Weird Equation

Calculus Level 4

e y + y ( x 2 ) = x 2 8 \large {e}^{y}+y(x-2)=x^{2}-8

The equation above defines y y implicitly as a function of x x near the point ( 3 , 0 ) (3,0) . Estimate the value of y y when x = 2.98 x=2.98 . Given the answer in 2 decimal places.


Clarification: e 2.71828 e \approx 2.71828 denotes the Euler's number .


The answer is -0.06.

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1 solution

Steven Chase
Apr 5, 2017

Use implicit differentiation to get the slope of the function:

e y + y x 2 y x 2 + 8 = 0 e y d y d x + y + x d y d x 2 d y d x 2 x = 0 d y d x ( e y + x 2 ) = 2 x y d y d x = 2 x y e y + x 2 \large{e^y + yx - 2y - x^2 + 8 = 0 \\ e^y \frac{dy}{dx} + y + x \frac{dy}{dx} - 2 \frac{dy}{dx} - 2x = 0 \\ \frac{dy}{dx} (e^y + x - 2) = 2x - y \\ \frac{dy}{dx} = \frac{2x - y}{e^y + x - 2} }

Write a computer program to start at ( 3 , 0 ) (3,0) and step backward until x = 2.98 x = 2.98 . Integrate the slope to update the y y value. When the program terminates, y 0.06 y \approx -0.06 .

import math

x = 3.0
y = 0.0

dx = 10.0**(-6.0)

while x >= 2.98:

    num = 2.0*x - y
    denom = math.exp(y) + x - 2.0

    yp = num / denom

    y = y + yp * (-dx)

    x = x - dx

print y


#### Result is -0.0613348355773

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