Question

Высшая математика, нужно подробное решение № 161

Answer 1

Ответ:

a)

${e}^{x} - x - {y}^{3} = 0 \\ ( {e}^{x} )'- x'- ( {y}^{3} )'= 0 \\ {e}^{x} - 1 - 3 {y}^{2} \times y'= 0 \\ 3 {y}^{2} y' = {e}^{x} - 1 \\ y' = \frac{ {e}^{x} - 1 }{3 {y}^{2} }$

_______________________

${e}^{x} - 1 - 3 {y}^{2} y'= 0 \\ {e}^{x} - 0 - (3 {y}^{2} )' \times y' - (y')'\times 3 {y}^{2} = 0 \\ {e}^{x} - 6y {(y')}^{2} - y'' \times 3 {y}^{2} = 0 \\ y'' \times 3 {y}^{2} = {e}^{x} - 6y {(y')}^{2} \\ y'' = \frac{ { {e}^{x} - 6y(y')}^{2} }{3 {y}^{2} } \\ \\ y'= \frac{ {e}^{x} - 1}{3 {y}^{2} } \\ \\ y'' = \frac{1}{3 {y}^{2} } \times ( {e}^{x} - 6y \times {( \frac{ {e}^{x} - 1}{3 {y}^{2} } )}^{2} ) = \\ = \frac{1}{3 {y}^{2} } ( {e}^{x} - 6y \times \frac{( {e}^{x} - 1) {}^{2} }{9 {y}^{4} } ) \\ y'' = \frac{ 9 {y}^{4} {e}^{x} - 6y( {e}^{x} -1) {}^{2} }{27 {y}^{6} } \\y''=\frac{3y^3e^x-2(e^x-1)^2}{9y^5}$

b)

$y'_x = \frac{y'_t}{x'_t}$

$y'_t = \frac{b}{ \cos {}^{2} (t) }$

$x'_t = \frac{a}{ \cos {}^{2} (t) }$

$y'_x = \frac{ \frac{b}{ \cos {}^{2} (t) } }{ \frac{a}{ \cos {}^{2} (t) } } = \frac{b}{ \cos {}^{2} (t) } \times \frac{ \cos {}^{2} (t) }{a} = \frac{b}{a}$

$y''_{xx }= \frac{(y'_x)'_t}{x'_t}$

$(y'_x)'_t = 0$

$y''_{xx }= 0$