Question

${\tan }^{-1}\frac{33\left(x^5-y^5\right)}{31\left(x^5+y^5\right)}=\frac{3\pi }{4}\left[\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\right],atx=1is$

Answer 1

${\tan }^{-1}\frac{33(x^5-y^5)}{31(x^5+y^5)}=\frac{3\pi }{4}$

$\Rightarrow \frac{33(x^5-y^5)}{31(x^5+y^5)}=\tan \frac{3\pi }{4}$

$\Rightarrow \frac{33(x^5-y^5)}{31(x^5+y^5)}=-1$

$\Rightarrow 33(x^5-y^5)=-31(x^5+y^5)$

$\Rightarrow 33x^5-33y^5=-31x^5-31y^5$

$\Rightarrow 33x^5+31x^5=33y^5-31y^5$

$\Rightarrow 64x^5=2y^5$

$\Rightarrow y^5=32x^5$

Differentiating both sides w.r.t $x$ we get

$\Rightarrow 5y^4\frac{dy}{dx}=160x^4$

$\Rightarrow y^4\frac{dy}{dx}=32x^4$ .......$\left(1\right)$

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{x=1}={\left[\frac{32x^4}{y^4}\right]}_{x=1}=\frac{32}{y^4}$

Differentiating $\left(1\right)$ both sides w.r.t $x$ we get

$\Rightarrow 4y^3{\left(\frac{dy}{dx}\right)}^2+y^4\frac{d^{2}y}{d{x}^2}=128x^3$

$\Rightarrow 4y^3{\left(\frac{dy}{dx}\right)}^2+y^4\frac{d^{2}y}{d{x}^2}=128x^3$ at $x=1$

$\Rightarrow 4y^3{\left(\frac{32}{y^4}\right)}^2+y^4\frac{d^{2}y}{d{x}^2}=128$

$\Rightarrow y^4\frac{d^{2}y}{d{x}^2}=128-4y^3{\left(\frac{32}{y^4}\right)}^2$

$\Rightarrow y^4\frac{d^{2}y}{d{x}^2}=128-4y^3\times \frac{1,024‬}{y^8}$

$\Rightarrow y^4\frac{d^{2}y}{d{x}^2}=128-\frac{4,096}{y^5}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{128y^5-4,096}{y^9}$

$\therefore \frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=\frac{128y^5-4,096}{y^9}+\frac{32}{y^4}$

Again when $x=1y^5=32x^5=32$

$\therefore y=2$

$\therefore \frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=\frac{128\times 2^5-4,096}{2^9}+\frac{32}{2^4}$

$\therefore \frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=\frac{128\times 2^5-4,096}{2^9}+\frac{32}{2^4}$

$\therefore \frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=\frac{4096-4,096}{2^9}+2=0+2=2$

Hence $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=2$