Question

If $y=(x^2+1)\sin x$ then ${\left(\frac{\pi }{2}\right)}^2-y_{20}\left(\frac{\pi }{2}\right)$ is equal to

Answer 1

$y=(x^2+1)\sin x$
$y_1=(x^2+1)\cos x+2x\sin x$
$y_2=-(x^2+1)\sin x+4x\cos x+2\sin x$
$\Rightarrow y_2=-y+4x\cos x+2\sin x$
$y_3=-y_1-4x\sin x+6\cos x$
$\Rightarrow y_3=-(x^2+1)\cos x-6x\sin x+6\cos x$
$y_4=-y_2-4x\cos x-10\sin x$
$\Rightarrow y_4=(x^2+1)\sin x-8x\cos x-12\sin x$
$\Rightarrow y_5=(x^2+1)\cos x+10x\sin x-20\cos x$
$\Rightarrow y_6=-(x^2+1)\sin x+12x\cos x+30\sin x$
$\Rightarrow y_7=-y_1-12x\sin x+42\cos x$
$\Rightarrow y_8=(x^2+1)\sin x-16x\cos x-56\sin x$
So, we can conclude,
$y_{4n}=(x^2+1)\sin x-2\left(4n\right)\cos x-\left(4n\right)(4n-1)\sin x$
Put $n=5$ in above eqn
$y_{20}=(x^2+1)\sin x-40\cos x-380\sin x$
$y_{20}\left(\frac{\pi }{2}\right)=\left(\right(\frac{\pi }{2}{)}^2+1)-380$
$\Rightarrow (\frac{\pi }{2}{)}^2-y_{20}(\frac{\pi }{2})=379$