Question

Let x dy/dx - y =x² (x cosx + sinx) is a differential equation. If f(π) = π then f" (π/2)+ f (π/2) =

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\displaystyle \: x \frac{dy}{dx} - y = {x}^{2} (x \cos x + \sin x)$

$\implies \: \displaystyle \: x {dy} - y \: dx = {x}^{2} (x \cos x + \sin x)dx$

$\implies \: \displaystyle \: \frac{ x {dy} - y \: dx}{ {x}^{2}} = (x \cos x + \sin x)dx$

$\implies \: \displaystyle \: d \bigg( \frac{y}{x} \bigg)= (x \cos x + \sin x)dx$

On integration

$\implies \: \displaystyle \: \frac{y}{x} = \displaystyle \int\limits_{}^{} (x \cos x + \sin x) \, dx$

$\implies \: \displaystyle \: \frac{y}{x} =x \sin x - \int\limits_{}^{} \sin x \, dx + \int\limits_{}^{} \sin x \, dx + c$

$\implies \: \displaystyle \: \frac{y}{x} =x \sin x + c \: \: ...(1)$

Where C is a constant

Now

$y = \pi \: \: when \: \: x = \pi \: \: gives$

$\displaystyle \: \frac{\pi}{\pi} =\pi \sin \pi + c$

$\therefore \: c = 1$

From Equation (1)

$\: \displaystyle \: \frac{y}{x} =x \sin x + 1$

$\implies \: \displaystyle \: {y} = {x}^{2} \sin x + x \: \:$

Differentiating both sides with respect to x two times

$\: \displaystyle \: \frac{dy}{dx} = {x}^{2} \cos x + 2x \sin x + 1 \: \:$

$\: \displaystyle \: \frac{ {d}^{2} y}{d {x}^{2} } = - {x}^{2} \sin x + 2xcosx + 2x \cos x + 2sinx\: \:$

$\implies \: \: \displaystyle \: \frac{ {d}^{2} y}{d {x}^{2} } = - {x}^{2} \sin x + 4xcosx + 2sinx\: \:$

$for \: \: x = \frac{\pi}{2}$

$\: \displaystyle \: \frac{dy}{dx} =\pi+ 1 \: \:$

And

$\: \displaystyle \: \frac{ {d}^{2} y}{d {x}^{2} } = - \frac{ {\pi}^{2} }{4} + 2$

$\:  \displaystyle \:     y  =    \frac{ {\pi}^{2} }{4}  +  \frac{\pi}{2}$

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So the required value is

$= \:  \displaystyle \:    +  \frac{\pi}{2} + 2$