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.