Математика: Найти y и y (производная 1 и 2-го порядка от y)

Найти y' и y'' (производная 1 и 2-го порядка от y)

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Ответ:

Korolevanet

19.01.2021 20:21

а)

arctg(y) = 4x + 5y

$\frac{1}{1 + {y}^{2} } \times y' = 4 + 5y' \\ \frac{y'}{1 + {y}^{2} } - 5y' = 4 \\ y'( \frac{1}{1 + {y}^{2} } - 5) = 4 \\ y' \times \frac{1 - 5(1 + {y}^{2}) }{1 + {y}^{2} } = 4 \\ y' = \frac{4(1 + {y)}^{2} }{1 - 5 - 5 {y}^{2} } \\ y' = \frac{4(1 + {y}^{2}) }{ - 4 - 5 {y}^{2} }$

$\frac{y'}{1 + {y}^{2} } = 4 + 5y'\\ \frac{y''(1 + {y}^{2}) - 2yy' \times y'}{ {(1 + {y}^{2}) }^{2} } = 5y''\\ \frac{y''}{1 + {y}^{2} } - \frac{2y {(y')}^{2} }{ {(1 + {y}^{2} )}^{2} } - 5y'' = 0 \\ y''( \frac{1}{1 + {y}^{2} } - 5) = \frac{2y {(y')}^{2} }{ {(1 + {y}^{2} )}^{2} } \\ y'' \times \frac{ - 4 - 5 {y}^{2} }{1 + {y}^{2} } = \frac{2y {(y')}^{2} }{ {(1 + {y}^{2}) }^{2} } \\ y'' = - \frac{2y {(y')}^{2} }{ {(1 + {y}^{2}) }^{2} } \times \frac{1 + {y}^{2} }{ 4 + 5 {y}^{2} } \\ y''= - \frac{2y {(y')}^{2} }{(1 + {y}^{2} )(4 + 5 {y}^{2}) }$

представим вместо у' найденную производную первого порядка

$y'' = - \frac{2y}{(1 + {y}^{2})(4 + 5 {y}^{2}) } \times \frac{16 {(1 + {y}^{2}) }^{2} }{ {( - (4 + 5 {y}^{2} ))}^{2} } \\ y'' = - \frac{2y}{4 + 5 {y}^{2} } \times \frac{16(1 + {y}^{2} )}{ {(4 + 5 {y}^{2}) }^{2} } \\ y''= - \frac{32(1 + {y}^{2} )}{ {(4 + 5 {y}^{2}) }^{3} }$

б)

$y'x = \frac{y't}{x't}\\$

$y''x = \frac{(y'x)'t}{x't}\\$

$y't = ( \frac{ {t}^{2} }{ {(t + 2)}^{2} } ) = \frac{2t {(t + 1)}^{2} - 2(1 + t) {t}^{2} }{ {(1 + t)}^{4} } = \\ = \frac{(t + 1)(2t(t + 1) - 2 {t}^{2}) }{ {(1 + t)}^{4} } = \\ = \frac{2 {t}^{2} + 2t - 2 {t}^{2} }{ {(t + 1)}^{3} } = \frac{2t}{ {(1 + t)}^{3} }$

$x't = - {(t + 2)}^{ - 2} = - \frac{1}{ {(t + 2)}^{2} }$

$y'x = \frac{2t}{ {(t + 1)}^{3} } \times ( - {(t + 2)}^{2} ) = \\ = - \frac{2t( {t}^{2} + 4t + 4) }{ {(t + 1)}^{3} } = \\ = \frac{ - 2 {t}^{3} - 8 {t}^{2} - 8t }{ {(t + 1)}^{3} }$

$(y'x)'t = \frac{( - 12 {t}^{2} - 16t - 8) {(t + 1)}^{3} - 3 {(t + 1)}^{2}( - 2 {t}^{3} - 8 {t}^{2} - 8t) }{ {(t + 1)}^{6} } = \\ = \frac{ {(t + 1)}^{2}((t + 1)( - 12 {t}^{2} - 16t - 8) - 3( - 2 {t}^{3} - 8 {t}^{2} - 8t) }{ {(t + 1)}^{6} } = \\ = \frac{ - 12 {t}^{3} - 16 {t}^{2} - 8t - 12 {t}^{2} - 16t - 8 + 6 {t}^{3} + 24 {t}^{2} + 24t}{ {(t + 1)}^{4} } = \\ = \frac{ - 6 {t}^{3} - 4 {t}^{2} - 8 }{ {(t + 1)}^{4} } = - \frac{6 {t}^{3} + 4 {t}^{2} + 8}{ {(t + 1)}^{4} }$

$y''x = - \frac{6 {t}^{3} + 4 {t}^{2} + 8}{ {(t + 1)}^{4} } \times ( - {(t + 2)}^{2} ) = \\ = \frac{(t + 2)(6 {t}^{3} + 4 {t}^{2} + 8)}{ {(t + 1)}^{4} }$