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Differential Equations: linearity principle

  1. Mar 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the linear system:
    dx/dt=x-y
    dy/dt=x+3y

    a. show that the function (x(t), y(t))=(te2t, -(t+1)e2t) is a solution to the differential equation (easy)
    b. Solve the initial value problem
    dx/dt=x-y
    dy/dt=x+3y

    y(0)=(0,2)

    need help with part b not a
    2. Relevant equations
    linarity principle
    y(t)=k1y1(t) +k2y2(t)


    3. The attempt at a solution
    Really don't even know where to start. I understand how to use the linearity principle with two given solutions but not one. I feel like I am missing a very fundamental idea about initial value problems. So any idea on where I should begin with just one given solution?
     
  2. jcsd
  3. Mar 15, 2016 #2

    Mark44

    Staff: Mentor

    This is confusing. y(0) is a single number. Did you write 0,2 to mean the same thing as 0.2? If not, what's the initial condition for x(0)?
    Which is what?
     
  4. Mar 15, 2016 #3
    please see the following to get an idea of the question-
    http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx
     
  5. Mar 16, 2016 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It looks like you are confusing two different notations. In your equations, dx/dt= x- y and dy/dt= x+ ,you have x and y as scalar functions but here you have y as a vector function with the previous x and y as components. In your original notation, you want x(0)= 0, y(0)= 2.
    Are you required to use the given solution? There are several different ways of solving such a system of equations. For one, you could write it as a matrix equation:
    [tex]\frac{d\begin{pmatrix}x \\ y \end{pmatrix}}{dx}= \begin{pmatrix}1 & -1 \\ 1 & 3 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}[/tex]
    and it is easy to "diagonalize" that matrix.

    Or, differentiating the first equation again, [itex]\frac{d^2x}{dt^2}= \frac{dx}{dt}- \frac{dy}{dt}[/itex] and replace that [itex]\frac{dy}{dx}[/itex] with the second equation- [itex]\frac{d^2x}{dt^2}= \frac{dx}{dt}- (x+ 3y)[/itex] and replace that "y" with [itex]y= x- \frac{dx}{dt}[/itex] from the first equation again.
    [itex]\frac{d^2x}{dt^2}= \frac{dx}{dt}- x- 3(x- \frac{dx}{dt})= 4\frac{dx}{dt}- 4x[/itex]. Can you solve [itex]\frac{d^2x}{dt^2}- 4\frac{dx}{dt}+ 4x= 0[/itex] with initial conditions x(0)= (0), x'(0)= x(0)- y(0)= 0- 2= -2?
     
  6. Mar 16, 2016 #5

    Mark44

    Staff: Mentor

    It's not so easy. There's a repeated eigenvalue with only one eigenvector.
     
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