First-order differential equation - Revision history

Vipul: /* Geometric description of solutions */

2012-07-07T00:09:22Z

Geometric description of solutions

← Older revision		Revision as of 00:09, 7 July 2012
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	A ''solution curve'' to a differential equation is a curve in the <math>xy</math>-plane corresponding to any solution to the differential equation. For instance, if <math>R(x,y) = 0</math> is a relational solution, then the curve <math>R(x,y) = 0</math> gives a solution curve for the differential equation.		A ''solution curve'' to a differential equation is a curve in the <math>xy</math>-plane corresponding to any solution to the differential equation. For instance, if <math>R(x,y) = 0</math> is a relational solution, then the curve <math>R(x,y) = 0</math> gives a solution curve for the differential equation.

	For first-order differential equations, we ''generally'' expect that specifying a point on the curve uniquely determines the solution curve (because that's an initial value specification). Thus, the solution curves are expected to be (mostly) non-intersecting curves that (hopefully) cover the plane or a large part of the plane.		For first-order differential equations, we ''generally'' expect that specifying a point on the curve uniquely determines the solution curve (because that's an initial value specification). Thus, the solution curves are expected to be (mostly) non-intersecting curves that (hopefully) cover the plane or a large part of the plane. This general expectation is likely to be met for first-order first-degree differential equations. For first-order differential equations of degree <math>n</math>, we generally expect up to <math>n</math> solution curves through a generic point.

	==Solution strategies==		==Solution strategies==

Vipul: /* Solution strategies in particular cases */

2012-07-05T18:38:34Z

Solution strategies in particular cases

← Older revision		Revision as of 18:38, 5 July 2012
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	\|-		\|-
	\| [[first-order exact differential equation]] <math>F(x,y,y') = 0</math> \|\| 1 \|\| Try to find a relation <math>R(x,y)</math> such that <math>F(x,y,y') = \frac{d}{dx}[R(x,y)]</math> using [[implicit differentiation]]. Finding the <math>R</math>, even if it does exist, can be tricky.		\| [[first-order exact differential equation]] <math>F(x,y,y') = 0</math> \|\| 1 \|\| Try to find a relation <math>R(x,y)</math> such that <math>F(x,y,y') = \frac{d}{dx}[R(x,y)]</math> using [[implicit differentiation]]. Finding the <math>R</math>, even if it does exist, can be tricky.
			\|-
			\| [[Bernoulli differential equation]] <math>y' + p(x)y = q(x)y^n</math> (<math>n \ne 0,1</math>) \|\| 1 \|\| Divide both sides by <math>y^n</math> (set aside possible stationary solution <math>y = 0</math>), then substitute <math>w = 1/y^{n-1}</math> to get a [[first-order linear differential equation]] with dependent variable <math>w</math> and independent variable <math>x</math>.
	\|-		\|-
	\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.		\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.

Vipul: /* Reduction methods */

2012-07-05T18:34:26Z

Reduction methods

← Older revision		Revision as of 18:34, 5 July 2012
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	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Method !! Can it reduce the degree of the differential equation, and potentially convert a higher~~-order~~ differential equation to first-~~order~~ differential equations? !! Description		! Method !! Can it reduce the degree of the differential equation, and potentially convert a higher degree differential equation to first-degree differential equations? !! Description
	\|-		\|-
	\| [[substitution method for solving differential equations]] (the substitutions of interest are zeroth-order substitutions) \|\| No \|\| We put <math>u = g(x,y)</math> and replace one of the variables <math>x,y</matH> with <math>u</math>. Then, after solving, we plug back <math>u = g(x,y)</math> to get the relational solutions between <math>x</math> and <math>y</math>.		\| [[substitution method for solving differential equations]] (the substitutions of interest are zeroth-order substitutions) \|\| No \|\| We put <math>u = g(x,y)</math> and replace one of the variables <math>x,y</matH> with <math>u</math>. Then, after solving, we plug back <math>u = g(x,y)</math> to get the relational solutions between <math>x</math> and <math>y</math>.

Vipul: /* Solution strategies in particular cases */

2012-07-05T18:33:49Z

Solution strategies in particular cases

← Older revision		Revision as of 18:33, 5 July 2012
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	\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.		\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.
	\|-		\|-
	\| [[Lagrange equation]] <math>y = f(y')x + g(y')</math> which is linear in <math>x</math> and <math>y</math> but not necessarily in <math>y'</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| ~~See~~ the ~~page~~ for ~~details~~		\| [[Lagrange equation]] <math>y = f(y')x + g(y')</math> which is linear in <math>x</math> and <math>y</math> but not necessarily in <math>y'</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| General solution is a family of curves, each described as a parametric curve with parameter the derivative <math>y'</math> (which we denote by <math>p</math>). There may be some special straight line solutions of the form <math>y = px + g(p)</math> for values <math>p</math> with <math>p = f(p)</math>.
	\|}		\|}

Vipul: /* Solution strategies in particular cases */

2012-07-05T18:32:40Z

Solution strategies in particular cases

← Older revision		Revision as of 18:32, 5 July 2012
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	\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.		\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.
	\|-		\|-
	\| [[Lagrange equation]] <math>f(y')x + g~~(y')y = h~~(y')</math> which is linear in <math>x</math> and <math>y</math> but not necessarily in <math>y'</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| See the page for details		\| [[Lagrange equation]] <math>y = f(y')x + g(y')</math> which is linear in <math>x</math> and <math>y</math> but not necessarily in <math>y'</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| See the page for details
	\|}		\|}

Vipul: /* Solution strategies in particular cases */

2012-07-05T17:52:56Z

Solution strategies in particular cases

← Older revision		Revision as of 17:52, 5 July 2012
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	\|-		\|-
	\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.		\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.
			\|-
			\| [[Lagrange equation]] <math>f(y')x + g(y')y = h(y')</math> which is linear in <math>x</math> and <math>y</math> but not necessarily in <math>y'</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| See the page for details
	\|}		\|}

Vipul: /* Reduction methods */

2012-07-05T17:49:58Z

Reduction methods

← Older revision		Revision as of 17:49, 5 July 2012
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	{\| class="sortable" border="1"		{\| class="sortable" border="1"
	! Method !! Can it reduce the degree of the differential equation? !! Description		! Method !! Can it reduce the degree of the differential equation, and potentially convert a higher-order differential equation to first-order differential equations? !! Description
	\|-		\|-
	\| [[substitution method for solving differential equations]] (the substitutions of interest are zeroth-order substitutions) \|\| No \|\| We put <math>u = g(x,y)</math> and replace one of the variables <math>x,y</matH> with <math>u</math>. Then, after solving, we plug back <math>u = g(x,y)</math> to get the relational solutions between <math>x</math> and <math>y</math>.		\| [[substitution method for solving differential equations]] (the substitutions of interest are zeroth-order substitutions) \|\| No \|\| We put <math>u = g(x,y)</math> and replace one of the variables <math>x,y</matH> with <math>u</math>. Then, after solving, we plug back <math>u = g(x,y)</math> to get the relational solutions between <math>x</math> and <math>y</math>.

Vipul: /* Solution strategies */

2012-07-05T17:49:22Z

Solution strategies

@@ Line 75: / Line 75: @@
 |-
 | [[separable differential equation]] which is of the form <math>y' = f(x)g(y)</math> (any [[first-order first-degree autonomous differential equation]] is separable, though there are separable differential equations that aren't autonomous) || 1 || Separate and solve as <math>\int \frac{dy}{g(y)} = \int f(x) \, dx</math>. Also find solutions corresponding to <math>y = k</math> where <math>g(k) = 0</math>.
 |-
 | [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> || need not be polynomial; if polynomial, may have any degree || <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.
 |}

Vipul: /* Solution strategies */

2012-07-01T19:09:28Z

Solution strategies

← Older revision		Revision as of 19:09, 1 July 2012
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	\| [[first-order linear differential equation]] which in simplified form looks like <math>y' + p(x)y = q(x)</math> \|\| 1 \|\| Use the [[integrating factor]] <math>e^{H(x)}</math> where <math>H'=p</math>. The general solution is <math>y = Ce^{-H(x)} + e^{-H(x)}\int p(x)e^{H(x)} \, dx</math>		\| [[first-order linear differential equation]] which in simplified form looks like <math>y' + p(x)y = q(x)</math> \|\| 1 \|\| Use the [[integrating factor]] <math>e^{H(x)}</math> where <math>H'=p</math>. The general solution is <math>y = Ce^{-H(x)} + e^{-H(x)}\int p(x)e^{H(x)} \, dx</math>
	\|-		\|-
	\| [[separable differential equation]] which is of the form <math>y' = f(x)g(y)</math> (any first-order [[autonomous differential equation]] is separable, though there are separable differential equations that aren't autonomous) \|\| 1 \|\| Separate and solve as <math>\int \frac{dy}{g(y)} = \int f(x) \, dx</math>. Also find solutions corresponding to <math>y = k</math> where <math>g(k) = 0</math>.		\| [[separable differential equation]] which is of the form <math>y' = f(x)g(y)</math> (any [[first-order first-degree autonomous differential equation]] is separable, though there are separable differential equations that aren't autonomous) \|\| 1 \|\| Separate and solve as <math>\int \frac{dy}{g(y)} = \int f(x) \, dx</math>. Also find solutions corresponding to <math>y = k</math> where <math>g(k) = 0</math>.
	\|-		\|-
	\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.		\| [[Clairaut's equation]] which is of the form <math>y = xy' + f(y')</math> \|\| need not be polynomial; if polynomial, may have any degree \|\| <math>y = Cx + f(C)</math> with <math>C \in \R</math> (all straight lines) and a single other solution explicitly described as the solution to <math>x + f(dy/dx) = 0</math>, given by <math>x = -f'(p), y = f(p) - pf'(p)</math> as a parametric curve in terms of <math>p</math>.
	\|}		\|}

Vipul at 17:17, 29 June 2012

2012-06-29T17:17:08Z

@@ Line 20: / Line 20: @@
 * '''Relational solution''': A relation <math>R(x,y) = 0</math> is termed a ''relational solution'' to the equation if <math>F(x,y,y') = 0</math> holds true for all <math>x,y</math> if we calculate the derivative <math>y'</math> using [[implicit differentiation]].
 ==Terminology==
@@ Line 36: / Line 48: @@
 | solution to initial value problem || a particular solution that satisfies the initial value condition. || A particular solution to <math>y + y' = 0</math> satisfying <math>y(0) = 1</math> is <math>y = e^{-x}</math>.
 |}
 ==Solution strategies==