How to solve the ODEy'=x-exx+ey

Get the answer to "How to solve the ODEy'=x-exx+ey"

tibukooinm

2022-03-31

How to solve the ODE
$y' = \frac{x - e^{x}}{x + e^{y}}$

Roy Brady

Beginner2022-04-01Added 19 answers

Step 1
The equation can be rewritten as

x y^{'} (x) + e^{y (x)} y^{'} (x) = x + e^{x} = x y^{'} (x) + {(e^{y})}^{'} (x) .

Let

z (x) = (e^{y}) (x),

,
hence,

z^{'} (x) = {(e^{y})}^{'} (x) = (e^{y}) (x) y^{'} (x) = z (x) y^{'} (x)

, implying that

y^{'} (x) = \frac{z^{'} (x)}{z (x)} .

This gets the equation to be

x \frac{z^{'} (x)}{z (x)} + z^{'} (x) = x + e^{x} .

Step 2
What this implies is that

x z^{'} (x) + z (x) z^{'} (x) = [x + z (x)] z^{'} (x) = (x + e^{x}) z (x) = [x + z (x)] ([1 + z^{'} (x)] - 1) = [x + z (x)] [1 + z^{'} (x)] - [x + z (x)] = (x + e^{x}) ([x + z (x)] - x) = (x + e^{x}) [x + z (x)] - x (x + e^{x}),

hence

[x + z (x)] [1 + z^{'} (x)] - (1 + x + e^{x}) [x + z (x)] = - x (x + e^{x}) .

Let

f (x) = x + z (x)

, thus

f^{'} (x) = 1 + z^{'} (x)

, simplifying the equation significantly to

f (x) f^{'} (x) - (1 + x + e^{x}) f (x) = - x (x + e^{x}) .

This is about the most you can do, if I understand correctly.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?