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    Find the solution to the differential equation (4-x^2)^1/2 dy/dx = (4-y^2) ^1/2 for which y= -1 and x=1.

    I just don't know how to integrate (4-y^2)^1/2 = (4-y^2)^1/2

    Can someone please help
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    (Original post by ChemBoy1)
    Find the solution to the differential equation (4-x^2)^1/2 dy/dx = (4-y^2) ^1/2 for which y= -1 and x=1.

    I just don't know how to integrate (4-y^2)^1/2 = (4-y^2)^1/2

    Can someone please help
    Not sure what the second line means.

    Use the substitution x=2sinu.
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    (Original post by ChemBoy1)
    Find the solution to the differential equation (4-x^2)^1/2 dy/dx = (4-y^2) ^1/2 for which y= -1 and x=1.

    I just don't know how to integrate (4-y^2)^1/2 = (4-y^2)^1/2

    Can someone please help
    \begin{aligned} (4-x^2)^{1/2} \dfrac{dy}{dx} = (4-y^2)^{1/2} & \Leftrightarrow (4-y^2)^{-1/2} .dy = (4-x^2)^{-1/2} .dx \\ & \Leftrightarrow \int (4-y^2)^{-1/2} .dy = \int (4-x^2)^{-1/2} .dx \end{aligned}

    Then just use two subs:

    x= 2 \sin u
    y= 2 \sin v
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    Update: I managed to get the equation as arcsin y/2 = arcsin x/2 +c. The question asks to express my answer in a form not involving trig functions. I don't see how I can do that.

    Thanks
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    (Original post by ChemBoy1)
    Update: I managed to get the equation as arcsin y/2 = arcsin x/2 +c. The question asks to express my answer in a form not involving trig functions. I don't see how I can do that.

    Thanks
    First of all, determine what c is.

    Then, write \dfrac{y}{2}=\sin \left( \arcsin \dfrac{x}{2} + c \right) and then apply the compound angle formula on the RHS.
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    I got c as pi/3. Ho do I convert it to an expression not involving trig functions?
    (Original post by RDKGames)
    First of all, determine what c is.

    Then, write \dfrac{y}{2}=\sin \left( \arcsin \dfrac{x}{2} + c \right) and then apply the compound angle formula on the RHS.
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    (Original post by ChemBoy1)
    I got c as pi/3. Ho do I convert it to an expression not involving trig functions?
    Read my post - I answered that question...
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    Sorry just saw that thanks. How can you use compound angle formula because i still got Sin (Sin-1 x) etc.
    (Original post by RDKGames)
    Read my post - I answered that question...
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    (Original post by ChemBoy1)
    Sorry just saw that thanks. How can you use compound angle formula because i still got Sin (Sin-1 x) etc.
    \sin \arcsin \frac{x}{2} = \frac{x}{2} (obviously) so, we just need \cos \arcsin \frac{x}{2}. If we denote \arcsin \frac{x}{2} = \theta then we have \sin \theta = \frac{x}{2} and we can use a right-angled triangle to determine what \cos \theta is.

    EDIT: Or just use \cos \theta = \sqrt{1-\sin^2 \theta} for this part.
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    Still confused with the right angle triangle thing. becuase the hyp = 2 and the opp = x. The adj would just be a separate variable and that would not work? Could you please help me out?
    (Original post by RDKGames)
    \sin \arcsin \frac{x}{2} = \frac{x}{2} (obviously) so, we just need \cos \arcsin \frac{x}{2}. If we denote \arcsin \frac{x}{2} = \theta then we have \sin \theta = \frac{x}{2} and we can use a right-angled triangle to determine what \cos \theta is.

    EDIT: Or just use \cos \theta = \sqrt{1-\sin^2 \theta} for this part.
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    (Original post by ChemBoy1)
    Still confused with the right angle triangle thing. becuase the hyp = 2 and the opp = x. The adj would just be a separate variable and that would not work? Could you please help me out?
    The adj would just be sqrt(hyp^2-opp^2) by Pythagoras
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    The book got y= 1/2 x - 1/2x (3(4-x^2))^1/2.

    I got no where near that answer, could you help elaborate.
    FYI values are y= -1 and x=1
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    (Original post by ChemBoy1)
    The book got y= 1/2 x - 1/2x (3(4-x^2))^1/2.

    I got no where near that answer, could you help elaborate.
    FYI values are y= -1 and x=1
    OK, first of all, c \neq \frac{\pi}{3}. It is in fact c= -\frac{\pi}{3}

    Next, we have
    \begin{aligned} \dfrac{y}{2} & = \sin \left( \arcsin \frac{x}{2} - \frac{\pi}{3} \right) \\ & = \sin \left( \arcsin \frac{x}{2} \right) \cos \left( \frac{\pi}{3} \right) - \sin \left( \frac{\pi}{3} \right) \cos \left( \arcsin \frac{x}{2} \right) \\ & = \frac{x}{2} \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \end{aligned}

    Finish it off.
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    Nearly got it, except for how they got 3(4-x^2)^1/2
    (Original post by RDKGames)
    OK, first of all, c \neq \frac{\pi}{3}. It is in fact c= -\frac{\pi}{3}

    Next, we have
    \begin{aligned} \dfrac{y}{2} & = \sin \left( \arcsin \frac{x}{2} - \frac{\pi}{3} \right) \\ & = \sin \left( \arcsin \frac{x}{2} \right) \cos \left( \frac{\pi}{3} \right) - \sin \left( \frac{\pi}{3} \right) \cos \left( \arcsin \frac{x}{2} \right) \\ & = \frac{x}{2} \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \end{aligned}

    Finish it off.
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    (Original post by ChemBoy1)
    Nearly got it, except for how they got 3(4-x^2)^1/2
    Factor out 1/4 inside the root, then bring that factor outside the root. Similarly, merge root(3) with the main root since \sqrt{a}\sqrt{b} = \sqrt{ab}
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    Thank you so much! You have helped a lot! Sorry if I have sounded very silly to you.
    (Original post by RDKGames)
    Factor out 1/4 inside the root, then bring that factor outside the root. Similarly, merge root(3) with the main root since \sqrt{a}\sqrt{b} = \sqrt{ab}
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    (Original post by ChemBoy1)
    Thank you so much! You have helped a lot! Sorry if I have sounded very silly to you.
    :thumbsup:
 
 
 
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