Clairaut's equation - Revision history

Vipul: /* Solution method and formula */

2012-07-24T16:47:30Z

Solution method and formula

← Older revision		Revision as of 16:47, 24 July 2012
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	* The solutions for the <math>\frac{d^2y}{dx^2} = 0</math> case are straight lines of the form <math>y = Cx + f(C)</math>. This is the solution family for the general solution.		* The solutions for the <math>\frac{d^2y}{dx^2} = 0</math> case are straight lines of the form <math>y = Cx + f(C)</math>. This is the solution family for the general solution.
	* The solution for the <math>x + f'\left(\frac{dy}{dx}\right) = 0</math> case is unique: it is a parametric curve given by <math>x = -f'(p), y = f(p) - pf'(p)</math> where <math>p = dy/dx</math> is the parameter moving along the curve. This is the singular solution. The curve representing this is the envelope of the general solution curves.		* The solution for the <math>x + f'\left(\frac{dy}{dx}\right) = 0</math> case is unique: it is a parametric curve given by <math>x = -f'(p), y = f(p) - pf'(p)</math> where <math>p = dy/dx</math> is the parameter moving along the curve. This is the singular solution. The curve representing this is the envelope of the general solution curves. In order to verify that this solution is correct, we can calculate <math>dy/dx</math> using parametric differentiation with respect to <math>p</math> and check that it is indeed equal to <math>p</math>.

	==Related equations==		==Related equations==

Vipul: /* Clairaut's after substitution */

2012-07-06T21:22:39Z

Clairaut's after substitution

← Older revision		Revision as of 21:22, 6 July 2012
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	<math>0 = x^2y' + \cos(y + xy')</math>		<math>0 = x^2y' + \cos(y + xy')</math>

	We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>xy' = u' - y</math>, so <math>x^2y' = x(xy') = x(u' - y)</math>, so:		We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>xy' = u' - y</math>, so <math>x^2y' = x(xy') = x(u' - y) = xu' - u</math>, so:

	<math>0 = ~~x(u~~' - y) + \cos(u')</math>		<math>0 = xu' - u + \cos(u')</math>

	Rearrange to get:		Rearrange to get:

	~~<math>xy = xu' + \cos(u')</math>~~

	~~Now plug <math>u = xy</math> on the left to get:~~

	<math>u = xu' + \cos(u')</math>		<math>u = xu' + \cos(u')</math>

Vipul: /* Clairaut's after substitution */

2012-07-06T21:17:22Z

Clairaut's after substitution

← Older revision		Revision as of 21:17, 6 July 2012
Line 53:		Line 53:
	<math>0 = x^2y' + \cos(y + xy')</math>		<math>0 = x^2y' + \cos(y + xy')</math>

	We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>x^2y' = x(u' - y)</math>, so:		We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>xy' = u' - y</math>, so <math>x^2y' = x(xy') = x(u' - y)</math>, so:

	<math>0 = x(u' - y) + \cos(u')</math>		<math>0 = x(u' - y) + \cos(u')</math>

Vipul: /* Clairaut's after substitution */

2012-07-06T21:14:51Z

Clairaut's after substitution

← Older revision		Revision as of 21:14, 6 July 2012
Line 53:		Line 53:
	<math>0 = x^2y' + \cos(y + xy')</math>		<math>0 = x^2y' + \cos(y + xy')</math>

	We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>x^2y' = x(u' - y) ~~= xu' - u~~</math>~~. We thus get~~:		We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>x^2y' = x(u' - y)</math>, so:

	<math>0 = xu' - u + \cos(u')</math>		<math>0 = x(u' - y) + \cos(u')</math>

	Rearrange to get:		Rearrange to get:

			<math>xy = xu' + \cos(u')</math>

			Now plug <math>u = xy</math> on the left to get:

	<math>u = xu' + \cos(u')</math>		<math>u = xu' + \cos(u')</math>

Vipul: /* Clairaut's after substitution */

2012-07-06T21:13:39Z

Clairaut's after substitution

← Older revision		Revision as of 21:13, 6 July 2012
Line 53:		Line 53:
	<math>0 = x^2y' + \cos(y + xy')</math>		<math>0 = x^2y' + \cos(y + xy')</math>

	We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>x^2y' = xu' - u</math>. We thus get:		We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>x^2y' = x(u' - y) = xu' - u</math>. We thus get:

	<math>0 = xu' - u + \cos(u')</math>		<math>0 = xu' - u + \cos(u')</math>

Vipul: /* Examples */

2012-07-05T18:31:44Z

Examples

← Older revision		Revision as of 18:31, 5 July 2012
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	* The solutions for the <math>\frac{d^2y}{dx^2} = 0</math> case are straight lines of the form <math>y = Cx + f(C)</math>. This is the solution family for the general solution.		* The solutions for the <math>\frac{d^2y}{dx^2} = 0</math> case are straight lines of the form <math>y = Cx + f(C)</math>. This is the solution family for the general solution.
	* The solution for the <math>x + f'\left(\frac{dy}{dx}\right) = 0</math> case is unique: it is a parametric curve given by <math>x = -f'(p), y = f(p) - pf'(p)</math> where <math>p = dy/dx</math> is the parameter moving along the curve. This is the singular solution. The curve representing this is the envelope of the general solution curves.		* The solution for the <math>x + f'\left(\frac{dy}{dx}\right) = 0</math> case is unique: it is a parametric curve given by <math>x = -f'(p), y = f(p) - pf'(p)</math> where <math>p = dy/dx</math> is the parameter moving along the curve. This is the singular solution. The curve representing this is the envelope of the general solution curves.

			==Related equations==

			* [[Lagrange equation]] is a more general setup that includes Clairaut's equation as a special case in terms of method, albeit a special case that is qualitatively somewhat different.

	==Examples==		==Examples==

Vipul: /* Examples */

2012-07-05T18:29:51Z

Examples

← Older revision		Revision as of 18:29, 5 July 2012
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	* The solution family for the general solution is <math>y = Cx + \cos C</math>, with <math>C \in \R</math>.		* The solution family for the general solution is <math>y = Cx + \cos C</math>, with <math>C \in \R</math>.
	* There is a special solution given parametrically by <math>x = \sin p, y = \cos p + p \sin p</math>, with <math>p \in \R</math>. Note that this is a ''single'' solution; the parameter <math>p</math> varies to cover the points of the solution, and it is different from the notion of parameter for a family of solutions.		* There is a special solution given parametrically by <math>x = \sin p, y = \cos p + p \sin p</math>, with <math>p \in \R</math>. Note that this is a ''single'' solution; the parameter <math>p</math> varies to cover the points of the solution, and it is different from the notion of parameter for a family of solutions.

			===Clairaut's after substitution===

			Consider the equation:

			<math>0 = x^2y' + \cos(y + xy')</math>

			We first note that the expression whose cosine is being taken is the derivative of <math>xy</math>, hence the natural choice of substitution is to try for <math>u = xy</math>. Our next step is to try and write <math>x^2y'</math> in terms of <math>u</math> and <math>u'</math>. Recall that <math>u = xy</math> and <math>u' = y + xy'</math>, so <math>x^2y' = xu' - u</math>. We thus get:

			<math>0 = xu' - u + \cos(u')</math>

			Rearrange to get:

			<math>u = xu' + \cos(u')</math>

			This is a Clairaut's equation with dependent variable <math>u</math> and independent variable <math>x</math>, so the solutions are:

			* The solution family for the general solution is <math>xy = Cx + \cos C</math>, with <math>C \in \R</math>. This can be rewritten as <math>y = C + (\cos C)/x</math>, with <math>C \in \R</math>.
			* There is a special solution given parametrically by <math>x = \sin p, xy = \cos p + p \sin p</math>, with <math>p \in \R</math>. This can be rewritten as <math>x = \sin p, y = \cot p + p</math>, with <math>p \in \R</math>.

Vipul at 18:23, 5 July 2012

2012-07-05T18:23:04Z

← Older revision		Revision as of 18:23, 5 July 2012
Line 29:		Line 29:
	* The solutions for the <math>\frac{d^2y}{dx^2} = 0</math> case are straight lines of the form <math>y = Cx + f(C)</math>. This is the solution family for the general solution.		* The solutions for the <math>\frac{d^2y}{dx^2} = 0</math> case are straight lines of the form <math>y = Cx + f(C)</math>. This is the solution family for the general solution.
	* The solution for the <math>x + f'\left(\frac{dy}{dx}\right) = 0</math> case is unique: it is a parametric curve given by <math>x = -f'(p), y = f(p) - pf'(p)</math> where <math>p = dy/dx</math> is the parameter moving along the curve. This is the singular solution. The curve representing this is the envelope of the general solution curves.		* The solution for the <math>x + f'\left(\frac{dy}{dx}\right) = 0</math> case is unique: it is a parametric curve given by <math>x = -f'(p), y = f(p) - pf'(p)</math> where <math>p = dy/dx</math> is the parameter moving along the curve. This is the singular solution. The curve representing this is the envelope of the general solution curves.

			==Examples==

			We will denote <matH>dy/dx</math> by <math>y'</math>.

			===Simple example===

			<math>y = xy' + \cos(y')</math>

			The solutions are:

			* The solution family for the general solution is <math>y = Cx + \cos C</math>, with <math>C \in \R</math>.
			* There is a special solution given parametrically by <math>x = \sin p, y = \cos p + p \sin p</math>, with <math>p \in \R</math>. Note that this is a ''single'' solution; the parameter <math>p</math> varies to cover the points of the solution, and it is different from the notion of parameter for a family of solutions.

Vipul: Created page with "==Definition== ===Form of the differential equation=== Clairaut's equation is a first-order differential equation of the form: $y = x \frac{dy}{dx} + f\left(\frac{..."$

2012-02-12T23:12:16Z

Created page with "==Definition== ===Form of the differential equation=== Clairaut's equation is a first-order differential equation of the form: <math>y = x \frac{dy}{dx} + f\left(\frac{..."

New page

==Definition==

===Form of the differential equation===

Clairaut's equation is a [[first-order differential equation]] of the form:

<math>y = x \frac{dy}{dx} + f\left(\frac{dy}{dx}\right)</math>

Here, <math>f</math> is a suitable function.

===Solution method and formula===

Differentiate both sides with respect to <math>x</math> and obtain:

<math>\frac{dy}{dx} = \frac{dy}{dx} + \left[x + f'\left(\frac{dy}{dx}\right)\right]\frac{d^2y}{dx^2}</math>

Cancel the common term from both sides and obtain:

<math> 0 = \left[x + f'\left(\frac{dy}{dx}\right)\right]\frac{d^2y}{dx^2}</math>

This gives two possible solution types:

<math>\frac{d^2y}{dx^2} = 0</math>

<math>x + f'\left(\frac{dy}{dx}\right) = 0</math>

Note that when we differentiated, we lost information, so it is not true that all solutions of the differentiated equation solve the original equation. Rather, we need to plug these solutions into the original equation to constrain them. We have the following:

* The solutions for the <math>\frac{d^2y}{dx^2} = 0</math> case are straight lines of the form <math>y = Cx + f(C)</math>. This is the solution family for the general solution.
* The solution for the <math>x + f'\left(\frac{dy}{dx}\right) = 0</math> case is unique: it is a parametric curve given by <math>x = -f'(p), y = f(p) - pf'(p)</math> where <math>p = dy/dx</math> is the parameter moving along the curve. This is the singular solution. The curve representing this is the envelope of the general solution curves.

Clairaut's equation - Revision history

Vipul: /* Solution method and formula */

Vipul: /* Clairaut's after substitution */

Vipul: /* Clairaut's after substitution */

Vipul: /* Clairaut's after substitution */

Vipul: /* Clairaut's after substitution */

Vipul: /* Examples */

Vipul: /* Examples */

Vipul at 18:23, 5 July 2012

Vipul: Created page with "==Definition== ===Form of the differential equation=== Clairaut's equation is a first-order differential equation of the form: y = x \frac{dy}{dx} + f\left(\frac{..."

Vipul: Created page with "==Definition== ===Form of the differential equation=== Clairaut's equation is a first-order differential equation of the form: $y = x \frac{dy}{dx} + f\left(\frac{..."$