If xy+8ey=8e, find the value of y at the point where x = 0.

Given,xy+8ey=8eTo find the value of y'' at the point x=0.We have,xy+8ey=8e (1)Differentiating with respect to x,⇒xy′+y+8eyy′=0 (2)Differentiating again with respect to x,⇒xy″+y′+y′+8eyy″+8ey′y′=0⇒(x+88ey)y″+2y′+8ey(y′)2=0 (3)Put x=0 in (1),(2) and (3),0+8ey=8e⇒ey=e∴y=1By (2),⇒0+1+8ey′=0⇒8ey′=−1∴y′=−18eBy (3),⇒(0+8e)y″+2(−18e)+8e(−18e)2=0⇒8ey″−28e+18e=0⇒8ey″−18e=0∴y″=18e(8e)=164e2

"If xy+8ey=8e, find the value of y at..." solved and explained

kursval7z

Answered question

2021-11-14

x y + 8 e^{y} = 8 e

, find the value of y at the point where x = 0.

Answer & Explanation

Jick1984

Beginner2021-11-15Added 12 answers

Given,
$x y + 8 e^{y} = 8 e$
To find the value of y'' at the point x=0.
We have,
$x y + 8 e^{y} = 8 e$ (1)
Differentiating with respect to x,
$\Rightarrow x y^{'} + y + 8 e^{y} y^{'} = 0$ (2)
Differentiating again with respect to x,
$\Rightarrow x y^{″} + y^{'} + y^{'} + 8 e^{y} y^{″} + 8 e^{y^{'}} y^{'} = 0$
$\Rightarrow (x + 88 e^{y}) y^{″} + 2 y^{'} + 8 e^{y} {(y^{'})}^{2} = 0$ (3)
Put $x = 0$ in (1),(2) and (3),
$0 + 8 e^{y} = 8 e$
$\Rightarrow e^{y} = e$
$∴ y = 1$
By (2),
$\Rightarrow 0 + 1 + 8 e y^{'} = 0$
$\Rightarrow 8 e y^{'} = - 1$
$∴ y^{'} = \frac{- 1}{8 e}$
By (3),
$\Rightarrow (0 + 8 e) y^{″} + 2 (- \frac{1}{8 e}) + 8 e {(- \frac{1}{8 e})}^{2} = 0$
$\Rightarrow 8 e y^{″} - \frac{2}{8 e} + \frac{1}{8 e} = 0$
$\Rightarrow 8 e y^{″} - \frac{1}{8 e} = 0$
$∴ y^{″} = \frac{1}{8 e (8 e)} = \frac{1}{64 e^{2}}$

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