Question

If $y=y(x)$ is solution of differential equation $\frac{d y}{d x}=-x \sin x ; y(0)=0$, then

• A. $y>0$ if $x \in\left(\frac{-\pi}{2}, 0\right)$.
• B. $y<0$ if $x \in\left(\frac{-\pi}{2}, 0\right)$.
• C. number of inflection points of $y=y(x)$ in $\left(0, \frac{\pi}{2}\right)$ is 0 .
• D. $\frac{d^2 y}{d x^2}+y=-2 \sin x$

Answer 1

Answer: (A), (C), (D)

$ d y=-x \sin x \cdot d x $
$y=x \cos x-\sin x+c$
$y=x \cos x-\sin x \{\Theta c=0\} $
$=\cos x(x-\tan x) $
$y^{\prime \prime}=x \cos x-\sin x=-(x \cos x+\sin x)<0 \text { in }\left(0, \frac{\pi}{2}\right)$
$\Rightarrow \text { No inflection point. }$