If the differential equation y d x-x d y+y2 cos x d x=0 satisfies the initial condition y((π/2))=1, then y (π) equals

Question

If the differential equation y d x-x d y+y2 cos x d x=0 satisfies the initial condition y((π/2))=1, then y (π) equals (A)  (2 π/π+2) (B) (2 π/π+3) (C) (2 π/π+1) (D) (π+2/2)

Tardigrade Admin · Accepted Answer

(y d x-x d y/y2)+ cos x d x=0 d((x/y))+ cos x d x=0 (x/y)+ sin x=c x+y sin x=c y (π/2)+1=c ⇒ x+y sin x=((π/2)+1) y π=((π/2)+1) y ⇒ y=(2 π/π+2)

Q. If the differential equation $y d x - x d y + y^{2} cos x d x = 0$ satisfies the initial condition $y (\frac{π}{2}) = 1$ , then $y (π)$ equals

$\frac{2 π}{π + 2}$

$\frac{2 π}{π + 3}$

$\frac{2 π}{π + 1}$

$\frac{π + 2}{2}$

$\frac{y d x - x d y}{y ^{2}} + cos x d x = 0$
$d (\frac{x}{y}) + cos x d x = 0$
$\frac{x}{y} + sin x = c$
$x + y sin x = cy$
$\frac{π}{2} + 1 = c$
$\Rightarrow x + y sin x = (\frac{π}{2} + 1) y$
$π = (\frac{π}{2} + 1) y \Rightarrow y = \frac{2 π}{π + 2}$

Q. If the differential equation ydx−xdy+y2cosxdx=0 satisfies the initial condition y(2π​)=1, then y(π) equals

π+22π​

π+32π​

π+12π​

2π+2​