Question

If = a cos mx + b sin mx, then show that y₂ + m² y = 0.

Answer 1

y = a cos mx + b sin mx 

y₁ = a \(\frac{d}{dx}\)(cos mx) + b \(\frac{d}{dx}\)(sin mx) 

[∵ \(\frac{d}{dx}\)(sin mx) = cos mx \(\frac{d}{dx}\)(mx) = (cos mx).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] 

= -m² y 

∴ y₂ + m² y = 0