12-267/Homework Assignment 1: Difference between revisions

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{{In Preparation}}
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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 15:30, 17 September 2012

In Preparation

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

Question 1. Show that if is a solution of , and is a solution of , then for any constant , is a solution of .

Question 2. Solve the following differential equations

  1. For , .
  2. with ; you may want to solve for first.
  3. .
  4. .
  5. , with .
  6. , where are constants.
  7. , where are constants.
  8. .