Question

If x=sint and y=sin mt ,prove that (1=x2)y2-x y1 + m2y= 0

Answer 1

In the questions there is a typing mistake.

LHS: (1-x²) d²y/dx² + x dy/dx + m² y

x = sin t,     dx/dt = cos t = √(1-x²)
y = sin mt,  dy/dt = m cos mt = m √(1-y²)
 =>  dy/dx = dx/dt ÷ dx/dt  
                 = m cos(mt) / cos t = m √(1-y²) / √(1-x²)
=> d²y/dx² = d/dx [dy/dx]
             = m [ √(1-x²) * {-y/√(1-y²)} * dy/dx - √(1-y²) {-x/√(1-x²) } ] / (1-x²)
             = - m² y / (1-x²)   +  m x √(1-y²) / (1-x²)³/²
                     (substituting the value of  dy/dx)...

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

Hence proved.