*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**

**f(y)·**d**y = g(x)·**d**x**which allows for separate integration of left and right sides.

This can be assured if our initial equation

**F(x,y,y')=0**

**y'=P(x)·Q(y)**Indeed, from the last equation follows

*d*

**y/**d**x = P(x)·Q(x)**and

*d*

**y/Q(y) = P(x)·**d**x**which can be integrated separately, left side - by

*and right side - by*

**y***.*

**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)***d*

**y/(1−y) = (x+1)**d**x**

**∫**d**y/(1−y) = ∫(x+1)**d**x**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)}= 0Perform the transformation:

**y' = (x+1)·e**^{x}·e^{y}

**y'·e**^{−y}= (x+1)·e^{x}

**e**d^{−y}·**y = (x+1)·e**d^{x}·**x**

**∫e**d^{−y}·**y = ∫(x+1)·e**d^{x}·**x**Integral on the left is straight forward.

Integral on the right can be calculated using the integration "by-part":

**−e**d^{−y}= (x+1)·e^{x}−∫e^{x}·**(x+1)**

**−e**^{−y}= (x+1)·e^{x}− e^{x}+ C

**−e**^{−y}= x·e^{x}+ C

**e**^{−y}= C−x·e^{x}

**y = −ln(C−x·e**^{x})*Example 3*

**ln(y') = x + y**Perform the transformation:

**y' = e**^{x}· e^{y}

**e**d^{−y}·**y = e**d^{x}·**x**

**∫e**d^{−y}·**y = ∫e**d^{x}·**x**

**−e**^{−y}= e^{x}+C

**e**^{−y}= −e^{x}−C

**y = −ln(C−e**^{x})*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)·cos(y)+cos(x)·sin(y)

**sin(x−y) =**

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

from which follows

**sin(x+y)+sin(x−y) =**

= 2·sin(x)·cos(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)·**d**y/cos(y) = 2·sin(x)·**d**x**Continue transformation:

**−**d**cos(y)/cos(y) = −2·**d**cos(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)| = e**^{2·cos(x)+C}

**y = arccos(e**^{2·cos(x)+C})
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