12-267/Homework Assignment 1: Difference between revisions

From Drorbn
Jump to navigationJump to search
No edit summary
No edit summary
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>.
Line 13: Line 15:
# <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.

Revision as of 16:31, 17 September 2012

In Preparation

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

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 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 arbitrary constants.
  7. , where are arbitrary constants.
  8. .