Question

Please solve!!

All the 3 questions.

Answer 1

Hope u like my process
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The linear form of differential equation is

$= > \bf \: \frac{dy}{dx} + py = q$
Thus the Integrating factor can be calculated by solving..

$= > \bf {e}^{ \int \: p \: dx}$
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Now,

i)

$= > \it \: x \frac{dy}{dx} - y = 2 {x}^{2} \sec {}^{2} (2x) \\ \\ = > \frac{dy}{dx} - \frac{1}{x} .y = 2x \sec {}^{2} (2x) \\ \\ thus \: \: \\ \\ \bf \: integrating \: \: factor \: \: will \: \: be \\ \\ = {e}^{ \int( - \frac{ 1 }{x}) dx} = {e}^{ - log(x) } \\ \\ = {e}^{ log( \frac{1}{x} ) } = \bf \underline{ \frac{1}{x} } \: \: \: \: ( \it \: answer)$

ii)

$= > y \frac{dx}{dy} - x = 2 {y}^{2} \\ \\ = > \bf \frac{dx}{dy} + ( - \frac{1}{y} )x = 2 {y}^{2} \\ \\ thus \: \: \: p = - \frac{1 }{y} \\ \\ \: \bf the \: \: integrating \: \: factor \: \: will \: \: be \\ \\ = > {e}^{ \int \: p \: dy} = {e}^{ \int( - \frac{ 1}{y}) dy} = {e}^{ - log(y) } \\ \\ = {e}^{ log( \frac{1}{y} ) } = \bf \underline{ \frac{1}{y} } \: \: \: \: ( \it \: answer)$

iii)

$= > (1 + {x}^{2} ) \frac{dy}{dx} + xy = \frac{ {x}^{4} }{(1 + {x}^{5} )} {( \sqrt{1 - { {x}^{2} }} } ) ^{3} \\ \\ = > \bf \frac{dy}{dx} + ( \frac{x}{1 + {x}^{2} } )y = \frac{ {x}^{4}( \sqrt{1 - {x}^{2} }) {}^{3} }{(1 + {x}^{5}) } \\ \\ thus \: \: \: p = \frac{x}{1 + {x}^{2} } = \frac{1}{2} ( \frac{2x}{1 + {x}^{2} } ) \\ \\ now \: \: \\ \\ \bf \: integrating \: \: \: factor \: \: wil \: \: be \\ \\ = {e}^{ \int \: p \: dx} = {e}^{ \frac{1}{2} \int \frac{2x}{1 + {x}^{2} } dx} = {e}^{ \frac{1}{2} log(1 + {x}^{2} ) } \\ \\ = {e}^{ log( {(1 + {x}^{2}) }^{ \frac{1}{2} } ) } = {e}^{ log( \sqrt{1 + {x}^{2} } ) } = \bf \underline{ \sqrt{1 + {x}^{2} } } \: \: \: \: \: ( \it \: answer)$
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Hope this is ur required answer

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