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यदि ` x = a cos theta + b sin theta ` और ` y = a sin theta - b cos theta ` हो , तो दर्शाइए कि -
` y ^2 ( d^ 2y ) / (dx ^2 ) - x ( dy ) / ( dx ) + y = 0 `

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दिया गया है -
` x = a cos theta + b sin theta " " `...(1)
` y = a sin theta - bcos theta " " `...(2)
समीकरण (1 ) का ` theta ` के सापेक्ष अवकलन करने पर,
` therefore (dx ) /( d theta ) = - a sin theta + bcos theta `
` = - (a sin theta - b cos theta ) = - y `
समीकरण (2 ) का ` theta ` के सापेक्ष अवकलन करने पर
और ` ( dy ) / ( d theta ) = a cos theta + b sin theta = x `
अब, ` ( d y ) /( d x ) = ( dy // d theta ) /( dx // d theta ) `
` rArr ( dy ) / ( dx ) = ( x ) /( - y ) `
` rArr y (dy ) / ( dx ) + x = 0 " " `...(3)
समी. (3 ) का x के सापेक्ष अवकलन करने पर,
` y ( d^ 2 y ) / ( dx^ 2 ) + ( ( dy ) /( dx ) )^ 2 + 1 = 0 `
` rArr y ( d^ 2y ) /( dx ^ 2 )) + ( ( dy ) / (dx )) ((dy ) / ( dx )) + 1 = 0 `
` rArr y ( d^2 y ) /( dx^ 2 ) + ( - ( x ) /( y ) ) ( dy ) /(dx ) + 1 = 0 `
`" "[ therefore (dy ) /(dx ) = - ( x ) /( y ) ] `
` rArr y ^ 2 ( d^ 2 y ) / (dx ^2 ) - x ( d y ) / ( dx ) + y = 0 `

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