Question

Let the solution curve $y=y(x)$ of the differential equation
$\frac{dy}{dx}-\frac{3x^5{\tan }^{-1}\left(x^3\right)}{{\left(1+x^6\right)}^{3/2}}y=2x\exp \left\{\frac{x^3-{\tan }^{-1}{x}^3}{\sqrt{\left(1+x^6\right)}}\right\}$ 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)

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