Initial Value Problem - Revision history

Khanh at 03:40, 6 November 2021

2021-11-06T03:40:56Z

Lila: /* Examples */

2021-10-18T20:00:11Z

Examples

Lila: /* =Examples */

2021-10-14T22:00:02Z

=Examples

Lila at 21:59, 14 October 2021

2021-10-14T21:59:26Z

Lila at 02:07, 18 September 2021

2021-09-18T02:07:28Z

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2021-09-18T02:07:03Z

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Lila: Created page with "With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general soluti..."

2021-09-18T02:06:57Z

Created page with "With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general soluti..."

New page

With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.

Examples:
* <math> y' = 2x </math>, <math> y(2) = 0 </math>. With this point and the general solution <math>y = x^2 + C</math>, we can calculate the constant C to be -4. Thus the particular solution is <math>y = x^2 - 4</math>.
* <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>.
* <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>.

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 * <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>.
-==Resources==
+== Licensing ==
-* [https://en.wikipedia.org/wiki/Initial_value_problem Initial value problem], Wikipedia
+Content obtained and/or adapted from:

@@ Line 13: / Line 13: @@
 Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. <math>y''(t)=f(t,y(t),y'(t))</math>.
-===Examples==
+===Examples===
 With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.
@@ Line 21: / Line 21: @@
 * <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>.
 * <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>.

← Older revision		Revision as of 21:59, 14 October 2021
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		+	== Definition ==
		+	An '''initial value problem''' is a differential equation
		+	:<math>y'(t) = f(t, y(t))</math> with <math>f\colon \Omega \subset \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n</math> where <math>\Omega</math> is an open set of <math>\mathbb{R} \times \mathbb{R}^n</math>,
		+	together with a point in the domain of <math>f</math>
		+	:<math>(t_0, y_0) \in \Omega,</math>
		+	called the initial condition.
		+
		+	A '''solution''' to an initial value problem is a function <math>y</math> that is a solution to the differential equation and satisfies
		+	:<math>y(t_0) = y_0.</math>
		+
		+	In higher dimensions, the differential equation is replaced with a family of equations <math>y_i'(t)=f_i(t, y_1(t), y_2(t), \dotsc)</math>, and <math>y(t)</math> is viewed as the vector <math>(y_1(t), \dotsc, y_n(t))</math>, most commonly associated with the position in space. More generally, the unknown function <math>y</math> can take values on infinite dimensional spaces, such as Banach spaces or spaces of distributions.
		+
		+	Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. <math>y''(t)=f(t,y(t),y'(t))</math>.
		+
		+	===Examples==
		+
	With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.		With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.

← Older revision		Revision as of 02:07, 18 September 2021
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−	With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.

−	~~Examples:~~
−	* <math> y' = 2x </math>, <math> y(2) = 0 </math>. With this point and the general solution <math>y = x^2 + C</math>, we can calculate the constant C to be -4. Thus the particular solution is <math>y = x^2 - 4</math>.
−	* <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>.
−	* <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>.

@@ Line 17: / Line 17: @@
 With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on.
-Examples:
+A simple example is to solve <math>y'(t) = 0.85 y(t)</math> and <math>y(0) = 19</math>.  We are trying to find a formula for <math>y(t)</math> that satisfies these two equations.
 * <math> y' = 2x </math>, <math> y(2) = 0 </math>. With this point and the general solution <math>y = x^2 + C</math>, we can calculate the constant C to be -4. Thus the particular solution is <math>y = x^2 - 4</math>.
 * <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>.