Let y = y ( x ) , y 0 be a solution curve of the differential equation ( 1 + x 2 ) ? d y = y ( x - y ) d x . If y ( 0 ) = 1 and y ( 2 2 ) =, then [2023]

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Solution

Q.
Let $y = y (x), y > 0$ be a solution curve of the differential equation $(1 + x^{2}) d y = y (x - y) d x .$ If $y (0) = 1$ and $y (2 \sqrt{2}) = β$ , then [2023]

1 $e^{3 β^{- 1}} = e (5 + \sqrt{2})$

2 $e^{3 β^{- 1}} = e (3 + 2 \sqrt{2})$

3 $e^{β^{- 1}} = e^{- 2} (3 + 2 \sqrt{2})$

4 $e^{β^{- 1}} = e^{- 2} (5 + \sqrt{2})$

Ans.
(2)

$Given differential equation is$

$(1 + x^{2}) d y = y (x - y) d x$

$\Rightarrow \frac{d y}{d x} = \frac{x y - y^{2}}{1 + x^{2}} \Rightarrow \frac{d y}{d x} = \frac{x y}{1 + x^{2}} - \frac{y^{2}}{1 + x^{2}}$

$\frac{d y}{d x} - \frac{x y}{1 + x^{2}} = \frac{- y^{2}}{1 + x^{2}}$

$\Rightarrow \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} \frac{x}{1 + x^{2}} = \frac{- 1}{1 + x^{2}}$ ...(i)

Put $\frac{1}{y} = z \Rightarrow \frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d z}{d x}$

So, (i) becomes

$\frac{d z}{d x} + \frac{x z}{1 + x^{2}} = \frac{1}{1 + x^{2}}$

Which is a linear differential equation.

$I.F. = e^{\int \frac{x}{1 + x^{2}} d x} = e^{\int \frac{x}{1 + x^{2}} d x} = e^{\ln \sqrt{1 + x^{2}}} = \sqrt{1 + x^{2}}$

$∴$ $Solution is z \sqrt{1 + x^{2}} = \int \frac{\sqrt{1 + x^{2}}}{1 + x^{2}} d x$

$z \sqrt{1 + x^{2}} = \int \frac{d x}{\sqrt{1 + x^{2}}}$

$z \sqrt{1 + x^{2}} = \log$ $| x + \sqrt{1 + x^{2}} |$ $+ C$

$\frac{\sqrt{1 + x^{2}}}{y} = \log$ $| x + \sqrt{1 + x^{2}} |$ $+ C$ ...(ii)

$We have, y (0) = 1$

So, from (ii), $1 = \log 1 + C \Rightarrow C = 1$

$∴$ $(ii) becomes$

$\Rightarrow \frac{\sqrt{1 + x^{2}}}{y} = \log$ $| x + \sqrt{1 + x^{2}} |$ $+ 1$

$\Rightarrow$ $\frac{\sqrt{1 + x^{2}}}{y} = \log$ $| x + \sqrt{1 + x^{2}} |$ $+ \log e$

$\Rightarrow$ $\frac{\sqrt{1 + x^{2}}}{y} = \log (e (x + \sqrt{1 + x^{2}}))$

$\Rightarrow$ Also, $y (2 \sqrt{2}) = β$

$\frac{3}{β} = \log (e (2 \sqrt{2} + 3)) \Rightarrow e^{\frac{3}{β}} = (e (3 + 2 \sqrt{2}))$

$\Rightarrow e^{3 β^{- 1}} = (e (3 + 2 \sqrt{2}))$

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