Question

y = sin (sinx) then prove that y′′ + (tanx) y′ + y cos²x = 0

Answer 1

Such expression can be easily proved using implict differention. 

⇒ y′ = cos (sin x) cos x 

⇒ sec x.y′ = cos (sin x) 

again differentiating w.r.t x, we can get 

secx y′′ + y′ sec x tan x = – sin (sin x) cos x 

⇒ tanx y′ = – y . cos² x 

⇒ y′′ +(tanx) y′ + y cos²x = 0