If x = t, y = pt, then - Vidyadip

MCQ

If $x = \sin t$, $y = \cos pt$, then

A
$(1 - {x^2}){y_2} + x{y_1} + {p^2}y = 0$
B
$(1 - {x^2}){y_2} + x{y_1} - {p^2}y = 0$
C
$(1 + {x^2}){y_2} - x{y_1} + {p^2}y = 0$
✓
$(1 - {x^2}){y_2} - x{y_1} + {p^2}y = 0$

✓

Answer

Correct option: D.

$(1 - {x^2}){y_2} - x{y_1} + {p^2}y = 0$

d
(d) $x = \sin t$, $y = \cos pt$

$\frac{{dx}}{{dt}} = \cos t$; $\frac{{dy}}{{dt}} = - p\sin pt$; $\frac{{dy}}{{dx}} = \frac{{ - p\sin pt}}{{\cos t}}$

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \cos t\,{p^2}\cos pt(dt/dx) - p\sin pt\sin t(dt/dx)}}{{{{\cos }^2}t}}$

==> $(1 - {x^2})\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} + {p^2}y = 0$

or $(1 - {x^2}){y_2} - x{y_1} + {p^2}y = 0$.

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