Homogeneous First Order Linear System

Type 1: Distinct Roots


Great news! If you remember how to find eigenvalues and eigenvectors for a matrix, then you can solve these systems.


Matrix Representation for this Linear System

y1=ay1+by2y2=cy1+dy2 ⟷  y⃗ ′=[abcd]y⃗\begin{array}{c} y_1 = ay_1 + by_2 \\ y_2 = cy_1 + dy_2 \end{array} \ \longleftrightarrow \ \ \vec{y}\ '= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \vec{y}
A=[abcd] A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}

How do you solve these?

Match the eigenvalues with the eigenvectors. If v⃗1\vec{v}_1 and v⃗2\vec{v}_2 are linearly independent eigenvectors with eigenvalues λ1\lambda_1 and λ2\lambda_2 respectively, then then we have this general solution.

y=C1e(λ1t)v⃗1+C2e(λ2t)v⃗2 y = C_1 e^{(\lambda_1 t)}\vec{v}_1 + C_2 e^{(\lambda_2 t)}\vec{v}_2

Why does this work?

Example: The the first order system y⃗ ′=Ay⃗\vec{y}\ '= A \vec{y} with matrix A=[2112]A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. To understand why these solutions work, take a look at the direction field for this system below. At points in the plane, arrows display normalized velocity vectors, showing the direction of movement at each point. From the equation, each velocity y⃗ ′\vec{y} \ ' is just the position y⃗\vec{y} put into the matrix AA. Take a look.

Direction Field Vectors: $y'=Ay$
Direction Field Vectors: y′=Ayy'=Ay

For the last window, fullscreen the interactive figure and move the point P around. Notice the velocity (direction) vector coming out of P.

Check the box to see the unique solution that passes through the point.



Practice

1)

Find the general solution to the homogeneous system:

y⃗ ′=[2112]y⃗.\vec{y} \ ' = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \vec{y}.
y⃗(t)=\vec{y}(t) =
C1e(C_1 e^(
t)^{t)}
+
C2e(C_2 e^(
t)^{t)}

2)

Find the general solution to the homogeneous system:

y⃗ ′=[3333]y⃗.\vec{y} \ ' = \begin{bmatrix} 3 & 3 \\ 3 & 3 \end{bmatrix} \vec{y}.
y⃗(t)=\vec{y}(t) =
C1e(C_1 e^(
t)^{t)}
+
C2e(C_2 e^(
t)^{t)}

3)

Find the general solution to the homogeneous system:

y⃗ ′=[1221]y⃗.\vec{y} \ ' = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \vec{y}.
y⃗(t)=\vec{y}(t) =
C1e(C_1 e^(
t)^{t)}
+
C2e(C_2 e^(
t)^{t)}

4)

Find the general solution to the homogeneous system:

y⃗ ′=[11−186−10]y⃗.\vec{y} \ ' = \begin{bmatrix} 11 & -18 \\ 6 & -10 \end{bmatrix} \vec{y}.
y⃗(t)=\vec{y}(t) =
C1e(C_1 e^(
t)^{t)}
+
C2e(C_2 e^(
t)^{t)}