12-267/Homework Assignment 2: Difference between revisions

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'''Solution to Task 2''' --[[User:Twine|Twine]] 18:44, 24 October 2012 (EDT)

'''Part 1'''

What we want is an integration factor <math>\mu</math> such that <math>(\mu M)_y = (\mu N)_x</math>. Let <math>xy = z</math>

When <math>\mu</math> is a function of z, <math>\frac{dv}{dx} = \frac{dv}{dz} \frac{dz}{dx} = \frac{dv}{dz} y</math>.

Similarly <math>\frac{dv}{dy} = \frac{dv}{dz} x</math>.

Then we have

<math>(\mu M)_y = (\mu N)_x</math>

<math>\iff \mu_{z}xM - \mu_{xy}yN = \mu (N_x - M_y)</math>

<math>\iff \frac{\mu_{z}}{mu} = \frac{N_x - M_y}{xM - yN}</math>

If the right hand side depends only on xy, we can get

<math>\mu (xy) = e^{\int \frac{N_x - M_y}{xM - yN} d(xy)}</math>

which satisfies the requirements of an integrating factor.

'''Part 2'''

As in Part 1, we need <math>(\mu M)_y = (\mu N)_x</math>. Let <math>z = x + y</math>

When <math>\mu</math> is a function of <math>x + y</math>, <math>\frac{d \mu}{dx} = \frac{d\mu}{dz} \frac{dz}{dx} = \frac{d\mu}{dz}</math>.

Similarly <math>\frac{d \mu}{dy} = \frac{d\mu}{dz}</math>

Then we have

<math>(\mu M)_y = (\mu N)_x</math>

<math>\iff \frac{\mu_{z}}{\mu} = \frac{N_x - M_y}{M-N}</math>

If the right hand side of this equation depends only on z (that is, only on (x+y)), then we have

<math>\mu (x + y) = e^{\int \frac{N_x - M_y}{M - N} d(x + y)}</math>

which satisfies the requirements of an integrating factor.


Solution to HW2: [[User:Mathstudent|Mathstudent]]
Solution to HW2: [[User:Mathstudent|Mathstudent]]

Latest revision as of 18:44, 24 October 2012

This assignment is due at the tutorial on Tuesday October 2. 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.

Task 0. Identify yourself in the Class Photo!

Task 1. Solve the following differential equations:

  1. (hint: try ).
  2. .
  3. .
  4. (hint: consider trying and for good , ).
  5. with .
  6. .

Task 2. Let and be differentiable functions of and .

  1. Show that if depends only on , then the differential equation has an integrating factor of the form , where is a function of a single variable.
  2. Find a condition on and that would imply that the differential equation would have an integrating factor of the form , where is a function of a single variable.

Task 3. The equation is called a "Bernoulli Equation".

  1. Explain why you already know how to solve the Bernoulli equation when and when .
  2. Show that if , then the substitution reduces the Bernoulli equation to an equation you already know how to solve.
  3. Solve the equation (for ).

Task 4. Find an example of a non-differentiable function which is nevertheless Lipschitz.

Dror's notes above / Student's notes below

Solution to Task 2 --Twine 18:44, 24 October 2012 (EDT)

Part 1

What we want is an integration factor such that . Let

When is a function of z, .

Similarly .

Then we have

If the right hand side depends only on xy, we can get

which satisfies the requirements of an integrating factor.

Part 2

As in Part 1, we need . Let

When is a function of , .

Similarly

Then we have

If the right hand side of this equation depends only on z (that is, only on (x+y)), then we have

which satisfies the requirements of an integrating factor.

Solution to HW2: Mathstudent