Question

⁕Find the particular solution of the differential equation
$\sf(1 + e ^{2x} )dy + (1 + y {}^{2} )e { }^{x} dx = 0 \:$
Given that y=1 when x=0

Solve Only If you know the answer
​

Answer 1

Solution:-

=> For particular solution First of all we have to find general equation

$\rm \implies \: (1 + {e}^{2x} )dy +( 1 + {y}^{2} ) {e}^{x} dx$

Now using variable separation method

$\rm \implies\: (1 + {e}^{2x} )dy = - (1 + {y}^{2} ) {e}^{x} dx$

$\rm \implies \: dy = \dfrac{ - (1 + {y}^{2}) {e}^{x} }{1 + {e}^{2x} } dx$

$\rm \implies \: \dfrac{dy}{1 + {y}^{2} } = - \dfrac{ {e}^{x} dx}{1 + {e}^{2x} }$

Now integrate on both side

$\rm \implies \: \int\dfrac{dy}{1 + {y}^{2} } = - \int\dfrac{ {e}^{x} dx}{1 + {e}^{2x} }$

Now take

$\rm \implies \int \dfrac{dy}{1 + {y}^{2} } = \tan {}^{ - 1} y$

Now integrate

$\rm \implies - \int \dfrac{ {e}^{x}dx }{1 + {e}^{2x} }$

Using substitution method

Let

$\rm \implies \: {e}^{x} = t$

$\rm \implies \: {e}^{x} dx = dt$

We can write

$\rm \implies - \int \: \dfrac{dt}{1 + {t}^{2} } = \tan {}^{ - 1} t$

$\rm \implies \tan {}^{ - 1} y = - \tan {}^{ - 1} t + c$

$\rm \implies \tan {}^{ - 1} y = - \tan {}^{ - 1} {e}^{x} + c$

Now its given y = 1 and x = 0

$\rm \implies \tan {}^{ - 1} 1 + \tan {}^{ - 1} {e}^{0} = c$

$\rm \implies \tan {}^{ - 1} 1 + \tan {}^{ - 1} 1 = c$

$\rm \implies \dfrac{\pi}{4} + \dfrac{\pi}{4} = c$

$\rm \implies \: c = \dfrac{\pi}{2}$

We get

$\rm \implies \tan {}^{ - 1} y + \tan {}^{ - 1} {e}^{x} = c$

$\rm \implies \tan {}^{ - 1} y + \tan {}^{ - 1} {e}^{x} = \dfrac{\pi}{2}$

using properties of inverse trigonometry

$\implies \rm \tan {}^{ - 1} x+ \cot {}^{ - 1}x = \dfrac{\pi}{2}$

$\rm \implies \tan {}^{ - 1} x = \cot {}^{ - 1} \dfrac{1}{x}$

Now

$\rm \implies \: \tan {}^{ - 1} {e}^{x} = \dfrac{\pi}{2} - \tan {}^{ - 1} y$

$\rm \implies \: \tan {}^{ - 1} {e}^{x} = \tan {}^{ - 1} \dfrac{1}{y}$

So put the value

$\rm \implies \tan {}^{ - 1} y + \tan {}^{ - 1} \dfrac{1}{y} = \dfrac{\pi}{2}$

$\rm \implies \: \tan {}^{ - 1} {e}^{x} = \tan {}^{ - 1} \dfrac{1}{y}$

$\rm \implies \: {e}^{x} = \dfrac{1}{y}$

$\rm \implies {e}^{x} y = 1$

Answer

$\rm \implies {e}^{x} y = 1$

Answer 2

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