12-267/Homework Assignment 1: Difference between revisions

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'''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 2.''' Solve the following differential equations
# For <math>x>0</math>, <math>xy'+2y=\sin x</math>.
# <math>\frac{dy}{dx}=\frac{1}{e^y-x}</math> with <math>y(1)=0</math>; you may want to solve for <math>x</math> first.
# <math>xy'=\sqrt{1-y^2}</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>\frac{dy}{dx}=\frac{ax+b}{cx+d}</math>, where <math>a,b,c,d</math> are constants.
# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are constants.
# <math>0=(e^x\sin y + 3y)dx - (3x-e^x\sin y)dy</math>.

Revision as of 14:30, 17 September 2012

In Preparation

The information below is preliminary and cannot be trusted! (v)

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

  1. For [math]\displaystyle{ x\gt 0 }[/math], [math]\displaystyle{ xy'+2y=\sin x }[/math].
  2. [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.
  3. [math]\displaystyle{ xy'=\sqrt{1-y^2} }[/math].
  4. [math]\displaystyle{ \frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y} }[/math].
  5. [math]\displaystyle{ xdx+ye^{-x}dy=0 }[/math], with [math]\displaystyle{ y(0)=1 }[/math].
  6. [math]\displaystyle{ \frac{dy}{dx}=\frac{ax+b}{cx+d} }[/math], where [math]\displaystyle{ a,b,c,d }[/math] are constants.
  7. [math]\displaystyle{ \frac{dy}{dx}=-\frac{ax+by}{bx+cy} }[/math], where [math]\displaystyle{ a,b,c }[/math] are constants.
  8. [math]\displaystyle{ 0=(e^x\sin y + 3y)dx - (3x-e^x\sin y)dy }[/math].