Question

If $x=a\cos \text{ }\\!\\!\theta\\!\\!\text{ }+b\sin \text{ }\\!\\!\theta\\!\\!\text{ }$ and $y=a\sin \text{ }\\!\\!\theta\\!\\!\text{ }-b\cos \text{ }\\!\\!\theta\\!\\!\text{ }$ then prove that${{y}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+y=0$

Answer 1

Since we are given two equations:
$\begin{aligned} & x=a\cos \text{ }\\!\\!\theta\\!\\!\text{ }+b\sin \text{ }\\!\\!\theta\\!\\!\text{ }......\text{(1)} \\\ & y=a\sin \text{ }\\!\\!\theta\\!\\!\text{ }-b\cos \text{ }\\!\\!\theta\\!\\!\text{ }......\text{(2)} \\\ \end{aligned}$
Differentiate both the equations with respect to $\text{ }\\!\\!\theta\\!\\!\text{ }$. We get the values of $\dfrac{dx}{d\theta }$ and $\dfrac{dy}{d\theta }$. Divide $\dfrac{dy}{d\theta }$ by $\dfrac{dx}{d\theta }$ to get the value of $\dfrac{dy}{dx}$. Now, differentiate $\dfrac{dy}{dx}$ with respect to x and get the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$. Substitute the values in the equation ${{y}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+y$ and get the result.

Complete step by step answer:
We have the following two equations:
$\begin{aligned} & x=a\cos \theta +b\sin \theta ......(1) \\\ & y=a\sin \theta -b\cos \theta ......(2) \\\ \end{aligned}$
Now, differentiation equation (1) and (2) with respect to $\theta$, we get:
$\begin{aligned} & \dfrac{dx}{d\theta }=-a\sin \theta +b\cos \theta ......(3) \\\ & \dfrac{dy}{d\theta }=a\cos \theta +b\sin \theta ......(4) \\\ \end{aligned}$
$\left[ \because \dfrac{d}{d\theta }\left( \sin \theta \right)=\cos \theta ;\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta \right]$
Now, we need to find the value of $\dfrac{dy}{dx}$. So, divide equation (4) by equation (3).
We get:
$\dfrac{dy}{dx}=\dfrac{a\cos \theta \text{ }+b\sin \theta }{-a\sin \theta +b\cos \theta }$
Taking minus common from the denominator, we can write:
$\dfrac{dy}{dx}=-\dfrac{a\cos \theta \text{ }+b\sin \theta }{a\sin \theta -b\cos \theta }......(5)$
Since from equation (1) and (2), we have:
$x=a\cos \theta +b\sin \theta$ and $y=a\sin \theta -b\cos \theta$
So, we can write equation (5) as:
$\dfrac{dy}{dx}=-\dfrac{x}{y}......\text{(6)}$
Now, to find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$, differentiate equation (6) with respect to x.
By applying the identity: $\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{\left( \dfrac{d}{dx}f(x) \right)g(x)-f(x)\left( \dfrac{d}{dx}g(x) \right)}{{{\left( g(x) \right)}^{2}}}$
We get:

& \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{\left( 1\times y \right)-\left( x\times \dfrac{dy}{dx} \right)}{{{y}^{2}}} \right) \\\ & =-\left( \dfrac{y-x\dfrac{dy}{dx}}{{{y}^{2}}} \right)......(7) \end{aligned}$$ Since we need to prove that: ${{y}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+y=0$ LHS = ${{y}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+y$ Substitute the values of $\dfrac{dy}{dx}$ and $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ in LHS, we get: $\begin{aligned} & \Rightarrow {{y}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+y \\\ & \Rightarrow {{y}^{2}}\left( -\dfrac{y-x\dfrac{dy}{dx}}{{{y}^{2}}} \right)-x\left( -\dfrac{x}{y} \right)+y \\\ & \Rightarrow -\left( y-x\dfrac{dy}{dx} \right)+\dfrac{{{x}^{2}}}{y}+y \\\ & \Rightarrow -y+x\left( -\dfrac{x}{y} \right)+\dfrac{{{x}^{2}}}{y}+y \\\ & \Rightarrow -y-\dfrac{{{x}^{2}}}{y}+\dfrac{{{x}^{2}}}{y}+y \\\ & \Rightarrow 0=RHS \\\ \end{aligned}$ **Hence, proved that ${{y}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+y=0$** **Note:** Since x and y are given in terms of $\text{ }\\!\\!\theta\\!\\!\text{ }$, so differentiate them with respect to $\text{ }\\!\\!\theta\\!\\!\text{ }$. Do not divide the both equations and try to find $\dfrac{dy}{dx}$. That is a wrong method, because x and y do not depend on each other, they are dependent on $\text{ }\\!\\!\theta\\!\\!\text{ }$. Also, while double-differentiating, do not differentiate $\dfrac{dy}{d\theta }$ and $\dfrac{dx}{d\theta }$ with respect to $\text{ }\\!\\!\theta\\!\\!\text{ }$ again because it will give you the value of ${{d}^{2}}y$ and ${{d}^{2}}x$. But we require the value of $d{{x}^{2}}$. So, firstly get a relation between x and y and the get the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.