Second order linear ODE - Revision history

Rice: /* Constant coefficient, homogeneous */

2024-06-10T04:10:33Z

Constant coefficient, homogeneous

← Older revision		Revision as of 04:10, 10 June 2024
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	\frac{1}{2i}(y_1-y_2)= e^{\lambda t}\sin (\mu t)		\frac{1}{2i}(y_1-y_2)= e^{\lambda t}\sin (\mu t)
	</math>		</math>

			where <math>\lambda</math> is the real term, and the <math>\mu</math> is the imaginary term.

	This gives us a set of real fundamental solutions.		This gives us a set of real fundamental solutions.

2024-05-22T05:10:25Z

Constant coefficient, homogeneous

← Older revision		Revision as of 05:10, 22 May 2024
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	This gives us a set of real fundamental solutions.		This gives us a set of real fundamental solutions.

			For repeated roots, use [[reduction of order]] to find a second solution. The second solution is of the form

	<math>		<math>
			te^{rt}
	</math>		</math>

	~~For repeated roots,~~

	== Fundamental set of solutions ==		== Fundamental set of solutions ==

2024-05-22T00:21:45Z

Constant coefficient, homogeneous

@@ Line 51: / Line 51: @@
 For 2 real roots, the two solutions are always independent so we get general solution easy.
-For complex conjugates,
+For complex conjugates, the following real solutions can be derived from [[Euler's formula]]. Since complex number + complex conjugate becomes real.
 For repeated roots,

2024-05-21T23:49:45Z

Solutions

← Older revision		Revision as of 23:49, 21 May 2024
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	Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>		Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>

			There are three cases: 2 real roots, complex conjugates, and repeated roots, depending on the discriminant.

			For 2 real roots, the two solutions are always independent so we get general solution easy.

			For complex conjugates,

			For repeated roots,

	== Fundamental set of solutions ==		== Fundamental set of solutions ==

2024-05-21T23:33:28Z

Fundamental set of solutions

← Older revision		Revision as of 23:33, 21 May 2024
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	The independence of the solutions can be checked using the [[Wronskian]]. The two solutions are called a '''fundamental set''' of solutions.		The independence of the solutions can be checked using the [[Wronskian]]. The two solutions are called a '''fundamental set''' of solutions.

			This fundamental set always exists according to the [[General existence theorem]].

2024-05-21T22:53:31Z

Constant coefficient, homogeneous

← Older revision		Revision as of 22:53, 21 May 2024
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	= Solutions =		= Solutions =
	== Constant coefficient, homogeneous ==		== Constant coefficient, homogeneous ==
			Whatever derived usually also works for variable coefficients.

	These are the simplest kind. They have the general form		These are the simplest kind. They have the general form

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	Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>		Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>

			== Fundamental set of solutions ==

	Given that two linearly independent solutions are given, the general solution is of the following form		Given that two linearly independent solutions are given, the general solution is of the following form
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	</math>		</math>

	The independence of the solutions can be checked using the [[Wronskian]].		The independence of the solutions can be checked using the [[Wronskian]]. The two solutions are called a '''fundamental set''' of solutions.

2024-05-21T22:17:12Z

Constant coefficient, homogeneous

← Older revision		Revision as of 22:17, 21 May 2024
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	Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>		Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>

	~~Based on [[Abel's theorem]]~~, the general solution is of the following form		Given that two linearly independent solutions are given, the general solution is of the following form

	<math>		<math>
	y(t) = c_1 y_1+c_2 y_2		y(t) = c_1 y_1+c_2 y_2
	</math>		</math>

			The independence of the solutions can be checked using the [[Wronskian]].

2024-05-19T18:59:05Z

← Older revision		Revision as of 18:59, 19 May 2024
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			[[Category:Differential Equations]]
	Second order linear ODEs are in the following form:		Second order linear ODEs are in the following form:

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	</math>		</math>

	~~=== Particular~~ solution ~~===~~		We can guess the form of the solution to be exponential, since exponential functions has the property of being the same after many differentiation.
	An exponential ~~function~~ has the property of being the same after many differentiation. ~~We take advantage of this property and guess the solution to be the form of~~

	<math>		<math>
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	Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>		Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>

	~~=== General solution ===~~		Based on [[Abel's theorem]], the general solution is of the following form

	~~For reasons that will be elaborated~~ on ~~in the future~~, the general solution is of the following form

	<math>		<math>
	y(t) = c_1 y_1+c_2 y_2		y(t) = c_1 y_1+c_2 y_2
	</math>		</math>


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

2024-05-18T00:09:51Z

General solution

← Older revision		Revision as of 00:09, 18 May 2024
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	=== General solution ===		=== General solution ===

			For reasons that will be elaborated on in the future, the general solution is of the following form

	<math>		<math>
	y(t) = c_1 y_1+c_2 y_2		y(t) = c_1 y_1+c_2 y_2

2024-05-01T21:34:25Z

Constant coefficient, homogeneous

← Older revision		Revision as of 21:34, 1 May 2024
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	</math>		</math>

			=== Particular solution ===
	An exponential function has the property of being the same after many differentiation. We take advantage of this property and guess the solution to be the form of		An exponential function has the property of being the same after many differentiation. We take advantage of this property and guess the solution to be the form of

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	ar^2+br+c&=0		ar^2+br+c&=0
	\end{aligned}		\end{aligned}
			</math>

			Depending on the constants, it will give us anywhere from zero to two solutions: <math>y_1</math> and <math>y_2</math>

			=== General solution ===
			<math>
			y(t) = c_1 y_1+c_2 y_2
	</math>		</math>


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