JEE 2023 — Mathematics PYQ

Solution

Integrating Factor ($I.F.$):

$I.F.=e^{\int P(x)dx}=\exp \left(\int -\frac{3x^5{\tan }^{-1}(x^3)}{(1+x^6{)}^{3/2}}dx\right)$

Let $x^3=\tan \theta ⟹3x^{2}dx={\sec }^2\theta d\theta$:

$\int -\frac{\tan \theta \cdot \theta \cdot {\sec }^2\theta }{{\sec }^3\theta }d\theta =\int -\theta \sin \theta d\theta =\theta \cos \theta -\sin \theta$

$I.F.=\exp \left(\frac{{\tan }^{-1}{x}^3}{\sqrt{1+x^6}}-\frac{x^3}{\sqrt{1+x^6}}\right)=\exp \left(\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}\right)$

Solution of D.E.:

$y\cdot e^{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}}=\int 2x\cdot e^{\frac{x^3-{\tan }^{-1}{x}^3}{\sqrt{1+x^6}}}\cdot e^{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}}dx$

$y\cdot e^{\frac{{\tan }^{-1}{x}^3-x^3}{\sqrt{1+x^6}}}=\int 2xdx=x^2+c$

As it passes through $(0,0)⟹c=0$:

$y=x^2\exp \left(\frac{x^3-{\tan }^{-1}{x}^3}{\sqrt{1+x^6}}\right)$

For $x=1$:

$y(1)=1^2\cdot \exp \left(\frac{1-{\tan }^{-1}1}{\sqrt{1+1}}\right)=\exp \left(\frac{1-\pi /4}{\sqrt{2}}\right)=\exp \left(\frac{4-\pi }{4\sqrt{2}}\right)$

Correct Option: (4)