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

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• Sep 17th 2012, 09:45 AM
sharmala
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
• Sep 17th 2012, 09:53 AM
Prove It
Re: Please help to solve for me.....
This is unreadable. Please use brackets where they're needed, or else learn some basic LaTeX.
• Sep 17th 2012, 10:50 AM
MarkFL
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 $\LaTeX$)?
• Sep 22nd 2012, 04:01 AM
MarkFL
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:

$\cos(ax)\frac{d^2y}{dx^2}=a^2\left(\frac{\cos(ax)} {1+\sin(ax)} \right)^2=a^2y^2$