## Friday, June 16, 2017

### Unizor - Ordinary Differential Equations - Separable Equations

Notes to a video lecture on http://www.unizor.com

Separable Ordinary Differential Equations

The process of "separation" as a method of solving differential equation of the first order F(x,y,y')=0 should result in the following equality:
f(y)·dy = g(x)·dx
which allows for separate integration of left and right sides.

This can be assured if our initial equation F(x,y,y')=0 can be transformed into y'=P(x)·Q(y).
Indeed, from the last equation follows
dy/dx = P(x)·Q(x)
and
dy/Q(y) = P(x)·dx
which can be integrated separately, left side - by y and right side - by x.

Examples below use exactly this approach.

Example 1

y' + x·y + y − x = 1

Perform the transformation:
y' = −x·y − y + x + 1
y' = −y·(x+1) + (x + 1)
y' = (1−y)·(x+1)
dy/(1−y) = (x+1)dx
dy/(1−y) = (x+1)dx
Both integrals are trivial.
d(y−1)/(y−1)=(x+1)d(x+1)
−ln(y−1) = (x+1)²/2 + C
y = 1 + C·e−(x+1)²/2

Example 2

y' − (x+1)·e(x+y) = 0

Perform the transformation:
y' = (x+1)·ex·ey
y'·e−y = (x+1)·ex
e−y·dy = (x+1)·ex·dx
e−y·dy = (x+1)·ex·dx
Integral on the left is straight forward.
Integral on the right can be calculated using the integration "by-part":
−e−y = (x+1)·ex −ex·d(x+1)
−e−y = (x+1)·ex − ex + C
−e−y = x·ex + C
e−y = C−x·ex
y = −ln(C−x·ex)

Example 3

ln(y') = x + y

Perform the transformation:
y' = ex · ey
e−y·dy = ex·dx
e−y·dy = ex·dx
−e−y = ex+C
e−y = −ex−C
y = −ln(C−ex)

Example 4

sin(y)·y' = sin(x+y) + sin(x−y)

First of all, recall the trigonometric identities
sin(x+y) =
= sin(x)·cos(y)+cos(x)·sin(y)

sin(x−y) =
= sin(x)·cos(y)−cos(x)·sin(y)

from which follows
sin(x+y)+sin(x−y) =
= 2·sin(x)·cos(y)

Perform the transformation of our equation using the last expression:
sin(y)·y' = 2·sin(x)·cos(y)
Now we can separate:
sin(y)·dy/cos(y) = 2·sin(x)·dx
Continue transformation:
dcos(y)/cos(y) = −2·dcos(x)
Easy to integrate now:
ln(|cos(y)|) = 2·cos(x) + C
Ignoring difficulties with absolute value and periodicity to shorten the presentation of an idea, it can be solved for y
|cos(y)| = e2·cos(x)+C
y = arccos(e2·cos(x)+C)