# If $x=a\:cos\theta+b\:sin\theta$ and $y=a\:sin\theta-b\:cos\theta$, then show that $y^2.\large\frac{d^2y}{dx^2}$$-x\large\frac{dy}{dx}$$+y=0$

$(a)\;1\qquad(b)\;0\qquad(c)\;2\qquad(d)\;4$

Given: $x=a\:cos\theta+b\:sin\theta$ and $y=a\:sin\theta-b\:cos\theta$
Step 1
Differentiating $y$ and $x$ with respect to $\theta$ we get
$\large\frac{dx}{d\theta}$$=-a\:sin\theta+b\:cos\theta and \large\frac{dy}{d\theta}$$=a\:cos\theta+b\:sin\theta$
$\large\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{a\:cos\theta+b\:sin\theta}{b\:cos\theta-a\:sin\theta}$
Step 2
$\large\frac{d^2y}{dx^2}=\frac{d}{d\theta}(\frac{dy}{dx}).\frac{d\theta}{dx}$
$\frac{d}{d\theta}(\frac{dy}{dx})=\frac{(b\:cos\theta-a\:sin\theta)(-a\:sin\theta+b\:cos\theta)-(a\:cos\theta+b\:sin\theta)(-b\:sin\theta-a\:cos\theta)}{(b\:cos\theta-a\:sin\theta)^2}$
$=\large\frac{(b\:cos\theta-a\:sin\theta)^2+(a\:cos\theta+b\:sin\theta)^2}{(b\:cos\theta-a\:sin\theta)^2}$
$=\large\frac{b^2\:cos^2\theta+a^2\:sin^2\theta+a^2\:cos^2\theta+b^2\:sin^2\theta}{(b\:cos\theta-a\:sin\theta)^2}$
$\Rightarrow\:\frac{d}{d\theta}(\frac{dy}{dx})=\large\frac{a^2+b^2}{(b\:cos\theta-a\:sin\theta)^2}$
Step 3
$\large\frac{d\theta}{dx}=\frac{1}{b\:cos\theta-a\:sin\theta}$
$\Rightarrow\:\frac{d}{d\theta}(\frac{dy}{dx}).\frac{d\theta}{dx}=\large\frac{a^2+b^2}{(b\:cos\theta-a\:sin\theta)^2}\times \frac{1}{b\:cos\theta-a\:sin\theta}=\frac{a^2+b^2}{(b\:cos\theta-a\:sin\theta)^3}$
$\Rightarrow\:\large\frac{d^2y}{dx^2}=\frac{a^2+b^2}{(b\:cos\theta-a\:cos\theta)^3}$
Step 4
$y^2=(a\:sin\theta-b\:cos\theta)^2$
$y^2.\large\frac{d^2y}{dx^2}$$=(a\:sin\theta-b\:cos\theta)^2\times \frac{a^2+b^2}{(b\:cos\theta-a\:sin\theta)^3} =\frac{a^2+b^2}{b\:cos\theta-a\:sin\theta} x.\large\frac{dy}{dx}$$=(a\:cos\theta+b\:sin\theta)\times \frac{a\:cos\theta+b\:sin\theta}{b\:cos\theta-a\:sin\theta}$
$=\frac{(a\:cos\theta+b\:sin\theta)^2}{b\:cos\theta-a\:sin\theta}$
Step 5
$y^2\large\frac{d^2y}{dx^2}$$-x\large\frac{dy}{dx}$$+y$=
$=\frac{a^2+b^2}{b\:cos\theta-a\:sin\theta}-\frac{(a\:cos\theta+b\:sin\theta)^2}{b\:cos\theta-a\:sin\theta}+$$(a\:sin\theta-b\:cos\theta) =\large\frac{a^2+b^2-(a\:cos\theta+b\:sin\theta)^2-(b\:cos\theta-a\:sin\theta)^2}{b\:cos\theta-a\:sin\theta} =\large\frac{a^2+b^2-(a^2\:cos^2\theta+b^2\:sin^2\theta+b^2\:cos^2\theta+a^2\:sin^2\theta}{b\:cos\theta-a\:sin\theta} =\large\frac{a^2+b^2-(a^2+b^2)}{b\:cos\theta-a\:sin\theta}$$=0$
Hence proved.
answered Mar 23, 2014