First-order linear differential equation - Revision history

Vipul at 23:42, 5 July 2012

2012-07-05T23:42:24Z

← Older revision		Revision as of 23:42, 5 July 2012
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	<math>y = e^{-H(x)}\left(e^{H(x_0)}y_0 + \int_{x_0}^x q(t)e^{H(t)} \, dt\right)</math>		<math>y = e^{-H(x)}\left(e^{H(x_0)}y_0 + \int_{x_0}^x q(t)e^{H(t)} \, dt\right)</math>

			==Examples==

			===Simple example===

			Consider the differential equation:

			<math>y' + y = e^{e^x}</math>

			Here, <math>p(x) = 1, q(x) = e^{e^x}</math>. Take <math>H(x) =x</math> and get:

			<math>y = e^{-x} \int e^{e^x}e^x \, dx</math>

			This gives:

			<math>y = e^{-x}(e^{e^x} + C), \qquad C \in \R</math>

			===Example that is better solved by subtitution===

			Consider:

			<math>xy' + y = \sin x</math>

			Divide both sides by <math>x</math> to get:

			<math>y' + \frac{y}{x} = \frac{\sin x}{x}</math>

			This is linear, with <math>p(x) = 1/x</math>, <math>q(x) = (\sin x)/x</math>. Take <math>H(x) = \ln x</math> and <math>e^{H(x)} = x</math> (see note):

			<math>y = \frac{1}{x} \int \frac{\sin x}{x} x \, dx</math>

			This gives:

			<math>y = \frac{C - \cos x}{x}, \qquad C \in \R</math>

			The linear method is unnecessary -- we divided and multiplied by <math>x</math>. A better solution would be to substitute <math>u = xy</math> and get a [[separable differential equation]].

			===Example where a particular solution is obtained by inspection===

			Consider:

			<math>y' + y = \tan x + \tan^2x</math>

			The linear method gives:

			<math>y = e^{-x} \int e^x(\tan x + \tan^2x) \, dx</math>

			The integration is not easy. So, instead of trying to do the integration directly, we note that the answer is:

			<math>y = \mbox{Particular solution} + Ce^{-x}</math>

			It thus suffices to find a particular solution. Inspection and guesswork gives a solution <math>y = \tan x - 1</math>. The general solution is thus:

			<math>y = \tan x - 1 + Ce^{-x}</math>

Vipul at 15:15, 29 June 2012

2012-06-29T15:15:43Z

← Older revision		Revision as of 15:15, 29 June 2012
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	The function <math>e^{H(x)}</math> is termed the [[integrating factor]] for the differential equation because multiplying by this turns the differential equation into an exact differential equation, i.e., a differential equation to which we can apply integration on both sides.		The function <math>e^{H(x)}</math> is termed the [[integrating factor]] for the differential equation because multiplying by this turns the differential equation into an exact differential equation, i.e., a differential equation to which we can apply integration on both sides.

			===Solution method and formula: definite integral version===

			Suppose we are given the initial value condition that at <math>x = x_0, y = y_0</math>.

			Let <math>H(x)</math> be an [[antiderivative]] for <math>p(x)</math>, so that <math>H'(x) = p(x)</math>. Then, we multiply both sides by <math>e^{H(x)}</math>. Simplifying, we get:

			<math>\frac{d}{dx}[e^{H(x)}y] = q(x)e^{H(x)}</math>

			Integrating from <math>x_0</math> to (arbitrary) <math>x</math>, we get:

			<math>e^{H(x)}y - e^{H(x_0)}y_0= \int_{x_0}^x q(t)e^{H(t)} \, dt</math>

			Thus, the general expression is:

			<math>y = e^{-H(x)}\left(e^{H(x_0)}y_0 + \int_{x_0}^x q(t)e^{H(t)} \, dt\right)</math>

Vipul at 22:54, 12 February 2012

2012-02-12T22:54:32Z

← Older revision		Revision as of 22:54, 12 February 2012
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	{{quotation\|<math>y = e^{-H(x)}\int q(x)e^{H(x)} \, dx</math> where <math>H</math> is an antiderivative of <math>p</math>.}}		{{quotation\|<math>y = e^{-H(x)}\int q(x)e^{H(x)} \, dx</math> where <math>H</math> is an antiderivative of <math>p</math>.}}

			In particular, we obtain that:

			<math>\mbox{General solution} = \mbox{Particular solution} + Ce^{-H(x)}, C \in \R</math>

	The function <math>e^{H(x)}</math> is termed the [[integrating factor]] for the differential equation because multiplying by this turns the differential equation into an exact differential equation, i.e., a differential equation to which we can apply integration on both sides.		The function <math>e^{H(x)}</math> is termed the [[integrating factor]] for the differential equation because multiplying by this turns the differential equation into an exact differential equation, i.e., a differential equation to which we can apply integration on both sides.

Vipul: Created page with "==Definition== ===Format of the differential equation=== A '''first-order linear differential equation''' is a differential equation of the form: $\frac{dy}{dx} + p(x)..."$

2012-02-12T22:53:43Z

Created page with "==Definition== ===Format of the differential equation=== A '''first-order linear differential equation''' is a differential equation of the form: <math>\frac{dy}{dx} + p(x)..."

New page

==Definition==

===Format of the differential equation===

A '''first-order linear differential equation''' is a differential equation of the form:

<math>\frac{dy}{dx} + p(x)y = q(x)</math>

where <math>p,q</math> are known functions.

===Solution method and formula: indefinite integral version===

Let <math>H(x)</math> be an [[antiderivative]] for <math>p(x)</math>, so that <math>H'(x) = p(x)</math>. Then, we multiply both sides by <math>e^{H(x)}</math>. Simplifying, we get:

<math>\frac{d}{dx}[e^{H(x)}y] = q(x)e^{H(x)}</math>

Integrating, we get:

<math>e^{H(x)}y = \int q(x)e^{H(x)} \, dx</math>

Rearranging, we get:

{{quotation|<math>y = e^{-H(x)}\int q(x)e^{H(x)} \, dx</math> where <math>H</math> is an antiderivative of <math>p</math>.}}

The function <math>e^{H(x)}</math> is termed the [[integrating factor]] for the differential equation because multiplying by this turns the differential equation into an exact differential equation, i.e., a differential equation to which we can apply integration on both sides.

First-order linear differential equation - Revision history

Vipul at 23:42, 5 July 2012

Vipul at 15:15, 29 June 2012

Vipul at 22:54, 12 February 2012

Vipul: Created page with "==Definition== ===Format of the differential equation=== A '''first-order linear differential equation''' is a differential equation of the form: \frac{dy}{dx} + p(x)..."

Vipul: Created page with "==Definition== ===Format of the differential equation=== A '''first-order linear differential equation''' is a differential equation of the form: $\frac{dy}{dx} + p(x)..."$