Integrating factor - Revision history

Rice at 19:46, 17 May 2024

2024-05-17T19:46:21Z

← Older revision		Revision as of 19:46, 17 May 2024
Line 1:		Line 1:
	[[Category:Differential Equations]]		[[Category:Differential Equations]]
	~~[[Category:Techniques for solving differential equations]]~~
	'''Integrating factor''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s. It takes advantage of the product rule of derivatives		'''Integrating factor''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s. It takes advantage of the product rule of derivatives

Rice at 19:45, 17 May 2024

2024-05-17T19:45:54Z

← Older revision		Revision as of 19:45, 17 May 2024
Line 1:		Line 1:
			[[Category:Differential Equations]]
			[[Category:Techniques for solving differential equations]]
	'''Integrating factor''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s. It takes advantage of the product rule of derivatives		'''Integrating factor''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s. It takes advantage of the product rule of derivatives

Line 29:		Line 31:

	<math>y(t) = \frac{1}{\mu(t)}\int \mu g dt + C </math>		<math>y(t) = \frac{1}{\mu(t)}\int \mu g dt + C </math>

	~~[[Category:Differential Equations]]~~

Rice at 19:28, 17 May 2024

2024-05-17T19:28:14Z

← Older revision		Revision as of 19:28, 17 May 2024
Line 29:		Line 29:

	<math>y(t) = \frac{1}{\mu(t)}\int \mu g dt + C </math>		<math>y(t) = \frac{1}{\mu(t)}\int \mu g dt + C </math>

			[[Category:Differential Equations]]

Rice at 23:30, 23 April 2024

2024-04-23T23:30:23Z

← Older revision		Revision as of 23:30, 23 April 2024
Line 4:		Line 4:

	and attempts to move all y terms into the same differential order.		and attempts to move all y terms into the same differential order.

			Note that [[Separation of variables]] is generally easier when both options are available.

	= Procedure =		= Procedure =

Rice at 23:28, 23 April 2024

2024-04-23T23:28:03Z

← Older revision		Revision as of 23:28, 23 April 2024
Line 1:		Line 1:
	'''Integrating factor''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s.		'''Integrating factor''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s. It takes advantage of the product rule of derivatives

	Consider		<math>(\mu y)' = \mu' y + \mu y'</math>

			and attempts to move all y terms into the same differential order.

			= Procedure =
			Consider a linear first order ODE.

	<math>y' + p(t) y = g(t)</math>		<math>y' + p(t) y = g(t)</math>

Rice at 23:26, 23 April 2024

2024-04-23T23:26:09Z

← Older revision		Revision as of 23:26, 23 April 2024
Line 1:		Line 1:
	'''Integrating ~~factors~~''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s.		'''Integrating factor''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s.

	Consider		Consider

Rice: Rice moved page Integrating factors to Integrating factor

2024-04-23T23:25:53Z

Rice moved page Integrating factors to Integrating factor

← Older revision		Revision as of 23:25, 23 April 2024
(No difference)

Rice at 21:52, 17 April 2024

2024-04-17T21:52:19Z

← Older revision		Revision as of 21:52, 17 April 2024
Line 8:		Line 8:

	<math>\mu' = \mu p</math>		<math>\mu' = \mu p</math>

			<math>\mu(t) = C \exp\left[ \int p dt \right]</math>

	Applying the integrating factor, we have		Applying the integrating factor, we have
Line 18:		Line 20:

	From there, simple integration and algebra will solve the equation.		From there, simple integration and algebra will solve the equation.

			<math>y(t) = \frac{1}{\mu(t)}\int \mu g dt + C </math>

Rice: Created page with "'''Integrating factors''' is a method to solve scalar Linear First Order ODEs. Consider $y' + p(t) y = g(t)$ We multiply bothsides by integrating factor $\mu(t)$ such that the left hand side to be the ''exact derivative of a product.'' For this to work, we need $\mu' = \mu p$ Applying the integrating factor, we have $\begin{aligned} \mu(t) y + \mu(t) p(t) y &= \mu(t) g(t) \\ \mu y' + \mu ' y &= \mu g \\ (\mu y)' &=..."$

2024-04-17T21:47:25Z

Created page with "'''Integrating factors''' is a method to solve scalar Linear First Order ODE<nowiki/>s. Consider <math>y' + p(t) y = g(t)</math> We multiply bothsides by integrating factor <math>\mu(t)</math> such that the left hand side to be the ''exact derivative of a product.'' For this to work, we need <math>\mu' = \mu p</math> Applying the integrating factor, we have <math>\begin{aligned} \mu(t) y + \mu(t) p(t) y &= \mu(t) g(t) \\ \mu y' + \mu ' y &= \mu g \\ (\mu y)' &=..."

New page

'''Integrating factors''' is a method to solve scalar [[Linear First Order ODE]]<nowiki/>s.

Consider

<math>y' + p(t) y = g(t)</math>

We multiply bothsides by integrating factor <math>\mu(t)</math> such that the left hand side to be the ''exact derivative of a product.'' For this to work, we need

<math>\mu' = \mu p</math>

Applying the integrating factor, we have

<math>\begin{aligned}
\mu(t) y + \mu(t) p(t) y &= \mu(t) g(t) \\
\mu y' + \mu ' y &= \mu g \\
(\mu y)' &= \mu g
\end{aligned}</math>

From there, simple integration and algebra will solve the equation.