Question

Let the solution curve $y=y(x)$ of the differential equation
$\frac{d y}{d x}-\frac{3 x^5 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^3-\tan ^{-1} x^3}{\sqrt{\left(1+x^6\right)}}\right\}$ pass through the origin. Then $y(1)$ is equal to :

• A. $\exp \left(\frac{4+\pi}{4 \sqrt{2}}\right)$
• B. $\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)$
• C. $\exp \left(\frac{1-\pi}{4 \sqrt{2}}\right)$
• D. $\exp \left(\frac{\pi-4}{4 \sqrt{2}}\right)$

Answer 1

Answer: (B)

$ \frac{d y}{d x}+\left(\frac{-3 x^5 \tan ^{-1} x^3}{\left(1+x^6\right)^{3 / 2}}\right) y=2 e^{\left\{\frac{x-\tan x}{\left.\sqrt{1+x^6}\right\}}\right\}} $
$ \text { I.F. }=e^{\int \frac{-3 x^5 \tan ^{-1} x^3}{\left(1+x^6\right)^{32}} d x} $
$ =e^{\frac{\tan ^{-1} x^3-x^3}{\sqrt{1+x^6}}}$
Solution of differential equation
$y \cdot e^{\frac{\tan ^{-1} x^3-x^3}{\sqrt{1+x^6}}}=\int 2 x e^{\left(\frac{x^3-\tan ^{-1} x^3}{\sqrt{1+x^6}}\right)} \cdot e^{\left(\frac{\tan ^{-1}\left(x^3\right)-x^3}{\sqrt{1+x^6}}\right)} d x$
$ =\int 2 x d x+c$
$y \cdot e^{\frac{\tan ^{-1} x^3-x^3}{\sqrt{1+x^6}}}=x^2+c$
Also it passes through origin
$c =0$
$ y(1) \cdot e^{\frac{\tan ^{-1}(1)-1}{\sqrt{2}}}=1 $
$ y(1) \cdot e^{\frac{\frac{\pi}{4}-1}{\sqrt{2}}}=1 $
$y(1) \cdot e^{\frac{\pi-4}{4 \sqrt{2}}}=1$
$ y(1)=\frac{1}{e^{\frac{\pi-4}{4 \sqrt{2}}}}=e^{\frac{4-\pi}{4 \sqrt{2}}}$