12-267/Homework Assignment 2

From Drorbn
Jump to: navigation, search

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. x^2y^3+x(1+y^2)y'=0 (hint: try \mu=x^\alpha y^\beta).
  2. dx+(\frac{x}{y}-\sin y)dy=0.
  3. (x^2+3xy+y^2)dx-x^2dy=0.
  4. \frac{dy}{dx}=\frac{2y-x+5}{2x-y-4} (hint: consider trying x_1=x+\alpha and y_1=y+\beta for good \alpha, \beta).
  5. y'=\frac{y^3}{1-2xy^2} with y(0)=1.
  6. \frac{dy}{dx}=\frac{2y+\sqrt{x^2-y^2}}{2x}.

Task 2. Let M and N be differentiable functions of x and y.

  1. Show that if \frac{N_x-M_y}{xM-yN} depends only on xy, then the differential equation Mdx+Ndy=0 has an integrating factor of the form \mu(xy), where \mu is a function of a single variable.
  2. Find a condition on M and N that would imply that the differential equation Mdx+Ndy=0 would have an integrating factor of the form \mu(x+y), where \mu is a function of a single variable.

Task 3. The equation y'+p(x)y=q(x)y^n is called a "Bernoulli Equation".

  1. Explain why you already know how to solve the Bernoulli equation when n=0 and when n=1.
  2. Show that if n\neq0,1, then the substitution v=y^{1-n} reduces the Bernoulli equation to an equation you already know how to solve.
  3. Solve the equation x^2y'+2xy-y^3=0 (for x>0).

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 \mu such that (\mu M)_y = (\mu N)_x. Let xy = z

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

Similarly \frac{dv}{dy} = \frac{dv}{dz} x.

Then we have

(\mu M)_y = (\mu N)_x

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

\iff \frac{\mu_{z}}{mu} = \frac{N_x - M_y}{xM - yN}

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

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

which satisfies the requirements of an integrating factor.

Part 2

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

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

Similarly \frac{d \mu}{dy} = \frac{d\mu}{dz}

Then we have

(\mu M)_y = (\mu N)_x

\iff \frac{\mu_{z}}{\mu} = \frac{N_x - M_y}{M-N}

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

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

which satisfies the requirements of an integrating factor.

Solution to HW2: Mathstudent