Question

if x=sint and y= sinpt then prove that(1-y2)d2y/dx2 - x dy/dx + p2y= 0

Answer 1

There is a mistake in the qn. typing error..
To prove that    LHS= (1-x²) d²y/dx² + x dy/dx + m² y  = 0 RHS

x = sin t,     dx/dt = cos t = √(1-x²)
y = sin pt,  dy/dt = p cos pt = p √(1-y²)

 => dy/dx = dx/dt ÷ dx/dt  
                = p cos(pt) / cos t = p √(1-y²) / √(1-x²)

=> d²y/dx² = d/dx [dy/dx]
             = p [ √(1-x²) * {-y/√(1-y²)} * dy/dx - √(1-y²) {-x/√(1-x²) } ] / (1-x²)
             = - p² y / (1-x²)   +  p x √(1-y²) / (1-x²)³/²
                     (substituting the value of  dy/dx)...

LHS= (1-x²) d²y/dx² + x dy/dx + m² y
       = [ - p² y  + p x √(1-y²) /√(1-x²) ] + [ p x √(1-y²) /√(1-x²) ]  + p² y
       = 0   as the terms cancel out.

Hence proved.