Second-order first-degree autonomous differential equation - Revision history

Vipul: /* Case C_1 = 0 */

2012-07-30T23:45:29Z

Case C_1 = 0

← Older revision		Revision as of 23:45, 30 July 2012
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	<math>x = \frac{-1}{t + C_2}</math>		<math>x = \frac{-1}{t + C_2}</math>

	To verify that this is a solution, we compute:		To verify that this is a solution, we compute, using the [[chain rule for differentiation]] and [[differentiation rule for power functions]]:

	<math>\frac{dx}{dt} = \frac{1}(t + C_2)^2}</math>		<math>\frac{dx}{dt} = \frac{1}{(t + C_2)^2}</math>

	and:		and:

	<math>\frac{d^2x}{dt^2} = \frac{-2}{(t + C_2)^3}</math>		<math>\frac{d^2x}{dt^2} = \frac{-2}{(t + C_2)^3}</math>

			We can now plug these into the original differential equation and verify.

	====Case <math>C_1 > 0</math>====		====Case <math>C_1 > 0</math>====

Vipul: /* Case C_1 = 0 */

2012-07-30T23:44:37Z

Case C_1 = 0

← Older revision		Revision as of 23:44, 30 July 2012
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	<math>x = \frac{-1}{t + C_2}</math>		<math>x = \frac{-1}{t + C_2}</math>

			To verify that this is a solution, we compute:

			<math>\frac{dx}{dt} = \frac{1}(t + C_2)^2}</math>

			and:

			<math>\frac{d^2x}{dt^2} = \frac{-2}{(t + C_2)^3}</math>

	====Case <math>C_1 > 0</math>====		====Case <math>C_1 > 0</math>====

Vipul: /* Case C_1 > 0 */

2012-07-30T23:43:33Z

Case C_1 > 0

← Older revision		Revision as of 23:43, 30 July 2012
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	<math>x = k \tan(k(t + C_2)), \qquad k > 0, C_2 \in \R</math>		<math>x = k \tan(k(t + C_2)), \qquad k > 0, C_2 \in \R</math>

	We can verify that this ~~satisfies~~ the original differential equation.		{{quotation\|Note that a priori, applying tan to both sides causes us to lose the information that <math>k(t + C_2)</math> is in the range of <math>\arctan</math>. However, this constraint is "artificial" anyway and can be got rid of by readjusting the value of <math>C_2</math>. Note that arc tangent constraints are artificial, but arc sine constraints are genuine, because the sine function repeats itself ''within'' a period but the tangent function does not.}}

			To verify that this is a solution, we can compute:

			<math>\frac{dx}{dt} = k^2 \sec^2(k(t + C_2))</math>

			<math>\frac{d^2x}{dt^2} = 2k^3 \sec^2(k(t + C_2)) \tan(k(t + C_2))</math>

			We can now plug in and verify the original differential equation.

	====Case <math>C_1 < 0</math>====		====Case <math>C_1 < 0</math>====

Vipul: /* Case C_1 = 0= */

2012-07-05T23:16:09Z

Case C_1 = 0=

← Older revision		Revision as of 23:16, 5 July 2012
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	We now make cases based on the sign of <math>C_1</math>.		We now make cases based on the sign of <math>C_1</math>.

	===Case <math>C_1 = 0</math>====		====Case <math>C_1 = 0</math>====

	In this case, the differential equation becomes:		In this case, the differential equation becomes:

Vipul: /* Example */

2012-07-05T17:02:12Z

Example

← Older revision		Revision as of 17:02, 5 July 2012
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	==Example==		==Example==

			===Separable example===

	Consider the following differential equation:		Consider the following differential equation:
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	We now make cases based on the sign of <math>C_1</math>.		We now make cases based on the sign of <math>C_1</math>.

	====Case <math>C_1 = 0</math>===		===Case <math>C_1 = 0</math>====

	In this case, the differential equation becomes:		In this case, the differential equation becomes:
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	<math>x = \frac{-1}{t + C_2}</math>		<math>x = \frac{-1}{t + C_2}</math>

	====Case <math>C_1 > 0</math>===		====Case <math>C_1 > 0</math>====

	Let <math>k = \sqrt{C_1}</math>. The differential equation becomes:		Let <math>k = \sqrt{C_1}</math>. The differential equation becomes:
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	<math>x = l\left(\frac{1 + \exp(2l(t + C_2))}{1 - \exp(2l(t + C_2))}\right) \qquad \mbox{ or } x = l\left(\frac{1 - \exp(2l(t + C_2))}{1 + \exp(2l(t + C_2))}\right) </math>		<math>x = l\left(\frac{1 + \exp(2l(t + C_2))}{1 - \exp(2l(t + C_2))}\right) \qquad \mbox{ or } x = l\left(\frac{1 - \exp(2l(t + C_2))}{1 + \exp(2l(t + C_2))}\right) </math>

			====All cases====

			The following are the solution families:

			* <math>x = C_0, C_0 \in \R</math> (this is a singular solution family)
			* <math>x = \frac{-1}{t + C_2}, C_2 \in \R</math>
			* <math>x = k \tan(k(t + C_2)), \qquad k > 0, C_2 \in \R</math>
			* <math>x = l\left(\frac{1 + \exp(2l(t + C_2))}{1 - \exp(2l(t + C_2))}\right), \qquad l > 0, C_2 \in \R</math>
			* <math>x = l\left(\frac{1 - \exp(2l(t + C_2))}{1 + \exp(2l(t + C_2))}\right), \qquad l > 0, C_2 \in \R</math>

Vipul: /* Case C_1 */

2012-07-05T16:58:50Z

Case C_1

← Older revision		Revision as of 16:58, 5 July 2012
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	The solutions are respectively:		The solutions are respectively:

	<math>x = l\frac{1 + \exp(2l(t + C_2))}{1 - \exp(2l(t + C_2))} \qquad \mbox{ or } x = l\frac{1 - \exp(2l(t + C_2))}{1 + \exp(2l(t + C_2))} </math>		<math>x = l\left(\frac{1 + \exp(2l(t + C_2))}{1 - \exp(2l(t + C_2))}\right) \qquad \mbox{ or } x = l\left(\frac{1 - \exp(2l(t + C_2))}{1 + \exp(2l(t + C_2))}\right) </math>

Vipul at 16:58, 5 July 2012

2012-07-05T16:58:17Z

← Older revision		Revision as of 16:58, 5 July 2012
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	This is now a [[first-order differential equation]]. If <math>H</math> can locally be inverted, we can write <math>dx/dt</math> as an explicit function of <math>x</math> (locally) and then solve the resultant [[separable differential equation]]. Otherwise, there may be some other method available.		This is now a [[first-order differential equation]]. If <math>H</math> can locally be inverted, we can write <math>dx/dt</math> as an explicit function of <math>x</math> (locally) and then solve the resultant [[separable differential equation]]. Otherwise, there may be some other method available.

			==Example==

			Consider the following differential equation:

			<math>\frac{d^2x}{dt^2} = 2x\frac{dx}{dt}</math>

			We describe how to solve this using the technique discussed here. First, set <math>v = dx/dt</math>. We get:

			<math>v \frac{dv}{dx} = 2xv</math>

			Thus, we get:

			<math>v = 0 \qquad \mbox{ or } \qquad \frac{dv}{dx} = 2x</math>

			The case <math>v = 0</math> solves to gives <math>x = C_0, C_0 \in \R</math>

			The case:

			<math>\frac{dv}{dx} = 2x</math>

			solves to give:

			<math>v = x^2 + C_1</math>

			Now, we plug back the original substitution to get:

			<math>\frac{dx}{dt} = x^2 + C_1</math>

			We now make cases based on the sign of <math>C_1</math>.

			====Case <math>C_1 = 0</math>===

			In this case, the differential equation becomes:

			<math>\frac{dx}{dt} = x^2</math>

			This has a stationary solution <math>x = 0</math> (which, incidentally, is ''already'' included in the prior solution family of all constant solutions) and the general solution can be obtained as:

			<math>\int \frac{dx}{x^2} = \int dt</math>

			This gives:

			<math>\frac{-1}{x} = t + C_2</math>

			This solves to give:

			<math>x = \frac{-1}{t + C_2}</math>

			====Case <math>C_1 > 0</math>===

			Let <math>k = \sqrt{C_1}</math>. The differential equation becomes:

			<math>\frac{dx}{dt} = x^2 + k^2</math>

			There are no stationary solutions. The general solution is given by:

			<math>\int \frac{dx}{x^2 + k^2} = \int dt</math>

			This solves to give:

			<math>\frac{1}{k} \arctan\left(\frac{x}{k}\right) = t + C_2</math>

			Rearranging, we get:

			<math>x = k \tan(k(t + C_2)), \qquad k > 0, C_2 \in \R</math>

			We can verify that this satisfies the original differential equation.

			====Case <math>C_1 < 0</math>====

			Let <math>l = \sqrt{-C_1}</math>. The differential equation becomes:

			<math>\frac{dx}{dt} = x^2 - l^2</math>

			There are stationary solutions corresponding to <math>x = l</math> and <math>x = -l</math>. However, all stationary solutions are already included in another solution family, so we can ignore these. We continue solving:

			<math>\int \frac{dx}{x^2 - l^2} = \int dt</math>

			We get:

			<math>\frac{1}{2l} \ln \left\|\frac{x - l}{x + l}\right\| = t + C_2</math>

			This rearranges to gives:

			<math>\ln \left\|\frac{x - l}{x + l}\right\| = 2l(t + C_2)</math>

			Exponentiate both sides to get:

			<math>\frac{x - l}{x + l} = \exp(2l(t + C_2)) \qquad \mbox{ or } \frac{x - l}{x + l} = -\exp(2l(t + C_2))</math>

			The solutions are respectively:

			<math>x = l\frac{1 + \exp(2l(t + C_2))}{1 - \exp(2l(t + C_2))} \qquad \mbox{ or } x = l\frac{1 - \exp(2l(t + C_2))}{1 + \exp(2l(t + C_2))} </math>

Vipul: Vipul moved page Second-order autonomous differential equation of degree one to Second-order first-degree autonomous differential equation

2012-07-05T16:43:13Z

Vipul moved page Second-order autonomous differential equation of degree one to Second-order first-degree autonomous differential equation

← Older revision	Revision as of 16:43, 5 July 2012
(No difference)

Vipul: Vipul moved page Second-order autonomous differential equation to Second-order autonomous differential equation of degree one

2012-06-30T16:19:29Z

Vipul moved page Second-order autonomous differential equation to Second-order autonomous differential equation of degree one

← Older revision	Revision as of 16:19, 30 June 2012
(No difference)

Vipul at 16:59, 29 June 2012

2012-06-29T16:59:44Z

← Older revision		Revision as of 16:59, 29 June 2012
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	===Form of the differential equation===		===Form of the differential equation===

	A (one-dimensional) '''second-order autonomous differential equation''' is a differential equation of the form:		A (one-dimensional and degree one) '''second-order autonomous differential equation''' is a differential equation of the form:

	<math>\frac{d^2x}{dt^2} = F(x, \frac{dx}{dt})</math>		<math>\frac{d^2x}{dt^2} = F(x, \frac{dx}{dt})</math>