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$$x'=-y+z-e{-t}$$ $$y'=3x+y-3z$$ $$z'=2x-y-z-2e^{-t}$$

First we have to find the homegene part of the system. We find the fundamental system of solutions: $$(e^{t},e^{-2t},te^{t})$$ from the eigenvector we have that $$u=(0,1,1)$$ which are the values of the constants (c2,c5,c8) of the system $$x=C1e^{t}+C2te^{-2t}+C3te^{t}$$ $$x=C4e^{t}+C5te^{-2t}+C6te^{t}$$ $$x=C7e^{t}+C8te^{-2t}+C9te^{t}$$ so I replace these equations in the original system, solve it and I find that: $$C1=C7;C2=0;C3=-C4=C9;C5=C8;C6=0$$ And so the solution of the homegene part of this equation is : $$x0=c1e^t+C3te^{-t}$$ $$y0=-C3e^t+C5e^{-2t}$$ $$z=c1e^t+C5e^{-2t}+C3te^{-t}$$

But that is not right what am I doing wrong?

After that I have to find the non-homegene part of this system we have that m=0;alpha=-1;beta=0;s=0 That means that the solution is: $$x=ae^{-t}$$ $$y=be^{-t}$$ $$z=ce^{-t}$$ and then we replace in the original system and we have: $$-a+b-c+1=0$$ $$-b-3a-b+3=0$$ $$-2a+b+2=0$$ $$=>a=1;b=c=0$$ and we find that the non homegene part is equal to: $$x1=1e^{-t}$$ $$x2=0e^{-t}$$ $$x3=0e^{-t}$$

But that is also not correct according to wolfram, what am I doing wrong here ?

1 Answers 1

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For the homogeneous part, the matrix has two eigenvalues, one is $-2$, with the eigenvector you found. The other is $1$, which you did not mention at all. You should find the eigenvector $k_1$ corresponding to the eigenvalue $\lambda=1$, and another generalized eigenvector $k_2$, since $\lambda=1$ has multiplicity $2$. The solution of the homogeneous part should be $$\vec{x}=c_1\vec{u}e^{-2t}+c_2(\vec{k}_1te^t+\vec{k}_2e^t).$$

Your set up for the nonhomogeneous part is correct. The matrix is also correct. The solution is not. You can just double check that.

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    when I do the eigenvalue thing for lamba = 1, we have that the w1=w3;w2=0 so w=(1,0,1)=k1, k2 should be = (2,0,2) and solution is still wrong I double checked the nonhomogene system, everything is all right – 2017-01-08
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    @Harton: $k_2$ cannot be a multiple of $k_1$. You have to solve $(A-\lambda I)k_2=k_1$. For the nonhomogeneous system, I just double checked my system, the second equation should be $-3a-2b+3c=0$. – 2017-01-08
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    -3a-2b+3x=0=-b-3a-b+3x; k2=k1/(a-lambda*I) which can not be done – 2017-01-08
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    @Harton: The first one, you missed $c$ in your equation $-b-3a-b+3=0$. If there is a $c$, the solution $(1,0,0)$ is not right. For the second one, I wonder how you got the eigenvector $u$. To solve $(A-\lambda I)k_2=k_1$ is similar as solving for $u$. You need to put $(A-\lambda I)$ and $k_1$ in a matrix, then do Gaussian elimination. – 2017-01-08
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    \begin{bmatrix} {-1} & {-1} & {1} & {1} \\ {3} & {0} & {-3} & {0} \\ {2} & {-1} & {-2} & {1} \end{bmatrix} – 2017-01-08
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    So we have to make the first 3x3 to be the I but when you start doing that gaussian thing, you seee that it's impossible – 2017-01-08
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    @Harton: No you don't make it $I$. $A-\lambda I$ is never invertible, so doing Gaussian elimination will give you infinitely many solutions. If you solve it, you will get $x_2=-1$, $x_1=x_3$. Picking one as $k_2$: $(1,-1,1)$ would work. – 2017-01-08
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    You are right about the system but then again the coneficients are not accurate. i get that a=3/7;b=-7/8;c=7 – 2017-01-08
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    @Harton: After you solve a system, you can always plug them back to double check. – 2017-01-08