If x=asin pt and y=bcos pt ,then show that (a?-x?)y d?y/dx? +b?=0.

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Solution

$D e a r S t u d e n t G i v e n x = a \sin p t y = b \cos p t S o \frac{d x}{d t} = a p \cos p t \frac{d y}{d t} = - b p \sin p t N o w, \frac{d y}{d x} = \frac{- b p \sin p t}{a p \cos p t} = - \frac{b}{a} \tan p t T h e r e f o r e, \frac{d^{2} y}{d x^{2}} = - \frac{b p s e c^{2} p t}{a} \frac{d t}{d x} = \frac{- b p s e c^{2} p t}{a} \times \frac{1}{p a \cos p t} = \frac{- b^{2}}{(a^{2} - x^{2}) y} T h e r e f o r e (a^{2} - x^{2}) y \frac{d^{2} y}{d x^{2}} + b^{2} = 0 R e g a r d s$

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$i f x = \sqrt{a^{s i n^{- 1} t}}, y = \sqrt{a^{c o s^{- 1} t}}, t h e n s h o w t h a t \frac{d y}{d x} = - \frac{y}{x} .$

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