1}{x^2} \dfrac{dy}{dt}; differentiating 1/x^2 is easy, and differentiating dy/dt (considered as a function of t) w.r.t x is just like differentiating y w.r.t. x, so it's essentially just doing what you did to relate dy/dx to dy/dt in the previous step.

    It should only be about 5 lines of working in total if you make no mistakes.
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    (Original post by DFranklin)
    What you've done doesn't look right, although I'm not entirely clear where one line starts and another begins at points.

    I think you would make life easier for yourself if you wrote dt/dx in terms of 'x', not t; then when you differentiate the 2nd time (w.r.t. x), you just have \dfrac{-1}{x^2} \dfrac{dy}{dt}; differentiating 1/x^2 is easy, and differentiating dy/dt (considered as a function of t) w.r.t x is just like differentiating y w.r.t. x, so it's essentially just doing what you did to relate dy/dx to dy/dt in the previous step.

    It should only be about 5 lines of working in total if you make no mistakes.
    The method your doing is different from the homestudy provider, have you got a video on this?
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    (Original post by ckfeister)
    The method your doing is different from the homestudy provider, have you got a video on this?
    No, why would I have a video? If you want to do it keeping things in terms of t, that's fine too - I just think it's a little more confusing in this case.
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    (Original post by DFranklin)
    No, why would I have a video? If you want to do it keeping things in terms of t, that's fine too - I just think it's a little more confusing in this case.
    How is this confusing... I also meant youtube......................... ....
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    (Original post by ckfeister)
    How is this confusing... I also meant youtube......................... ....
    I understood you. I just think it's a rather sad situation when a student's first response, made within minutes of a post, is "have you got a video on that?".
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    (Original post by DFranklin)
    I understood you. I just think it's a rather sad situation when a student's first response, made within minutes of a post, is "have you got a video on that?".
    Ok, as you would like to be brutal, I will be brutal. Nothing you said made sense, very sad isn't it
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    FWIW, as far as I can see, your first error is in the line starting (fg)' (it's a sign error). There are other errors later.
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    (Original post by DFranklin)
    I understood you. I just think it's a rather sad situation when a student's first response, made within minutes of a post, is "have you got a video on that?".
    As it is so " sad " would you also justify why this video proves my way was actually correct? d/dx means to differentiate, you can't differentiate them so you use the product rule.
    https://www.youtube.com/watch?v=bTx3xegXgA4
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    (Original post by ckfeister)
    As it is so " sad " would you also justify why this video proves my way was actually correct? d/dx means to differentiate, you can't differentiate them so you use the product rule.
    Are you seriously lecturing a maths graduate on what d/dx means? Really?

    I also explicitly said that your way was "fine", so I'm really not sure what point you're trying to prove here. I certainly can't be bothered to watch a video on the topic.
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    (Original post by ckfeister)
    As it is so " sad " would you also justify why this video proves my way was actually correct? d/dx means to differentiate, you can't differentiate them so you use the product rule.
    https://www.youtube.com/watch?v=bTx3xegXgA4
    Your/the video's method is correct but it's not the most efficient. Normally the ExamSolutions videos are excellent but I don't like his method for this question:

    \frac{dt}{dx} = 2t^\frac{1}{2}

    And since x=t^\frac{1}{2}, we have that \frac{dt}{dx} = 2x.

    Writing the derivative in terms of x makes the rest of the working simpler with less room for errors.

    DFranklin has shown you the mistake you made using your method. It's possible you wouldn't have made the mistake if you used a "better" method but who knows.
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    (Original post by Notnek)
    Your/the video's method is correct but it's not the most efficient. Normally the ExamSolutions videos are excellent but I don't like his method for this question:

    \frac{dt}{dx} = 2t^\frac{1}{2}

    And since x=t^\frac{1}{2}, we have that \frac{dt}{dx} = 2x.

    Writing the derivative in terms of x makes the rest of the working simpler with less room for errors.

    DFranklin has shown you the mistake you made using your method. It's possible you wouldn't have made the mistake if you used a "better" method but who knows.
    I was offering to but he didn't explain I asked for video if any quickly on youtube but apparently, it's sad... I also got this information on cloud learn the study provider and I could only find this on youtube that was relevent.. if you could put me in the right direction to find on youtube it would be great.
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    (Original post by ckfeister)
    I was offering to but he didn't explain I asked for video if any quickly on youtube but apparently, it's sad... I also got this information on cloud learn the study provider and I could only find this on youtube that was relevent.. if you could put me in the right direction to find on youtube it would be great.
    It's not easy to find a video explaining a style of tackling a question. But I don't think you need a video and you can work this out yourself. For your question:

    x=t^{-1} then x^2 = t^{-2}

    So you had \frac{dt}{dx}=-t^{-2} which can be written as \frac{dt}{dx}=-x^2. Does this make sense?

    Now try continuing from here so the next step is to find \frac{dy}{dx} and then \frac{d^2y}{dx^2}.


    Or you can continue using your method - that isn't the biggest problem unless you find that you are making a lot of mistakes. You have been given the mistake you made using your method so try to correct that.
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    (Original post by Notnek)
    It's not easy to find a video explaining a style of tackling a question. But I don't think you need a video and you can work this out yourself. For your question:

    x=t^{-1} then x^2 = t^{-2}

    So you had \frac{dt}{dx}=-t^{-2} which can be written as \frac{dt}{dx}=-x^2. Does this make sense?

    Now try continuing from here so the next step is to find \frac{dy}{dx} and then \frac{d^2y}{dx^2}.


    Or you can continue using your method - that isn't the biggest problem unless you find that you are making a lot of mistakes. You have been given the mistake you made using your method so try to correct that.
    Ohh now I get it, thanks.
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    Your solution could be clearer but my recommendations are:

    - Check your answer for  \dfrac{\mathrm{d}^{2}y}{\mathrm{  d}x^{2}}, (you have missed out some terms),

    -Your attempt to simplify  \mathrm{d}^{2}y / \mathrm{d}x^{2}, (which you did not use in the end) is not correct and is not a step I would take.
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    Please look at the part a answer

    (Original post by ckfeister)
    (FP2 AQA)

    Name:  question5.PNG
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    Have I messed up yet? I'm repeating this question where I got it wrong (there is also no answer for this on the solutions)
    Attachment 697278697280
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    (Original post by Notnek)
    It's not easy to find a video explaining a style of tackling a question. But I don't think you need a video and you can work this out yourself. For your question:

    x=t^{-1} then x^2 = t^{-2}

    So you had \frac{dt}{dx}=-t^{-2} which can be written as \frac{dt}{dx}=-x^2. Does this make sense?

    Now try continuing from here so the next step is to find \frac{dy}{dx} and then \frac{d^2y}{dx^2}.


    Or you can continue using your method - that isn't the biggest problem unless you find that you are making a lot of mistakes. You have been given the mistake you made using your method so try to correct that.

    For
    \frac{dt}{dx}=-x^2.

    I get
    \frac{dt}{dx}=-x^-^2. as \frac{dx}{dt}=-x^2. Correct or wrong?
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    (Original post by gurungg48)
    Please look at the part a answer

    QUOTE=ckfeister;74195320](FP2 AQA)

    Name:  question5.PNG
Views: 15
Size:  52.8 KB

    Have I messed up yet? I'm repeating this question where I got it wrong (there is also no answer for this on the solutions)
    Attachment 697278697280
    [/QUOTE]

    Where does the  \frac{d^2y}{dt^2} t part come from I thought it was  t^-^2
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    (Original post by ckfeister)
    For
    \frac{dt}{dx}=-x^2.

    I get
    \frac{dt}{dx}=-x^-^2. as \frac{dx}{dt}=-x^2. Correct or wrong?
    Correct.
 
 
 
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