12-267/Homework Assignment 1: Difference between revisions
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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. |
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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>. |
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'''Solution to Question 1.''' [[User:Twine|Twine]] |
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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 |
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<math>(cy_1 + y_2)' + p(x)(cy_1 + y_2) = g(x)</math> |
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<math>c(y_1' + p(x)y_1) + (y_2' + p(x)y_2) = g(x)</math> |
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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>. |
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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>. |
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Solutions to HW1: [[User:Mathstudent|Mathstudent]] |
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Latest revision as of 16:05, 24 October 2012
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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]\displaystyle{ y=y_1(x) }[/math] is a solution of [math]\displaystyle{ y'+p(x)y=0 }[/math], and [math]\displaystyle{ y=y_2(x) }[/math] is a solution of [math]\displaystyle{ y'+p(x)y=g(x) }[/math], then for any constant [math]\displaystyle{ c }[/math], [math]\displaystyle{ y=cy_1+y_2 }[/math] is a solution of [math]\displaystyle{ y'+p(x)y=g(x) }[/math].
Question 2. Solve the following differential equations
- For [math]\displaystyle{ x\gt 0 }[/math], [math]\displaystyle{ xy'+2y=\sin x }[/math].
- [math]\displaystyle{ \frac{dy}{dx}=\frac{1}{e^y-x} }[/math] with [math]\displaystyle{ y(1)=0 }[/math]; you may want to solve for [math]\displaystyle{ x }[/math] first.
- [math]\displaystyle{ xy'=\sqrt{1-y^2} }[/math].
- [math]\displaystyle{ \frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y} }[/math].
- [math]\displaystyle{ xdx+ye^{-x}dy=0 }[/math], with [math]\displaystyle{ y(0)=1 }[/math].
- [math]\displaystyle{ \frac{dy}{dx}=\frac{ay+b}{cx+d} }[/math], where [math]\displaystyle{ a,b,c,d }[/math] are arbitrary constants.
- [math]\displaystyle{ \frac{dy}{dx}=-\frac{ax+by}{bx+cy} }[/math], where [math]\displaystyle{ a,b,c }[/math] are arbitrary constants.
- [math]\displaystyle{ 0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy }[/math].
| Dror's notes above / Student's notes below |
Solution to Question 1. Twine
Take y defined by [math]\displaystyle{ y=cy_1+y_2 }[/math] and plug it into [math]\displaystyle{ y'+p(x)y=g(x) }[/math]. We get
[math]\displaystyle{ (cy_1 + y_2)' + p(x)(cy_1 + y_2) = g(x) }[/math]
[math]\displaystyle{ c(y_1' + p(x)y_1) + (y_2' + p(x)y_2) = g(x) }[/math]
Based on our assumptions about [math]\displaystyle{ y_1 }[/math] and [math]\displaystyle{ y_2 }[/math], we have [math]\displaystyle{ y_1' + p(x)y_1 = 0 }[/math] and we have [math]\displaystyle{ y_2' + p(x)y_2 = g(x) }[/math], and so the above equation holds [math]\displaystyle{ \forall c \in \mathbb{R} }[/math].
Hence, [math]\displaystyle{ \forall c \in \mathbb{R} }[/math], [math]\displaystyle{ cy_1 + y_2 }[/math] is a solution of [math]\displaystyle{ y'+p(x)y=g(x) }[/math].
Solutions to HW1: Mathstudent
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