12-267/Homework Assignment 1 - Revision history

Twine: Added student divider

2012-10-24T21:05:13Z

Added student divider

← Older revision		Revision as of 17:05, 24 October 2012
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	# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.		# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

			{{Template:12-267:Dror/Students Divider}}
	Disclamer: The solutions below are by students, for students.

	'''Solution to Question 1.''' [[User:Twine\|Twine]]		'''Solution to Question 1.''' [[User:Twine\|Twine]]

Twine: Changed format of Mathstudent's solutions to a gallery

2012-10-24T21:00:32Z

Changed format of Mathstudent's solutions to a gallery

@@ Line 16: / Line 16: @@
 Disclamer: The solutions below are by students, for students.
 '''Solution to Question 1.''' [[User:Twine|Twine]]
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 Hence, <math>\forall c \in \mathbb{R}</math>, <math>cy_1 + y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

Twine: Corrected some formatting, added solution for question 1

2012-10-24T20:50:53Z

Corrected some formatting, added solution for question 1

← Older revision		Revision as of 16:50, 24 October 2012
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	# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.		# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

			Disclamer: The solutions below are by students, for students.
	[http://drorbn.net/index.php?title=Image:001.jpg Solution to HW1, page 1] [[User:Mathstudent\|Mathstudent]]

	[http://drorbn.net/index.php?title=Image:002.jpg Solution to HW1, page 2] [[User:Mathstudent\|Mathstudent]]
	[http://drorbn.net/index.php?title=Image:~~003~~.jpg ~~Solution~~ to ~~HW1~~, page 3] [[User:Mathstudent\|Mathstudent]]		Solutions to HW1: [http://drorbn.net/index.php?title=Image:001.jpg page 1], [http://drorbn.net/index.php?title=Image:002.jpg page 2], [http://drorbn.net/index.php?title=Image:003.jpg page 3], [http://drorbn.net/index.php?title=Image:004.jpg page 4] [[User:Mathstudent\|Mathstudent]]

	[http://drorbn.net/index.php?title=Image:004.jpg Solution to HW1, page 4] [[User:Mathstudent\|Mathstudent]]
			'''Solution to Question 1.''' [[User:Twine\|Twine]]

			Take y defined by <math>y=cy_1+y_2</math> and plug it into <math>y'+p(x)y=g(x)</math>. We get

			<math>(cy_1 + y_2)' + p(x)(cy_1 + y_2) = g(x)</math>

			<math>c(y_1' + p(x)y_1) + (y_2' + p(x)y_2) = g(x)</math>

			Based on our assumptions about <math>y_1</math> and <math>y_2</math>, we have <math>y_1' + p(x)y_1 = 0</math> and we have <math>y_2' + p(x)y_2 = g(x)</math>, and so the above equation holds <math>\forall c \in \mathbb{R}</math>.

			Hence, <math>\forall c \in \mathbb{R}</math>, <math>cy_1 + y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

Mathstudent at 23:22, 23 October 2012

2012-10-23T23:22:58Z

← Older revision		Revision as of 19:22, 23 October 2012
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	# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.		# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.
	# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.		# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

			[http://drorbn.net/index.php?title=Image:001.jpg Solution to HW1, page 1] [[User:Mathstudent\|Mathstudent]]
			[http://drorbn.net/index.php?title=Image:002.jpg Solution to HW1, page 2] [[User:Mathstudent\|Mathstudent]]
			[http://drorbn.net/index.php?title=Image:003.jpg Solution to HW1, page 3] [[User:Mathstudent\|Mathstudent]]
			[http://drorbn.net/index.php?title=Image:004.jpg Solution to HW1, page 4] [[User:Mathstudent\|Mathstudent]]

Drorbn at 12:30, 18 September 2012

2012-09-18T12:30:18Z

← Older revision		Revision as of 08:30, 18 September 2012
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	{{12-267/Navigation}}		{{12-267/Navigation}}

	This assignment is due at the tutorial on ~~Thursday~~ September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.		This assignment is due at the tutorial on Tuesday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

	'''Question 1.''' Show that if <math>y=y_1(x)</math> is a solution of <math>y'+p(x)y=0</math>, and <math>y=y_2(x)</math> is a solution of <math>y'+p(x)y=g(x)</math>, then for any constant <math>c</math>, <math>y=cy_1+y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.		'''Question 1.''' Show that if <math>y=y_1(x)</math> is a solution of <math>y'+p(x)y=0</math>, and <math>y=y_2(x)</math> is a solution of <math>y'+p(x)y=g(x)</math>, then for any constant <math>c</math>, <math>y=cy_1+y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

Drorbn at 11:23, 18 September 2012

2012-09-18T11:23:29Z

← Older revision		Revision as of 07:23, 18 September 2012
Line 1:		Line 1:
	{{12-267/Navigation}}		{{12-267/Navigation}}
	{{In Preparation}}

	This assignment is due at the tutorial on Thursday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.		This assignment is due at the tutorial on Thursday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

Drorbn at 21:31, 17 September 2012

2012-09-17T21:31:13Z

← Older revision		Revision as of 17:31, 17 September 2012
Line 1:		Line 1:
	{{12-267/Navigation}}		{{12-267/Navigation}}
	{{In Preparation}}		{{In Preparation}}

		⚫	This assignment is due at the tutorial on Thursday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

	'''Question 1.''' Show that if <math>y=y_1(x)</math> is a solution of <math>y'+p(x)y=0</math>, and <math>y=y_2(x)</math> is a solution of <math>y'+p(x)y=g(x)</math>, then for any constant <math>c</math>, <math>y=cy_1+y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.		'''Question 1.''' Show that if <math>y=y_1(x)</math> is a solution of <math>y'+p(x)y=0</math>, and <math>y=y_2(x)</math> is a solution of <math>y'+p(x)y=g(x)</math>, then for any constant <math>c</math>, <math>y=cy_1+y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.
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	# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.		# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.
	# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.		# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

⚫	This assignment is due at the tutorial on Thursday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

Drorbn at 21:30, 17 September 2012

2012-09-17T21:30:41Z

← Older revision		Revision as of 17:30, 17 September 2012
Line 13:		Line 13:
	# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.		# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.
	# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.		# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

			This assignment is due at the tutorial on Thursday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

Drorbn at 20:28, 17 September 2012

2012-09-17T20:28:36Z

← Older revision		Revision as of 16:28, 17 September 2012
Line 10:		Line 10:
	# <math>\frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y}</math>.		# <math>\frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y}</math>.
	# <math>xdx+ye^{-x}dy=0</math>, with <math>y(0)=1</math>.		# <math>xdx+ye^{-x}dy=0</math>, with <math>y(0)=1</math>.
	# <math>\frac{dy}{dx}=\frac{ax+b}{cx+d}</math>, where <math>a,b,c,d</math> are arbitrary constants.		# <math>\frac{dy}{dx}=\frac{ay+b}{cx+d}</math>, where <math>a,b,c,d</math> are arbitrary constants.
	# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.		# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.
	# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.		# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

Drorbn at 19:57, 17 September 2012

2012-09-17T19:57:16Z

← Older revision		Revision as of 15:57, 17 September 2012
Line 10:		Line 10:
	# <math>\frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y}</math>.		# <math>\frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y}</math>.
	# <math>xdx+ye^{-x}dy=0</math>, with <math>y(0)=1</math>.		# <math>xdx+ye^{-x}dy=0</math>, with <math>y(0)=1</math>.
	# <math>\frac{dy}{dx}=\frac{ax+b}{cx+d}</math>, where <math>a,b,c,d</math> are constants.		# <math>\frac{dy}{dx}=\frac{ax+b}{cx+d}</math>, where <math>a,b,c,d</math> are arbitrary constants.
	# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are constants.		# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.
	# <math>0=(e^x\sin y + 3y)dx - (~~3x-~~e^x\~~sin~~ y)dy</math>.		# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.