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यदि ` y = Pe^ ( ax ) + Q e ^( bx ) ` है , तो दर्शाइए कि ` ( d ^2 y ) / ( dx ^ 2 ) - ( a+ b ) ( dy ) / ( dx ) + aby = 0 `

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यहाँ
` y = Pe ^ ( ax ) + Qe ^( bx ) " " `...(1)
दोनों पक्षों का x के सापेक्ष अवकलां करने पर,
` ( d y ) /( dx ) = a Pe ^ ( ax ) + b Q e ^( b x ) " " `...(2 )
दोनों पक्षों का x के सापेक्ष पुनः अवकलन करने पर,
`( d ^2 y ) /( dx ^ 2 ) = a P ( ae^ ax ) + bQ ( be ^( bx )) `
` rArr ( d^ 2 y ) /( dx ^ 2 ) = a ^ 2 P e ^( ax ) + b^ 2 Q e^ ( bx ) " " `...(3)
अब, ` ( d ^2 y ) /( dx ^ 2 ) - ( a + b ) ( d y ) /( dx ) + aby `
` = a ^ 2 P e ^( ax) + b ^ 2 Q e ^ ( bx ) - ( a + b ) ( a P e ^( ax) + b Q e ^( bx )) + ab ( P e ^( ax ) + Q e^ ( bx ) ) `
[समी. (1 ) , (2 ) और (3 ) से ]
` = a ^ 2 P e ^( ax) + b ^ 2 Q e ^ ( bx ) - a ^2 P e ^( ax ) - ab Q e ^( bx ) - ab P e ^ ( ax ) - b ^ 2 Q e ^( bx ) + ab P e ^( ax ) + ab Qe ^( bx ) `
` = 0 `
अतः ` ( d ^2 y ) /( dx ^ 2 ) - ( a + b) ( dy ) /(dx ) + ab y = 0 `

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