FP2 1st Order Differential Equations - Revision history

IDI: /* Notes */

2014-06-10T14:04:55Z

‎Notes

2014-06-10T14:03:33Z

‎Objectives

2014-06-10T14:02:30Z

2014-06-10T14:02:16Z

2014-06-10T13:27:40Z

‎Objectives

2014-06-10T13:27:22Z

2013-12-16T10:54:41Z

‎Ex 4E q.17

2013-12-16T10:54:24Z

‎Ex 4E q.17

2013-12-16T10:53:38Z

‎Ex 4E q.17

2013-12-16T10:51:51Z

‎Ex 4E q.17

← Older revision		Revision as of 14:04, 10 June 2014
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	== Assessment ==		== Assessment ==
	== Notes ==		== Notes ==
		+	* Families of solution curves
		+	* Particular solutions for given initial conditions

@@ Line 5: / Line 5: @@
 * Solution of exact (result of product rule) first order D.E.s
 * Solution of first order D.E.s of the form <math>\frac{\mathrm{d}y}{\mathrm{d}{x}}+Py=Q</math>, where <math>P</math> and <math>Q</math> are functions of <math>x</math> by multiplying by the integrating factor <math>e^{\int{P\mathrm{d}{x}}}</math>
 </onlyinclude>

@@ Line 4: / Line 4: @@
 * Formation and solution of first order differential equations with separable variables
 * Solution of exact (result of product rule) first order D.E.s
-* Solution of first order D.E.s of the form <math>\frac{\mathrm{d}y}{\mathrm{d}{x}}+Py=Q</math>, where <math>P</math> and <math>Q</math> are functions of <math>x</math> by multiplying by the integrating factor <math>e^{\int{P\mathrm{d}{x}}</math>
+* Solution of first order D.E.s of the form <math>\frac{\mathrm{d}y}{\mathrm{d}{x}}+Py=Q</math>, where <math>P</math> and <math>Q</math> are functions of <math>x</math> by multiplying by the integrating factor <math>e^{\int{P\mathrm{d}{x}}}</math>
 </onlyinclude>

@@ Line 2: / Line 2: @@
 == Objectives ==
 <onlyinclude>
-* The manipulation and solution of algebraic inequalities and inequations, including those involving the modulus sign.
+* Formation and solution of first order differential equations with separable variables
 </onlyinclude>
@@ Line 10: / Line 12: @@
 == Homework ==
 == Key questions ==
 == Assessment ==
 == Notes ==

@@ Line 2: / Line 2: @@
 == Objectives ==
 <onlyinclude>
-* The manipulation and solution of algebraic
+* The manipulation and solution of algebraic inequalities and inequations, including those involving the modulus sign.
 </onlyinclude>
 == Preparation ==
 == Supporting activities ==

@@ Line 1: / Line 1: @@
-=== By substitution ===
+Part of [[FP2]]
-==== Ex 4E q.17 ====
+== Objectives ==
-Use the substitution <math>u=x+y</math> to transform the differential equation <math>\frac{dy}{dx}=(x+y+1)(x+y-1)</math> into a linear equation.
+<onlyinclude>
+* The manipulation and solution of algebraic
-Hence obtain the general solution of the original equation.
+inequalities and inequations, including those
+involving the modulus sign.
-<math>\dfrac{du}{dx}=1+\dfrac{dy}{dx}</math> so <math>\dfrac{dy}{dx}=\dfrac{du}{dx}-1</math>.
+</onlyinclude>
+== Preparation ==
-This yields the transformed equation: <math>\dfrac{du}{dx}-1=(u+1)(u-1)</math>.
+== Supporting activities ==
+== Consolidation ==
-Hence, <math>\dfrac{du}{dx}=u^2</math>.
+== Homework ==
+== Key questions ==
-Separating the variables: <math>\int{u^{-2}}du=\int{1}dx</math>.
+For example
+*<math>\dfrac{1}{x-4}>\dfrac{x}{x-2}</math>
-So
+*<math>|x^2-1|>2(x+1)</math>
+== Assessment ==
-<math>
+== Notes ==

@@ Line 13: / Line 13: @@
 Separating the variables: <math>\int{u^{-2}}du=\int{1}dx</math>.
-So <math>
+So
 \begin{align}
 -u^{-1}&=x+c\\

@@ Line 13: / Line 13: @@
 Separating the variables: <math>\int{u^{-2}}du=\int{1}dx</math>.
-So <math>\begin{align}
+So <math>
 -u^{-1}&=x+c\\
 \frac{-1}{x+y}&=x+c\\
@@ Line 19: / Line 20: @@
 yx+yc&=-1-x^2\\
 y&=\frac{-1-x^2}{x+c}\\
-\end{align}</math>
+\end{align}