Solve a given nth order linear differential equation

Find bases for the vector spaces of solutions to the following differential equations.

  1. y^{(3)} - 3y^{(1)} + 2y = 0
  2. y^{(4)} + 4y^{(3)} + 6y^{(2)} + 4y^{(1)} + y = 0

Let y_1 = y^{(0)}, y_2 = y^{(1)}, and y_3 = y^{(2)}, and consider the following linear system of first order differential equations.

\begin{array}{ccccccc} \frac{d}{dt} y_1 & = & & & y_2 & & \\ \frac{d}{dt} y_2 & = & & & & & y_3 \\ \frac{d}{dt} y_3 & = & -2y_1 & + & 3y_2 & & \end{array}

Which we can express as a matrix equation by

\dfrac{d}{dt} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & 3 & 0 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}

Now let A denote this coefficient matrix. Let P = \frac{1}{9} \begin{bmatrix} 1 & -2 & 1 \\ -4 & 8 & 5 \\ -6 & 3 & 3 \end{bmatrix}; evidently, P^{-1}AP = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} is in Jordan canonical form. (Computations performed by WolframAlpha.) Now P\left(\mathsf{exp}([-2]t) \oplus \mathsf{exp}(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}t\right) = P\begin{bmatrix} e^{-2t} & 0 & 0 \\ 0 & e^t & te^t \\ 0 & 0 & e^t \end{bmatrix} is a fundamental matrix of our linear system. Reading off the first row of this matrix (and multiplying by 9), we see that e^{-2t}, -2e^t, and -2te^t + e^t are solutions of our original 3rd order differential equation. Moreover, it is clear that these are linearly independent.

For the second equation, we follow a similar strategy and see that the corresponding coefficient matrix is A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & -4 & -6 & -4 \end{bmatrix}. Let Q = \begin{bmatrix} -1 & -3 & -6 & -10 \\ 1 & 2 & 3 & 4 \\ -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 \end{bmatrix}. Evidently, Q^{-1}AQ = \begin{bmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \end{bmatrix} = B is in Jordan canonical form. (Computation performed by WolframAlpha.) Taking the first row of Q\mathsf{exp}(Bt) (and multiplying by -1), we see that e^{-t}, e^{-t}(3t-2), e^{-t}(\frac{1}{2}t^2 + 3t + 6), and e^{-t}(\frac{1}{6}t^3 + \frac{3}{2}t^2 + 6t + 10) are linearly independent solutions of the original 4th order differential equation.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: