# General solution for system of differential equations

## Homework Statement

Find the general solution to the following differential equations
y'1 = -12y1 + 13y2 +10y3
y'2 = 4y1 - 3y2 - 4y3
y'3 = -21y1 +21y2 +19y3

## The Attempt at a Solution

I'm a little unsure about what to do at the end, or what form to put it in.
The eigenvalues are
λ1 = 5, λ2 = -2, λ3 = 1
and the eigenvectors are
v1 = [1, -1, 3], v2 = [1, 0, 1], v3 = [1, 1, 0] (respectively)
so once I have that, what do I do? I need to put them into an exponential form like
a*eλ2*t*v1 + b*eλ2*t*v2 + c*eλ3*t*v3 ???
I don't understand why, or what this means.

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Mark44
Mentor

## Homework Statement

Find the general solution to the following differential equations
y'1 = -12y1 + 13y2 +10y3
y'2 = 4y1 - 3y2 - 4y3
y'3 = -21y1 +21y2 +19y3

## The Attempt at a Solution

I'm a little unsure about what to do at the end, or what form to put it in.
The eigenvalues are
λ1 = 5, λ2 = -2, λ3 = 1
and the eigenvectors are
v1 = [1, -1, 3], v2 = [1, 0, 1], v3 = [1, 1, 0] (respectively)
so once I have that, what do I do? I need to put them into an exponential form like
a*eλ2*t*v1 + b*eλ2*t*v2 + c*eλ3*t*v3 ???
I don't understand why, or what this means.
Your system of diff. equations is coupled, meaning that the three derivative components all involve the three y components. It's much easier to solve an uncoupled system such as
z1' = a1z1
z2' = a2z2
z3' = a3z3

and this is the primary motivation for diagonalizing a system of diff. equations.

Your system can be represented as a matrix equation as:
Y' = AY

If we can find a diagonal matrix D that is similar to A, then we can find a related system of equations that is uncoupled, so easier to solve.

Since you have already found that there are three distinct eigenvalues, that means your matrix A is diagonalizable. You have also found three eigenvectors for the three eigenvalues.

Form a matrix P whose columns are the eigenvectors. The order doesn't matter, but the order will determine which values appear on the diagonal of matrix D (the diagonal matrix).

Let Z = P-1Y, or equivalently, Y = PZ.

Then Z' = P-1Y' = P-1AY = P-1APZ = DZ, where D is the diagonal matrix. As already noted, this system is easy to solve, and its solution is given by:

$$Z = Z(t) = \begin{bmatrix}e^{\lambda_1 t} & 0 & 0 \\ 0 & e^{\lambda_2 t} & 0 \\ 0&0&e^{\lambda_3 t} \end{bmatrix} \begin{bmatrix}c_1 \\ c_2 \\ c_3 \end{bmatrix}$$

We're not quite done. We're interested in Y = Y(t), not Z(t).

Since Y = PZ, then our solution is
$$Y = Y(t) = PZ = P\begin{bmatrix}e^{\lambda_1 t} & 0 & 0 \\ 0 & e^{\lambda_2 t} & 0 \\ 0 & 0 & e^{\lambda_3 t}\end{bmatrix}\begin{bmatrix}c_1 \\ c_2 \\ c_3 \end{bmatrix}$$

OK thank you, that makes sense! But how come the diagonal matrix's entries are in the form e? And the matrix Z is given here by [c1, c2, c3]?

Mark44
Mentor
What's the general solution of z1' = λ1z1? Once you answer that, you might see that the matrix equation summarizes the solution for the vector Z.

Of course! Thank you, you are beautiful.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find the general solution to the following differential equations
y'1 = -12y1 + 13y2 +10y3
y'2 = 4y1 - 3y2 - 4y3
y'3 = -21y1 +21y2 +19y3

## The Attempt at a Solution

I'm a little unsure about what to do at the end, or what form to put it in.
The eigenvalues are
λ1 = 5, λ2 = -2, λ3 = 1
and the eigenvectors are
v1 = [1, -1, 3], v2 = [1, 0, 1], v3 = [1, 1, 0] (respectively)
so once I have that, what do I do? I need to put them into an exponential form like
a*eλ2*t*v1 + b*eλ2*t*v2 + c*eλ3*t*v3 ???
I don't understand why, or what this means.
If you think of a solution of the form y1 = a*exp(r*x), y2 = b*exp(r*x), y3 = c*exp(r*x) [same r in all three] you will get a cubic equation in r --- essentially, the characteristic equation in the matrix representation. This cubic has 3 roots (r1 = -2, r2 = 1, r3 = 5 in this case), and so your solution will have the form y1 = a1*exp(r1*x) + a2*exp(r2*x) + a3*exp(r3*x), etc. You get equations for the ai, bi and ci by substituting these expressions into the DE; the solutions are, essentially, constants times eigenvectors of the matrix. Finally, initial conditions serve to identify the constants.

RGV