Question

If = a cos mx + b sin mx, then show that y₂ + m²y = 0.

Answer 1

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