Question

If y=sinpt,x=sint prove (1-x²)y²-xy₁+p²y=0

Answer 1

Step-by-step explanation:

It is given that

$y = sin \: pt...eq1 \\ \\ x = sin \: t ...eq2$

differentiate eq 1 and eq2 with respect to t

$\frac{dy}{dt} = p \: cos \: pt \\ \\ \frac{dx}{dt} = cos \: t$

$\frac{dy}{dx} ={y}^{'}= \frac{dy}{dt} \times \frac{dt}{dx} = \frac{p \: cos \: pt}{cos \: t} ...eq1 \\ \\ \frac{ {d}^{2}y }{d {x}^{2} } = \frac{ - cos \: t \: . {p}^{2}.sin \: pt - p.cos \: pt.( - sin \: t)}{ {cos}^{2}t } \times \frac{dt}{dx} \\ \\ \frac{ {d}^{2}y }{d {x}^{2} } = {y}^{''}= \frac{ - cos \: t \: . {p}^{2}.sin \: pt + p.cos \: pt.( sin \: t)}{ {cos}^{2}t } \times \frac{1}{cos \: t}...eq2$

Now put the values of eq1 and eq2 in the equation to prove

$(1 - {x}^{2} ) {y}^{''} - x {y}^{'} + {p}^{2} y = 0 \\ \\ (1 - {x}^{2} ) \Big(\frac{ - cos \: t \: . {p}^{2}.sin \: pt + p.cos \: pt.( sin \: t)}{ {cos}^{3}t }\Big) - x \Big(\frac{p \: cos \: pt}{cos \: t}\Big) + {p}^{2} y = 0 \\ \\ (1 - {sin}^{2}t ) \Big(\frac{ - cos \: t \: . {p}^{2}.sin \: pt + p.cos \: pt.( sin \: t)}{ {cos}^{3}t }\Big) - x \Big(\frac{p \: cos \: pt}{cos \: t}\Big) + {p}^{2} y = 0 \\ \\ {cos}^{2} t \Big(\frac{ - cos \: t \: . {p}^{2}.sin \: pt + p.cos \: pt.( sin \: t)}{ {cos}^{3}t }\Big) - x \Big(\frac{p \: cos \: pt}{cos \: t}\Big) + {p}^{2} y = 0 \\ \\ \Big(\frac{ - cos \: t \: . {p}^{2}.sin \: pt + p.cos \: pt.( sin \: t)}{ {cos}t }\Big) - x \Big(\frac{p \: cos \: pt}{cos \: t}\Big) + {p}^{2} y = 0 \\ \\ - cos \: t \: . {p}^{2}.sin \: pt + p.cos \: pt.( sin \: t) - x.p.cos \: pt + {p}^{2} cos \: t \: y = 0 \\ \\ - cos \: t \: . {p}^{2}.y +x.p.cos \: pt. - x.p.cos \: pt + {p}^{2} cos \: t \: .y = 0 \\ \\ since \: all \: terms \: cancels \: each \: other$

Hope it helps you.