If xy+6ey=6e find the value of y" at the point where x=0

Step 1Ans. xy+6ey=6exdydx+y+6ey+dydx=0xd2ydx2+dydx+6eyd2ydx2+6ey(dydx)2=0at point x=00d2ydx2+2(dydx)+6eyd2ydx2+6ey(dydx)2=0Step 26eyy"=−2y′−6ey(y′)2y"=−2y′6ey−6ey(y′)26eyy″=−13(y′ey)−(y′)2

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High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

impresijuzj

Answered question

2021-11-28

x y + 6 e^{y} = 6 e

find the value of
y"
at the point where

x = 0

Answer & Explanation

Ida Perry

Beginner2021-11-29Added 20 answers

Step 1
Ans. $x y + 6 e^{y} = 6 e$
$x \frac{d y}{d x} + y + 6 e^{y} + \frac{d y}{d x} = 0$
$x \frac{d^{2} y}{{d x}^{2}} + \frac{d y}{d x} + 6 e^{y} \frac{d^{2} y}{{d x}^{2}} + 6 e^{y} {(\frac{d y}{d x})}^{2} = 0$
at point x=0
$0 \frac{d^{2} y}{{d x}^{2}} + 2 (\frac{d y}{d x}) + 6 e^{y} \frac{d^{2} y}{{d x}^{2}} + 6 e^{y} {(\frac{d y}{d x})}^{2} = 0$
Step 2
$6 e^{y} y " = - 2 y^{'} - 6 e^{y} (y^{'})^{2} y " = \frac{- 2 y^{'}}{6 e^{y}} - \frac{6 e^{y} (y^{'})^{2}}{6 e^{y}}$
$y^{″} = - \frac{1}{3} (\begin{matrix} y^{'} \\ e^{y} \end{matrix}) - (y^{'})^{2}$

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