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    Hi , im stuck on this question :
    Find the gradient of the lines joining the following pairs of points:
    A=(p+3,q-7)
    B=(p+5,3-q )
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    y2-y1/x2-x1
    you have your co-ordinates, put it into this equation, you will have your answer
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    The Gradient of the line joining 2 points is the change in Y value divided by the change in X value.

    To find the change of either value, just take one away from the other. (y1-y2)/(x1-x2)

    So,

    (P+3)-(P+5) = P + 3 - P - 5 = -2. Change in X = -2

    (Q-7)-(3-Q) = Q - 7 - 3 + Q = 2Q -10 Change in Y= 2Q-10

    Therefore

    Gradient = (2Q-10) / (-2)

    Becoming

    -Q + 10
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    (Original post by Workaholism)
    (FULL SOLUTION)
    That's some nice working out. Can you solve this for me? I need the general solution. Thanks.

    \displaystyle 4x\frac{d^2y}{dx^2}+2(1-2\sqrt{x})\frac{dy}{dx}+y=3\sqrt  {x} using substitution x=t^2?


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    (Original post by RDKGames)
    That's some nice working out. Can you solve this for me? I need the general solution. Thanks.

    \displaystyle 4x\frac{d^2y}{dx^2}+2(1-2\sqrt{x})\frac{dy}{dx}+y=3\sqrt  {x} using substitution x=t^2?
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     y=\displaystyle A\sqrt xe^{\sqrt x}+Be^{\sqrt x}-3\sqrt x+6 .
    Solve  \displaystyle x^3\frac{d^3y}{dx^3}+2x^2\frac{d  ^2y}{dx^2}+x\frac{dy}{dx}-y=0 .
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    (Original post by RDKGames)
    That's some nice working out. Can you solve this for me? I need the general solution. Thanks.

    \displaystyle 4x\frac{d^2y}{dx^2}+2(1-2\sqrt{x})\frac{dy}{dx}+y=3\sqrt  {x} using substitution x=t^2?

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    Don't post full solutions.
    Huh? Why not?
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    (Original post by Workaholism)
    Huh? Why not?
    Well apart from it being against the posting rules, you are also doing the question for the person. Just give them hints like the first reply and let them figure it out on their own.

    (Original post by B_9710)
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     y=\displaystyle A\sqrt xe^{\sqrt x}+Be^{\sqrt x}-3\sqrt x+6 .
    Solve  \displaystyle x^3\frac{d^3y}{dx^3}+2x^2\frac{d  ^2y}{dx^2}+x\frac{dy}{dx}-y=0 .
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     \displaystyle y=Ax+Bcos(log(x))+Csin(log(x))
    Solve: \displaystyle \frac{d^4y}{dx^4}-\frac{d^2y}{dx^2}-x=0
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    (Original post by RDKGames)
    Well apart from it being against the posting rules, you are also doing the question for the person. Just give them hints like the first reply and let them figure it out on their own.
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     \displaystyle y=Ax+Bcos(log(x))+Csin(log(x))
    Solve: \displaystyle \frac{d^4y}{dx^4}-\frac{d^2y}{dx^2}-x=0
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    Your solution isn't quite right.
     \displaystyle y=Ae^x+Be^{-x}+Cx+D-\frac{1}{6}x^3 .
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    (Original post by B_9710)
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    Your solution isn't quite right.
     \displaystyle y=Ae^x+Be^{-x}+Cx+D-\frac{1}{6}x^3 .
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    Is it
    \displaystyle y=Ax+Bcos(ln(x))+Csin(ln(x)) ?
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    (Original post by RDKGames)
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    Is it
    \displaystyle y=Ax+Bcos(ln(x))+Csin(ln(x)) ?
    No.
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    (Original post by B_9710)
    No.
    Well damn, if Wolfram Alpha doesn't give me the right answer I don't know what will.
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    Alright, enough with the off topic DE's, don't you think?
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    (Original post by Zacken)
    Alright, enough with the off topic DE's, don't you think?
    They're spoilered so it's fine.
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    (Original post by Workaholism)
    The Gradient of the line joining 2 points is the change in Y value divided by the change in X value.

    To find the change of either value, just take one away from the other. (y1-y2)/(x1-x2)

    So,

    (P+3)-(P+5) = P + 3 - P - 5 = -2. Change in X = -2

    (Q-7)-(3-Q) = Q - 7 - 3 + Q = 2Q -10 Change in Y= 2Q-10

    Therefore

    Gradient = (2Q-10) / (-2)

    Becoming

    -Q + 10
    Your final answer should be -Q + 5, as (-10)/(-2) = 5.
 
 
 
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