Suppose (cos)2 y⋅ (d y/d x)=sinâ¡(x + y)+sinâ¡(x - y),|x|≤ (π /2) and |y|≤ (π /2) . If y((π /3))=-(π /2) , then y((π /2)) is

Question

Suppose (cos)2 y⋅ (d y/d x)=sinâ¡(x + y)+sinâ¡(x - y),|x|≤ (π /2) and |y|≤ (π /2) . If y((π /3))=-(π /2) , then y((π /2)) is (A) 0 (B) (π /4) (C) -(π /4) (D) (π /2)

Tardigrade Admin · Accepted Answer

cos2 y(d y/d x)=2sinâ¡xcosâ¡y ⇒ cos ydy=2sin â¡ xdx ⇒ sin y+2cos â¡ x=c Now, y((π /3))=-(π /2)⇒ c=0 So sin y=-2cos â¡ x At x=(π /2),sin y=0⇒ y((π /2))=0

Q. Suppose $(cos)^{2} y \cdot \frac{d y}{d x} = s in (x + y) + s in (x - y), ∣ x ∣ \leq \frac{π}{2}$ and $∣ y ∣ \leq \frac{π}{2}$ . If $y (\frac{π}{3}) = - \frac{π}{2}$ , then $y (\frac{π}{2})$ is

$0$

$\frac{π}{4}$

$- \frac{π}{4}$

$\frac{π}{2}$

$co s^{2} y \frac{d y}{d x} = 2 s in x cos y$
$\Rightarrow cosy d y = 2 s in x d x$
$\Rightarrow s in y + 2 cos x = c$
Now, $y (\frac{π}{3}) = - \frac{π}{2} \Rightarrow c = 0$
So $s in y = - 2 cos x$
At $x = \frac{π}{2}, s in y = 0 \Rightarrow y (\frac{π}{2}) = 0$

Q. Suppose (cos)2y⋅dxdy​=sin⁡(x+y)+sin⁡(x−y),∣x∣≤2π​ and ∣y∣≤2π​ . If y(3π​)=−2π​ , then y(2π​) is

0

4π​

−4π​

2π​

cos2ydxdy​=2sin⁡xcos⁡y ⇒cosydy=2sin⁡xdx ⇒siny+2cos⁡x=c Now, y(3π​)=−2π​⇒c=0 So siny=−2cos⁡x At x=2π​,siny=0⇒y(2π​)=0

Q. Suppose $(cos)^{2} y \cdot \frac{d y}{d x} = s in (x + y) + s in (x - y), ∣ x ∣ \leq \frac{π}{2}$ and $∣ y ∣ \leq \frac{π}{2}$ . If $y (\frac{π}{3}) = - \frac{π}{2}$ , then $y (\frac{π}{2})$ is

$0$

$\frac{π}{4}$

$- \frac{π}{4}$

$\frac{π}{2}$

$co s^{2} y \frac{d y}{d x} = 2 s in x cos y$
$\Rightarrow cosy d y = 2 s in x d x$
$\Rightarrow s in y + 2 cos x = c$
Now, $y (\frac{π}{3}) = - \frac{π}{2} \Rightarrow c = 0$
So $s in y = - 2 cos x$
At $x = \frac{π}{2}, s in y = 0 \Rightarrow y (\frac{π}{2}) = 0$