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

Hint: Since we are given two equations:
 $\begin{align}
  & x=a\cos \text{ }\!\!\theta\!\!\text{ }+b\sin \text{ }\!\!\theta\!\!\text{ }......\text{(1)} \\
 & y=a\sin \text{ }\!\!\theta\!\!\text{ }-b\cos \text{ }\!\!\theta\!\!\text{ }......\text{(2)} \\
\end{align}$
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:
$\begin{align}
  & x=a\cos \theta +b\sin \theta ......(1) \\
 & y=a\sin \theta -b\cos \theta ......(2) \\
\end{align}$
Now, differentiation equation (1) and (2) with respect to $\theta $, we get:
$\begin{align}
  & \dfrac{dx}{d\theta }=-a\sin \theta +b\cos \theta ......(3) \\
 & \dfrac{dy}{d\theta }=a\cos \theta +b\sin \theta ......(4) \\
\end{align}$
\[\left[ \because \dfrac{d}{d\theta }\left( \sin \theta \right)=\cos \theta ;\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta \right]\]
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:
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{\left( 1\times y \right)-\left( x\times \dfrac{dy}{dx} \right)}{{{y}^{2}}} \right) \\
 & =-\left( \dfrac{y-x\dfrac{dy}{dx}}{{{y}^{2}}} \right)......(7)
\end{align}\]
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:
$\begin{align}
  & \Rightarrow {{y}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+y \\
 & \Rightarrow {{y}^{2}}\left( -\dfrac{y-x\dfrac{dy}{dx}}{{{y}^{2}}} \right)-x\left( -\dfrac{x}{y} \right)+y \\
 & \Rightarrow -\left( y-x\dfrac{dy}{dx} \right)+\dfrac{{{x}^{2}}}{y}+y \\
 & \Rightarrow -y+x\left( -\dfrac{x}{y} \right)+\dfrac{{{x}^{2}}}{y}+y \\
 & \Rightarrow -y-\dfrac{{{x}^{2}}}{y}+\dfrac{{{x}^{2}}}{y}+y \\
 & \Rightarrow 0=RHS \\
\end{align}$

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 $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.