Year 13 Maths Help Thread

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    (Original post by kiiten)
    Now i got 3 + sqrt 17 instead of 3 + sqrt 13???
    Did you make 2x^2 equal the DOUBLE of the path?
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    (Original post by RDKGames)
    Did you make 2x^2 equal the DOUBLE of the path?
    No...

    so would it be

    4x^2 = (2x+2)(x+2) - 2x^2
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    (Original post by kiiten)
    No...

    so would it be

    4x^2 = (2x+2)(x+2) - 2x^2
    Almost. I see you took a different approach, which is still technically right.

    Try: 2x^2=2[(2x+2)(x+2)-2x^2]

    Im not sure where you pulled that 4x^2 from.
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    (Original post by RDKGames)
    Almost. I see you took a different approach, which is still technically right.

    Try: 2x^2=2[(2x+2)(x+2)-2x^2]

    Im not sure where you pulled that 4x^2 from.
    i multiplied the 2x^2 by 2 because it says the area of the lawn (2x^2) is twice that of the path hence 4x^2 ?

    EDIT: i misunderstood .... again. Thanks
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    (Original post by RDKGames)
    Almost. I see you took a different approach, which is still technically right.

    Try: 2x^2=2[(2x+2)(x+2)-2x^2]

    Im not sure where you pulled that 4x^2 from.
    Sorry i keep going on about this question xD but shouldnt the answer by multiplied by 2 (2x) ??
    its asking for the length of the lawn.
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    (Original post by kiiten)
    Sorry i keep going on about this question xD but shouldnt the answer by multiplied by 2 (2x) ??
    its asking for the length of the lawn.
    Yep.
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    (Original post by RDKGames)
    Yep.
    So its 6 +2 sqrt 13 (not 3 + sqrt 13)
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    (Original post by kiiten)
    So its 6 +2 sqrt 13 (not 3 + sqrt 13)
    Yep.
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    (Original post by RDKGames)
    Yep.
    yay finally i got something right
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    Someone explain what how am I supposed to go about this?

    Question asks for a complementary function and a particular integral of \frac{dy}{dx}-3y=6 and I've gotten the CF to be y=ae^{3x} but I'm unsure where to go from there because subbing it into the original differential doesn't exactly favour maths as 0\not= 6 so there's no value for a that will make this work, is there? What does this mean for the PI?
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    (Original post by RDKGames)
    Someone explain what how am I supposed to go about this?

    Question asks for a complementary function and a particular integral of \frac{dy}{dx}-3y=6 and I've gotten the CF to be y=ae^{3x} but I'm unsure where to go from there because subbing it into the original differential doesn't exactly favour maths as 0\not= 6
    Try a PI of the form  p where  p is just a real constant. You don't need to sub your complementary function into the ODE once you have found it. The arbitrary constant  a has to be there for the general solution and you can only find the value of  a if you're given boundary conditions. your general solution will just be  y=\text{A} e^{3x} + PI .
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    (Original post by B_9710)
    Try a PI of the form  p where  p is just a real constant. You don't need to sub your complementary function into the ODE once you have found it. The arbitrary constant  a has to be there for the general solution and you can only find the value of  a if you're given boundary conditions. your general solution will just be  y=\text{A} e^{3x} + PI .
    Ah yes I remember that now, y=CF+PI
    So CF=Ae^{3x} and since it cancels, it only leaves -3p=6
    and therefore p=-2=PI?
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    (Original post by RDKGames)
    Ah yes I remember that now, y=CF+PI
    So CF=Ae^{3x} and since it cancels, it only leaves -3p=6
    and therefore p=-2=PI?
    Yes that's right.
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    (Original post by B_9710)
    Yes that's right.
    Can you explain why the CF of \frac{dy}{dx}-2y=e^{2x} is y_c=Ce^{2x} and not, as I found it, y_c=Cxe^{2x}? Or are they wrong?
    Because when I used mine I get PI to be xe^{2x} which is correct on their answers.
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    (Original post by RDKGames)
    Can you explain why the CF of \frac{dy}{dx}-2y=e^{2x} is y_c=Ce^{2x} and not, as I found it, y_c=Cxe^{2x}? Or are they wrong?
    Because when I used mine I get PI to be xe^{2x} which is correct on their answers.
    The CF is  pxe^{2x} .
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    (Original post by B_9710)
    The CF is  pxe^{2x} .
    This book smh... :facepalm:
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    Am i doing this wrong?

    Solve: 5+2sin(2x+1)=6 for 0<=x<=720

    rearranged to sin(2x+1)= 1/2

    let u = 2x+1
    sinu=1/2
    u=1/6 pi

    change the bounds to 1 <=u<=4pi +1
    using a cast circle i got the first answer to be 2.05 but it should be 0.81 ???

    *sorry if this doesnt make sense - let me know and ill post the full working
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    Not quite sure where I'm going wrong.

    \frac{du}{dx}+\frac{2u}{x}=\frac  {1}{x^2}

    IF=e^{\int \frac{2}{x} .dx}=x^2

    but that doesn't seem right as I wouldn't get the left side from \frac{d}{dx}(ux^2)
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    (Original post by kiiten)
    Am i doing this wrong?

    Solve: 5+2sin(2x+1)=6 for 0<=x<=720

    rearranged to sin(2x+1)= 1/2

    let u = 2x+1
    sinu=1/2
    u=1/6 pi

    change the bounds to 1 <=u<=4pi +1
    using a cast circle i got the first answer to be 2.05 but it should be 0.81 ???

    *sorry if this doesnt make sense - let me know and ill post the full working
    sin(x)=\alpha
    Then:
    x=2n\pi + arcsin(\alpha)

    and due to symmetry of sine:

    x=2n\pi + \pi - arcsin(\alpha)

    For any integer n (try a few around 0 until your solutions go outside the range)

    Try that.
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    (Original post by RDKGames)
    sin(x)=\alpha
    Then:
    x=2n\pi + arcsin(\alpha)

    and due to symmetry of sine:

    x=2n\pi + \pi - arcsin(\alpha)

    For any integer n (try a few around 0 until your solutions go outside the range)

    Try that.
    ??? I have no idea what you're talking about.
    What is 2n and is arcsin inverse of sin?
 
 
 
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