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Math Help - Please help to solve for me.....

  1. #1
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    Please help to solve for me.....

    Hi please help me to solve this question....if y=cos ax/1+sin ax,where a is positive integer,show that cos ax d^2y/dx^2=a^2 y^2
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  2. #2
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    Re: Please help to solve for me.....

    This is unreadable. Please use brackets where they're needed, or else learn some basic LaTeX.
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  3. #3
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    Re: Please help to solve for me.....

    You are given:

    y=\frac{\cos(ax)}{1+\sin(ax)}

    You are asked to show:

    \cos(ax)\frac{d^2y}{dx^2}=a^2y^2

    Tools you will need:

    Quotient rule: \frac{d}{dx}\left(\frac{f(x)}{g(x)} \right)=\frac{\frac{df}{dx}g-f\frac{dg}{dx}}{g^2}

    Derivative of sine with chain rule: \frac{d}{dx}\left(\sin(u(x)) \right)=\cos(u(x))\frac{du}{dx}

    Derivative of cosine with chain rule: \frac{d}{dx}\left(\cos(u(x)) \right)=-\sin(u(x))\frac{du}{dx}

    Can you show your attempt (using bracketing symbols or \LaTeX)?
    Thanks from sharmala
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Please help to solve for me.....

    As an alternative, you could write:

    u(x)=y\left(\frac{\pi}{2a}-x \right)=\frac{\sin(ax)}{1+\cos(ax)}=\tan\left( \frac{ax}{2} \right)

    \frac{du}{dx}=\frac{a}{2}\sec^2\left(\frac{ax}{2} \right)

    \frac{d^2u}{dx^2}=\frac{a^2}{2}\sec^2\left(\frac{a  x}{2} \right)\tan\left(\frac{ax}{2} \right)=\frac{a^2}{2} \cdot\frac{2}{1+\cos(ax)} \cdot\frac{ \sin(ax)}{1+\cos(ax)}

    Hence:

    \cos(ax)\frac{d^2y}{dx^2}=a^2\left(\frac{\cos(ax)}  {1+\sin(ax)} \right)^2=a^2y^2
    Last edited by MarkFL; September 22nd 2012 at 07:46 AM.
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