Question

If x = sint and y = sinpt prove that : (1 - xsquare)d2y/dx2 - xdy/dx + psquarey = 0

Answer 1

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 + p² 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.

Answer 2

x = sint and y = sinpt

t = sin^-1x and t = 1/psin^-1y

sin^-1x = 1/psin^-1y

differentiate wrt x
1/√( 1 -x²) = 1/P 1/√( 1 - y²) dy/dx

take square both sides

( 1 - y²) P² = (1 -x²)(dy/dx)²

again differentiate wrt x

( -2y)P²dy/dx = ( 1- x²)2dy/dx.d²y/dx² + (-2x)(xy/dx)²

-P²ydy/dx ={( 1 - x²)d²ydx -xdy/dx }dy/dx

-P²y = ( 1 - x²)d²y/dx² - xdy/dx

( 1 -x²)d²y/dx² -xdy/dx +P²y =0

hence proved