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    Zacken Euclidean

    Any more hard/interesting questions then?
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    When I see a person helping out someone with their Maths problem...

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    (Original post by BBeyond)
    Zacken Euclidean

    Any more hard/interesting questions then?
    I'll think up some when I get home. :-)
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    (Original post by BBeyond)
    Zacken Euclidean

    Any more hard/interesting questions then?
    Try solving this:

    \displaystyle

\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} + 4xe^{-x^2}\sqrt{y+3} = 0\end{equation*}

    subject to the conditions x \geq 0 and that y=6 when x=0. Express your solution in the form y = f(x) and find \lim_{x \to \infty} y.
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    (Original post by Zacken)
    Try solving this:

    \displaystyle

\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} + 4xe^{-x^2}\sqrt{y+3} = 0\end{equation*}

    subject to the conditions x \geq 0 and that y=6 when x=0. Express your solution in the form y = f(x) and find \lim_{x \to \infty} y.
    I'll try after my dinner, cheers!
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    (Original post by BBeyond)
    I'll try after my dinner, cheers!
    Bon appetit.
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    (Original post by Zacken)
    Try solving this:

    \displaystyle

\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} + 4xe^{-x^2}\sqrt{y+3} = 0\end{equation*}

    subject to the conditions x \geq 0 and that y=6 when x=0. Express your solution in the form y = f(x) and find \lim_{x \to \infty} y.
    Kinda rushed this as I have a few things to do, so there may be a mistake somewhere :rofl:
    Spoiler:
    Show
    -\dfrac{dy}{dx} = 4xe^{-x^{2}}\sqrt{y+3}

    -\displaystyle \int \dfrac{1}{\sqrt{y+3}}\ dy = 4\displaystyle \int xe^{-x^{2}}\ dx

    Gonna go ahead and assume we know how to integrate both - substitution of u=y+3 for \dfrac{1}{\sqrt{y+3}} will do the trick
    And u = -x^{2} will do for xe^{-x^{2}}

    -2\sqrt{y+3} = -2e^{-x^{2}} + C

     y = 6,\ x = 0

    -6 = -2 + C

    \therefore C = -4

    -2\sqrt{y+3} = -2e^{-x^{2}} -4

    \sqrt{y+3} = e^{-x^{2}} + 2

    y = \left[ e^{-x^{2}} +2\right]^{2} - 3
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    (Original post by edothero)
    Kinda rushed this as I have a few things to do, so there may be a mistake somewhere :rofl:
    Yeaaah. It's y+3 but you seem to have switched to y-3 half way through which gets you ugly ass square roots.
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    (Original post by Zacken)
    Yeaaah. It's y+3 but you seem to have switched to y-3 half way through which gets you ugly ass square roots.
    Oh ffs

    There we go, I've changed it
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    (Original post by Zacken)
    Try solving this:

    \displaystyle

\begin{equation*} \frac{\mathrm{d}y}{\mathrm{d}x} + 4xe^{-x^2}\sqrt{y+3} = 0\end{equation*}

    subject to the conditions x \geq 0 and that y=6 when x=0. Express your solution in the form y = f(x) and find \lim_{x \to \infty} y.
    Spoiler:
    Show
    Name:  ImageUploadedByStudent Room1457119623.449167.jpg
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Size:  109.7 KB

    My attempt


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    (Original post by edothero)
    [Awesome work]
    All correct. Now what's \lim_{x \to \infty} y?
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    (Original post by BBeyond)
    ...
    (Original post by edothero)
    ...
    All correct, both of you. Good work, too bad we're only warming up. Let's get to the meaty bit: Let y be any solution to:

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }y}{\mathrm{d}x} - xe^{6x^2}(y+3)^{1-k} = 0 \quad (x\geq 0)\end{equation(}

    Find a k such that we have e^{-3x^2}y tending to a finite non-zero limit that you should determine as x \to \infty.
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    (Original post by Zacken)
    All correct. Now what's \lim_{x \to \infty} y?
    As x tends to infinity, e^x tends to 0, as we end up with e^{-2x^{2}} + 4e^{-x^{2}} + 1 then \lim_{x \to \infty} y would = 1 ?
    Is this the right way of going about that?
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    (Original post by edothero)
    As x tends to infinity, e^x tends to 0, as we end up with e^{-2x^{2}} + 4e^{-x^{2}} + 1 then \lim_{x \to \infty} y would = 1 ?
    Is this the right way of going about that?
    Yep, you needn't expand though: x \to \infty \Rightarrow e^{-x^2} \to 0 \Rightarrow y \to (0+2)^2 - 3 = 2^2 - 3 = 4-3=1.
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    (Original post by Zacken)
    Yep, you needn't expand though: x \to \infty \Rightarrow e^{-x^2} \to 0 \Rightarrow y \to (0+2)^2 - 3 = 2^2 - 3 = 4-3=1.
    Ah yes makes more sense I'll attempt that second question later tonight after revision. Seems tough!
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    (Original post by Zacken)
    \displaystyle

\begin{equation*}\frac{\mathrm{d  }y}{\mathrm{d}x} - xe^{6x^2}(y+3)^{1-k} = 0 \quad (x\geq 0)\end{equation(}

    Find a k such that we have e^{-3x^2}y tending to a finite non-zero limit that you should determine as x \to \infty.
    I think it's time you give me a hint
    Spoiler:
    Show
    \dfrac{dy}{dx} = \dfrac{xe^{6x^{2}}}{(y+3)^{k-1}}

    \displaystyle \int (y+3)^{k-1}\ dy = \displaystyle \int xe^{6x^{2}}\ dx

    (Same method, u=6x^{2}. And the (y+3) integral is standard (k-1)+1

    \dfrac{1}{k}(y+3)^{k} = \dfrac{1}{12}e^{6x^{2}} + C
    ..
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    (Original post by edothero)
    I think it's time you give me a hint
    Spoiler:
    Show
    \dfrac{dy}{dx} = \dfrac{xe^{6x^{2}}}{(y+3)^{k-1}}

    \displaystyle \int (y+3)^{k-1}\ dy = \displaystyle \int xe^{6x^{2}}\ dx

    (Same method, u=6x^{2}. And the (y+3) integral is standard (k-1)+1

    \dfrac{1}{k}(y+3)^{k} = \dfrac{1}{12}e^{6x^{2}} + C
    ..
    Do the obvious thing: isolate, multiply both sides by what you want, simplify. Look at limits.
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    (Original post by edothero)
    I think it's time you give me a hint
    Spoiler:
    Show
    \dfrac{dy}{dx} = \dfrac{xe^{6x^{2}}}{(y+3)^{k-1}}

    \displaystyle \int (y+3)^{k-1}\ dy = \displaystyle \int xe^{6x^{2}}\ dx

    (Same method, u=6x^{2}. And the (y+3) integral is standard (k-1)+1

    \dfrac{1}{k}(y+3)^{k} = \dfrac{1}{12}e^{6x^{2}} + C
    ..
    That's about as far as I made it as well ahahah
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    (Original post by Zacken)
    Do the obvious thing: isolate, multiply both sides by what you want, simplify. Look at limits.
    Not getting anywhere I'm afraid
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    (Original post by edothero)
    Not getting anywhere I'm afraid
    I'm in bed now, will put up a solution in the morning.
 
 
 
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