If y=y(x), y ∈[0, (π/2)) is the solution of the differential equation sec y (d y/d x)- sin (x +y)- sin (x-y)=0, with y(0)=0 then 5 y prime((π/2)) is equal to.

Question

If y=y(x), y ∈[0, (π/2)) is the solution of the differential equation sec y (d y/d x)- sin (x +y)- sin (x-y)=0, with y(0)=0 then 5 y prime((π/2)) is equal to. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

sec y (d y/d x)=2 sin x cos y sec 2 y d y=2 sin x d x tan y=-2 cos x+c c=2 tan y=-2 cos x+2 ⇒ at x=(π/2) tan y=2 sec 2 y (d y/d x)=2 sin x 5 (d y/d x)=2

Q. If $y = y (x), y \in [0, \frac{π}{2})$ is the solution of the differential equation $sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0$ , with $y (0) = 0$ then $5 y^{'} (\frac{π}{2})$ is equal to______.

$sec y \frac{d y}{d x} = 2 sin x cos y$
$sec^{2} y d y = 2 sin x d x$
$tan y = - 2 cos x + c$
$c = 2$
$tan y = - 2 cos x + 2$
$\Rightarrow$ at $x = \frac{π}{2}$
$tan y = 2$
$sec^{2} y \frac{d y}{d x} = 2 sin x$
$5 \frac{d y}{d x} = 2$

Q. If y=y(x),y∈[0,2π​) is the solution of the differential equation secydxdy​−sin(x+y)−sin(x−y)=0, with y(0)=0 then 5y′(2π​) is equal to______.

secydxdy​=2sinxcosy sec2ydy=2sinxdx tany=−2cosx+c c=2 tany=−2cosx+2 ⇒ at x=2π​ tany=2 sec2ydxdy​=2sinx 5dxdy​=2

Q. If $y = y (x), y \in [0, \frac{π}{2})$ is the solution of the differential equation $sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0$ , with $y (0) = 0$ then $5 y^{'} (\frac{π}{2})$ is equal to______.

$sec y \frac{d y}{d x} = 2 sin x cos y$
$sec^{2} y d y = 2 sin x d x$
$tan y = - 2 cos x + c$
$c = 2$
$tan y = - 2 cos x + 2$
$\Rightarrow$ at $x = \frac{π}{2}$
$tan y = 2$
$sec^{2} y \frac{d y}{d x} = 2 sin x$
$5 \frac{d y}{d x} = 2$