Difference between revisions of "Separation of Variables"

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[https://youtu.be/nzqLplhh0DU Differential Equations: Separation of Variables] by James Sousa
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== Definition ==
  
[https://youtu.be/OF9K9MK0Yvg Ex.1 Differential Equations: Separation of Variables] by James Sousa
+
A ''separable ODE'' is an equation of the form
 +
:<math>x'(t) = g(t) f(x(t))</math>
 +
for some functions <math>g: \mathbb R \to \mathbb R</math>, <math>f: \mathbb R^n \to \mathbb R^n</math>. In this chapter, we shall only be concerned with the case <math>n = 1</math>.
  
[https://youtu.be/A_RdFJJUyEg Ex.2 Differential Equations: Separation of Variables] by James Sousa
+
We often write for this ODE
 +
:<math>x' = g(t) f(x)</math>
 +
for short, omitting the argument of <math>x</math>.
  
[https://youtu.be/0-dy9aaNdZ8 Ex.3 Differential Equations: Separation of Variables] by James Sousa
+
[Note that the term "separable" comes from the fact that an important class of differential equations has the form
 +
:<math>x' = h(t, x)</math>
 +
for some <math>h: \mathbb R \times \mathbb R^n \to \mathbb R</math>; hence, a separable ODE is one of these equations, where we can "split" the <math>h</math> as <math>h(t, x) = g(t) f(x)</math>.]
  
[https://youtu.be/KNQFAusQ_KM Ex 1: Initial Value Problem - Separation of Variables] by James Sousa
+
== Informal derivation of the solution ==
  
[https://youtu.be/fBb68MsuAAk Ex 2: Initial Value Problem - Separation of Variables] by James Sousa
+
Using Leibniz' notation for the derivative, we obtain an informal derivation of the solution of separable ODEs, which serves as a good mnemonic.
  
[https://youtu.be/nNHlSB6b1HU Solving Separable First Order Differential Equations - Ex 1] by patrickJMT
+
Let a separable ODE
 +
:<math>x' = g(t) f(x)</math>
 +
be given. Using Leibniz notation, it becomes
 +
:<math>\frac{dx}{dt} = g(t) f(x)</math>.
 +
We now formally multiply both sides by <math>dt</math> and divide both sides by <math>f(x)</math> to obtain
 +
:<math>\frac{dx}{f(x)} = g(t) dt</math>.
 +
Integrating this equation yields
 +
:<math>\int \frac{dx}{f(x)} = \int g(t) dt</math>.
 +
Define
 +
:<math>F(x) := \int \frac{dx}{f(x)}</math>;
 +
this shall mean that <math>F</math> is a primitive of <math>\frac{1}{f(x)}</math>. If then <math>F</math> is invertible, we get
 +
:<math>x = F^{-1}\left( \int g(t) dt \right) = F^{-1} \circ G</math>,
 +
where <math>G</math> is a primitive of <math>g</math>; that is, <math>x(s) = F^{-1}(G(s))</math>, now inserting the variable of <math>x</math> back into the notation.
  
[https://youtu.be/XExEixAPK6s Solving Separable First Order Differential Equations - Ex 2] by patrickJMT
+
Now the formulae in this derivation don't actually mean anything; it's only a formal derivation. But below, we will prove that it actually yields the right result.
  
[https://youtu.be/uS_5bmRUYEI Solving a Separable Differential Equation, Another Example #1] by patrickJMT
+
== General solution ==
  
[https://youtu.be/Vc6MwLVdCuM Solving a Separable Differential Equation, Another Example #2] by patrickJMT
+
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
:'''Theorem 2.1''':
 +
:Let a separable, one-dimensional ODE
 +
::<math>x'(t) = g(t) f(x(t))</math>
 +
:be given, where <math>f</math> is never zero. Let <math>F</math> be an antiderivative of <math>f</math> and <math>G</math> an antiderivative of <math>g</math>. If <math>F</math> is invertible, the function
 +
::<math>x(t) := F^{-1}(G(t))</math>
 +
:solves the ODE under consideration.
 +
</blockquote>
 +
 
 +
'''Proof''':
 +
 
 +
By the inverse and chain rules,
 +
:<math>\frac{d}{dt} F^{-1}(G(t)) = \frac{1}{\frac{1}{f(F^{-1}(G(t)))}} G'(t) = f(F^{-1}(G(t))) g(t)</math>;
 +
since <math>f</math> is never zero, the fraction occuring above involving <math>f</math> is well-defined.
 +
 
 +
==Resources==
 +
*[https://youtu.be/nzqLplhh0DU Differential Equations: Separation of Variables] by James Sousa
 +
 
 +
*[https://youtu.be/OF9K9MK0Yvg Ex.1 Differential Equations: Separation of Variables] by James Sousa
 +
 
 +
*[https://youtu.be/A_RdFJJUyEg Ex.2 Differential Equations: Separation of Variables] by James Sousa
 +
 
 +
*[https://youtu.be/0-dy9aaNdZ8 Ex.3 Differential Equations: Separation of Variables] by James Sousa
 +
 
 +
*[https://youtu.be/KNQFAusQ_KM Ex 1: Initial Value Problem - Separation of Variables] by James Sousa
 +
 
 +
*[https://youtu.be/fBb68MsuAAk Ex 2: Initial Value Problem - Separation of Variables] by James Sousa
 +
 
 +
*[https://youtu.be/nNHlSB6b1HU Solving Separable First Order Differential Equations - Ex 1] by patrickJMT
 +
 
 +
*[https://youtu.be/XExEixAPK6s Solving Separable First Order Differential Equations - Ex 2] by patrickJMT
 +
 
 +
*[https://youtu.be/uS_5bmRUYEI Solving a Separable Differential Equation, Another Example #1] by patrickJMT
 +
 
 +
*[https://youtu.be/Vc6MwLVdCuM Solving a Separable Differential Equation, Another Example #2] by patrickJMT
 +
 
 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikibooks.org/wiki/Ordinary_Differential_Equations/Separable_equations:_Separation_of_variables Separation of Variables, Wikibooks: Ordinary Differential Equations/Separable equations] under a CC BY-SA license

Latest revision as of 11:14, 29 October 2021

Definition

A separable ODE is an equation of the form

for some functions , . In this chapter, we shall only be concerned with the case .

We often write for this ODE

for short, omitting the argument of .

[Note that the term "separable" comes from the fact that an important class of differential equations has the form

for some ; hence, a separable ODE is one of these equations, where we can "split" the as .]

Informal derivation of the solution

Using Leibniz' notation for the derivative, we obtain an informal derivation of the solution of separable ODEs, which serves as a good mnemonic.

Let a separable ODE

be given. Using Leibniz notation, it becomes

.

We now formally multiply both sides by and divide both sides by to obtain

.

Integrating this equation yields

.

Define

;

this shall mean that is a primitive of . If then is invertible, we get

,

where is a primitive of ; that is, , now inserting the variable of back into the notation.

Now the formulae in this derivation don't actually mean anything; it's only a formal derivation. But below, we will prove that it actually yields the right result.

General solution

Theorem 2.1:
Let a separable, one-dimensional ODE
be given, where is never zero. Let be an antiderivative of and an antiderivative of . If is invertible, the function
solves the ODE under consideration.

Proof:

By the inverse and chain rules,

;

since is never zero, the fraction occuring above involving is well-defined.

Resources

Licensing

Content obtained and/or adapted from: