Author Topic: TUT0602 QUIZ 1  (Read 529 times)

Fenglun Wu

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TUT0602 QUIZ 1
« on: September 27, 2019, 02:00:02 PM »
$$
\frac{dy}{dx} = \frac{x - e^{-x}}{y + e^y}
$$
Rewrite the equation:

$$
(e^{-x} - x) + (y + e^y)\frac{dy}{dx} = 0
$$

The equation is of the form

$$ M(x) +N(y)\frac{dy}{dx} = 0 $$

Hence, it is separable.

$$
\int (y + e^y) dy = \int (x - e^{-x}) dx
$$
$$
\frac{y^2}{2} + e^y = \frac{x^2}{2} + e^{-x} + c'
$$

Let $$c = 2c'$$

Solution:
$$
y^2 - x^2 + 2(e^y - e^{-x}) = c
$$

where $$y + e^y \neq 0$$