Let the solution curve y = y ( x ) of the differential equation d y d x - 3 x 5 tan - 1 ( x 3 ) ( 1 + x 6 ) 3 / 2 y = 2 x exp { x 3 - tan - 1 x 3 ( 1 + x 6 ) } pass through the origin. Then y ( 1 ) is equal to [2023]

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Solution

Q.
Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ pass through the origin. Then $y (1)$ is equal to [2023]

1 $\exp {\frac{4 + π}{4 \sqrt{2}}}$

2 $\exp {\frac{4 - π}{4 \sqrt{2}}}$

3 $\exp {\frac{1 - π}{4 \sqrt{2}}}$

4 $\exp {\frac{π - 4}{4 \sqrt{2}}}$

Ans.
(2)

Given differentiate equation is

$\frac{d y}{d x} - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}} y = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{(1 + x^{6})}}}$ ,

Which is in the form $\frac{d y}{d x} + P y = Q$

Here, $P = - \frac{3 x^{5} \tan^{- 1} (x^{3})}{{(1 + x^{6})}^{3 / 2}}, Q = 2 x \exp {\frac{x^{3} - \tan^{- 1} x^{3}}{\sqrt{1 + x^{6}}}}$

Let $I = - \int \frac{3 x^{5} \tan^{- 1} x^{3}}{{(1 + x^{6})}^{3 / 2}} d x$

Put $x^{3} = \tan θ \Rightarrow 3 x^{2} d x = \sec^{2} θ d θ$

$I = - \int \frac{\tan θ \cdot θ}{\sec^{3} θ} \times \sec^{2} θ d θ = - \int θ \cdot \sin θ d θ$

$= - [θ \int \sin θ d θ - \int (\int \sin θ d θ) d θ]$

$= - [- θ \cdot \cos θ + \sin θ] = - [\frac{x^{3}}{\sqrt{1 + x^{6}}} - \tan^{- 1} (x^{3}) \cdot \frac{1}{\sqrt{1 + x^{6}}}]$

$∴$ $I.F. = e^{\int p d x} = e^{I}$

$Solution of the D.E. is$

$y \cdot e^{I} = \int e^{I} \cdot 2 x \cdot \exp {\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} d x$

$= \int 2 x \cdot e^{\frac{- x^{3} + \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} \cdot e^{\frac{x^{3} - \tan^{- 1} (x^{3})}{\sqrt{1 + x^{6}}}} d x$

$= \int 2 x \cdot d x = x^{2} + c$ $∴$ $y e^{I} = x^{2} + c$

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

This curve passes through the origin.

So, $0 = 0 + c \Rightarrow c = 0$

The required solution is

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

At $x = 1$ $y (1) = \exp (\frac{1 - \frac{π}{4}}{\sqrt{2}}) = \exp (\frac{4 - π}{4 \sqrt{2}})$

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