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प्रश्न
If = a cos mx + b sin mx, then show that y2 + m2y = 0.
उत्तर
y = a cos mx + b sin mx
y1 = a `"d"/"dx"` (cos mx) + b `"d"/"dx"` (sin mx)
`[ ∵ "d"/"dx" (sin "m"x) = cos "m"x "d"/"dx" ("m"x) = (cos "m"x) . "m"]`
= a(-sin mx) . m + b(cos mx) . m
= -am sin mx + bm cos mx
y2 = -am(cos mx) . m + bm(-sin mx) . m
= -am2 cos mx – bm2 sin mx
= -m2 [a cos mx + b sin mx]
= -m2y
∴ y2 + m2y = 0
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