Question

If x = sin t and, y = sin pt, then prove that (1 - x²)d²y/dx² - xdy/dx + p²y = 0.

Comments

x=sin t
y=sin pt
dx/dt=cost=√(1-sin²t)=√(1-x²)
dy/dt=pcos pt=p√(1-sin²pt)=p√(1-y²)
dy/dx=p√(1-x²)/√(1-y²)  ------eq1
again differentiate wrt x:
(1-x²)d²y/dx² =p{ √(1-x²) × -2y/2√(1-y²) - √(1-y²)×-2x/2√(1-x²)}
(1-x²)d²y/dx²=-py√(1-x²)/√(1-y²)  + xp√(1-y²)/√(1-x²)}
from eq1:
(1-x²)d²y/dx²=-p²y  + xdy/dx
(1-x²)d²y/dx² -  xdy/dx + p²y -------proved

Answer 1

Given:

x = sin t and y = sin pt

Differentiating both sides with respect to t, we get

\(\frac{dx}{dt} = \cos t\) and \(\frac{dy}{dt} = p \cos pt\)

⇒ \(\frac{dy}{dx} = \cfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{p \cos pt}{\cos t}\)

Differentiating both sides with respect to x, we get

Answer 2

Given, x = sin t and y = sin pt

On differentiating both sides w.r.t. t, we get

Now on differentiating both sides w.r.t. x, we get

Answer 3

x = sin t

dx/dt = cost

y = sin pt

dy/dt =pcospt

so dy/dx = (pcospt)/cost =pcosptsect

=>d²y/dx² = [(pcospt)*sect*tant - p²sect*sinpt]dt/dx

=>d²y/dx² = [(pcospt)*sect*sint*sect - p²sect*sinpt]1/cost

=>cos²td²y/dx² = [(pcospt)*sect*sint*sect*cos²t - p²sect*sinpt*cos²t]1/cost

=>(1-sin²t)d²y/dx² = [(pcospt)*sint- p²sinpt*cost]1/cost

=>(1-sin²t)d²y/dx² = [(pcospt)cost*sint- p²sinpt

=>(1-x²)d²y/dx² = xdy/dx- p²y

^{=>(1-x2)d2y/dx2 - xdy/dx+p2y=0}