Differential equations with eigenvalues.

[WendysRules]

Homework Statement

Find all solutions of the given differential equations: $\frac{dx}{dt} = \begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix} x$

Homework Equations

The Attempt at a Solution

So, we just take the determinate of A-I$\lambda$ and set it equal to 0 to get the eigenvalues of 3, and 4. Then, we plug our eigenvalues one at a time into A-I$\lambda$ and solve it to equal the 0 vector.

For 3, I get $\begin{bmatrix} 3 & -3 \\ 2 & -2 \end{bmatrix}$ Which implies that $x_1 = x_2$ so I can just pick
$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ to be my vector? This is the part I am confused about, I'm not sure that's what I actually do! Do I just say for $\lambda$ = 3, my solution is $e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Similarly for $\lambda = 4$ i find my matrix to be
$\begin{bmatrix} 2 & -3 \\ 2 & -3 \end{bmatrix}$ which i see that $2x_1=3x_2$ this, i can just pick
$\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ Thus, my solution for $\lambda = 4$ is just $e^{4t} \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

Thanks for all the help, I'm just unsure this is the method I should be going about to solve these types of problems! (Also, if anyone knows of an algebraic calculator in which I could check my solutions for these types of problems, I would also appreciate that)

 

[vela]

Homework Statement

Find all solutions of the given differential equations: $\frac{dx}{dt} = \begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix} x$

Homework Equations

The Attempt at a Solution

So, we just take the determinate determinant of A-I$\lambda$ and set it equal to 0 to get the eigenvalues of 3 and 4. Then, we plug our eigenvalues one at a time into A-I$\lambda$ and solve it to equal the 0 vector.

For 3, I get $\begin{bmatrix} 3 & -3 \\ 2 & -2 \end{bmatrix}$ Which implies that $x_1 = x_2$ so I can just pick
$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ to be my vector? This is the part I'm confused about. I'm not sure that's what I actually do! Do I just say for $\lambda$ = 3, my solution is $e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Yup, that's right.

Similarly for $\lambda = 4$ i find my matrix to be
$\begin{bmatrix} 2 & -3 \\ 2 & -3 \end{bmatrix}$ which i see that $2x_1=3x_2$ this, i can just pick
$\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ Thus, my solution for $\lambda = 4$ is just $e^{4t} \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

You got the 2 and 3 backwards.

Thanks for all the help, I'm just unsure this is the method I should be going about to solve these types of problems! (Also, if anyone knows of an algebraic calculator in which I could check my solutions for these types of problems, I would also appreciate that)

So now that you have to two solutions, what's the general solution?

 

[Ray Vickson]

Homework Statement

Find all solutions of the given differential equations: $\frac{dx}{dt} = \begin{bmatrix} 6 & -3 \\ 2 & 1 \end{bmatrix} x$

Homework Equations

The Attempt at a Solution

So, we just take the determinate of A-I$\lambda$ and set it equal to 0 to get the eigenvalues of 3, and 4. Then, we plug our eigenvalues one at a time into A-I$\lambda$ and solve it to equal the 0 vector.

For 3, I get $\begin{bmatrix} 3 & -3 \\ 2 & -2 \end{bmatrix}$ Which implies that $x_1 = x_2$ so I can just pick
$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ to be my vector? This is the part I am confused about, I'm not sure that's what I actually do! Do I just say for $\lambda$ = 3, my solution is $e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Similarly for $\lambda = 4$ i find my matrix to be
$\begin{bmatrix} 2 & -3 \\ 2 & -3 \end{bmatrix}$ which i see that $2x_1=3x_2$ this, i can just pick
$\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ Thus, my solution for $\lambda = 4$ is just $e^{4t} \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

Thanks for all the help, I'm just unsure this is the method I should be going about to solve these types of problems! (Also, if anyone knows of an algebraic calculator in which I could check my solutions for these types of problems, I would also appreciate that)

You are correct.

However, when in doubt, check to see if your alleged "solutions" actually satisfy the DE---a step you should always do, anyway, as part of good working practice.

 

[Eclair_de_XII]

What you want to do is merge those two eigenvectors into a matrix, and left-multiply it by the solution to the diagonal system of differential equations to get the general solution for your original matrix.

 

[WendysRules]

So now that you have to two solutions, what's the general solution?

Should just be the two of the added together? So, $x(t) = C_1 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + C_2 e^{4t} \begin{bmatrix} 3 \\ 2 \end{bmatrix}$ where C is a constant.

However, when in doubt, check to see if your alleged "solutions" actually satisfy the DE---a step you should always do, anyway, as part of good working practice.

Do I check them vs the original equations, or just the new "eigenedited" matrices? I.E for $\lambda = 4$ I'd use the matrix $\begin{bmatrix} 2 & -3 \\ 2 & -3 \end{bmatrix}$ and multiple it by
$\begin{bmatrix} 3 \\ 2 \end{bmatrix}$ to get
$\begin{bmatrix} 6+-6 \\ 6+-6 \end{bmatrix}$ =
$\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ Thus it works.

 

[vela]

Should just be the two of the added together? So, $x(t) = C_1 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + C_2 e^{4t} \begin{bmatrix} 3 \\ 2 \end{bmatrix}$ where C is a constant.

Yes, that's correct.

Do I check them vs the original equations, or just the new "eigenedited" matrices?

You want to see if your solution $x(t)$ satisfies the differential equation, so plug it into the differential equation.