I am stuck in solving the following system: $$y_1'=-2y_1+y_2-2y_3$$ $$y_2'=y_1-2y_2+2y_3$$ $$y_3'=3y_1-3y_2+5y_3$$


My book gives the following solutions: $$y_1(x)=-C_1e^{3x}+(C_2-2C_3)e^{-x}$$ $$y_2(x)=C_1e^{3x}+C_2e^{-x}$$ $$y_3(x)=3C_1e^{3x}+C_3e^{-x}$$

Here is my attempt:

We differentiate the first equation: $$y_1''=-2y_1'+y_2'-2y_3'$$$$=-y_1+2y_2-4y_3$$

Now we have a $2x2$ system for $y_2,y_3$, using the first equation from the system and the last equation for $y_1''$:

$$y_1'=-2y_1+y_2-2y_3$$ $$y_1''=-y_1+2y_2-4y_3$$

Multiplying the first equation by $-2$ and adding these two equations gives: $$y_1''-2y_1'-3y_1=0$$

This gives:

$$\lambda_1=3,\lambda_2=-1\Rightarrow y_1(x)=C_1e^{3x}+C_2e^{-x}$$

This doesn't match with the above solution.

Is there a mistake?

I will show my full attempt with this (maybe wrong) first solution.

Now, for finding $y_2,y_3$ we have a new $2x2$ system such that $y_1$ is known: $$y_2'=y_1-2y_2+2y_3$$ $$y_3'=3y_1-3y_2+5y_3$$

Again, the same procedure. We differentiate the first equation: $$y_2''=y_1'-2y_2'+2y_3'$$ $$=y_1'-2y_1+4y_2-4y_3$$

Now, for finding $y_3$ we have a new $2x2$ system such that $y_1,y_2$ are known:

$$y_2'=y_1-2y_2+2y_3$$ $$y_2''=y_1'-2y_1+4y_2-4y_3$$

By adding these two, we get: $$y_3=-\frac{1}{2}y_2''-\frac{1}{2}y_2'+\frac{1}{2}y_1'-\frac{1}{2}y_1+y_2$$

Now we differentiate the equation $$y_2''=y_1'-2y_1+4y_2-4y_3$$ to obtain $$y_2'''=y_1''-2y_1'+4y_2'-4y_3'$$

By plugin previously evaluated $y_3$ we get: $$y_2'''+4y_2''+4y_2'=y_1''+2y_1'$$

We already have $y_1$ (note that this may be wrong), so: $$y_1'=3C_1e^{3x}-C_2e^{-x}$$ $$y_1''=9C_1e^{3x}+C_2e^{-x}$$

We have: $$y_2'''+4y_2''+4y_2'=15C_1e^{3x}-C_2e^{-x}$$

Now we will use variation of parameters:

$$\lambda^3+4\lambda^2+4\lambda=0$$ $$\lambda_1=0,\lambda_2=\lambda_3=-2$$

This gives multiple roots so it might again be a mistake.

I will stop here.

Could someone check this?

Where are mistakes so far?

EDIT: Could someone show the approach with linear algebra, eigenvalues/eigenvectors?

  • 1
    $\begingroup$ $C_2 = (C_2-2C_3)$ since these are just constants. Why not use eigenvalues / eigenvectors? Your approach works, but seems very algebra error prone . $\endgroup$
    – Moo
    Apr 1, 2017 at 22:42
  • $\begingroup$ @Moo, I have also assumed that, but please check the full post. $\endgroup$
    – user300045
    Apr 1, 2017 at 22:43
  • 3
    $\begingroup$ You can add all constants together. Until you are given initial conditions, they are just some constant. $\endgroup$
    – Moo
    Apr 1, 2017 at 22:56
  • 2
    $\begingroup$ See for example: (people.math.gatech.edu/~xchen/teach/ode/ExpMatrix.pdf) $\endgroup$
    – Jean Marie
    Apr 1, 2017 at 23:23
  • 1
    $\begingroup$ You can’t just look at one third of the solution and decide that it doesn’t match the book’s. As Moo said, the coefficients are arbitrary constants, so $D_1e^{3x}+D_2e^{-x}$ is just as good as $C_1e^{3x}+(C_2-C_3)e^{-x}$. You might find, if you work it all the way through, that you have $y_2=D_1e^{3x}+(D_2+2D_3)e^{-x}$ and $y_3=3D_1e^{3x}+D_3e^{-1}$, which is in fact equivalent to the book solution. $\endgroup$
    – amd
    Apr 1, 2017 at 23:39

1 Answer 1


The solution to the vector differential equation $\mathbf y'(x)=A\mathbf y(x)$ is, not surprisingly, $e^{xA}\mathbf C$, where $\mathbf C$ is a vector of constants determined by boundary conditions. The exponential of a matrix is defined via a power series, but in practice one doesn’t use that to compute it.

If $A$ is diagonalizable, it can be decomposed as $B\Lambda B^{-1}$, where $B=\begin{bmatrix}\mathbf b_1,\cdots,\mathbf b_n\end{bmatrix}$ with eigenvectors of $A$ as its columns and $\Lambda=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_j$ the eigenvalues corresponding to $\mathbf b_j$. Just as $A^k=B\Lambda^kB^{-1}=B\operatorname{diag}(\lambda_1^k,\dots,\lambda_n^k)B^{-1}$, so, too $e^{xA}=Be^{x\Lambda}B^{-1}=B\operatorname{diag}(e^{\lambda_1x},\dots,e^{\lambda_nx})B^{-1}$. So, the solution to the equation is $$e^{xA}\mathbf C=Be^{x\Lambda}B^{-1}\mathbf C.$$ Since $B$ has full rank, $B^{-1}\mathbf C$ is also a vector of arbitrary constants, so we can expand the above expression as $$C_1e^{\lambda_1x}\mathbf b_1+\cdots+C_ne^{\lambda_nx}\mathbf b^n.\tag1$$

In this problem, $$A=\begin{bmatrix}-2&1&-2\\1&-2&2\\3&-3&5\end{bmatrix}.$$ Its eigenvalues can be found to be $3$, $-1$ and $-1$. We compute the kernel of $3I-A$ via row-reduction: $$\begin{bmatrix}5&-1&2\\-1&5&-2\\-3&3&-2\end{bmatrix}\to\begin{bmatrix}1&0&\frac13\\0&1&-\frac13\\0&0&0\end{bmatrix}$$ so an eigenvector of $3$ is $(-1,1,3)^T$. Moving to $A+I$, $$\begin{bmatrix}-1&1&-2\\1&-1&2\\3&-3&6\end{bmatrix}\to\begin{bmatrix}1&-1&2\\0&0&0\\0&0&0\end{bmatrix}$$ from which we get $(1,1,0)^T$ and $(-2,0,1)^T$ as linearly independent eigenvectors. Plugging these values into (1), the solution to the equation is therefore $$C_1e^{3x}\begin{bmatrix}-1\\1\\3\end{bmatrix}+C_2e^{-x}\begin{bmatrix}1\\1\\0\end{bmatrix}+C_3e^{-x}\begin{bmatrix}-2\\0\\1\end{bmatrix}=\begin{bmatrix}-C_1e^{3x}+(C_2-2C_3)e^{-x}\\C_1e^{3x}+C_2e^{-x}\\3C_1e^{3x}+C_3e^{-x}\end{bmatrix}.$$


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