We have been given an inhomogeneous system $$y'=\begin{bmatrix}5 & -3 & 2 \\ 15 & -9 & 6 \\ 10 & -6 & 4\end{bmatrix}y+\begin{pmatrix}0\\0\\x\end{pmatrix}, y(0) = (1,1,0)$$ to solve.
What I have learned is to eliminate until I have only one equation, which I can solve. I don't think I can eliminate though, the matrix $A$ has only rank 1. This also means that I can't use eigenvectors as far as I know.
How can I bring the system into an easily solvable form? Can I use the fact that $A$ is of rank 1 somehow?
$$y'=\begin{bmatrix}5 & -3 & 2 \\ 15 & -9 & 6 \\ 10 & -6 & 4\end{bmatrix}y+\begin{pmatrix}0\\0\\x\end{pmatrix}, y(0) = (1,1,0)$$ I substitued $y=(r,s,t)$: $$\pmatrix {r \\ s \\ t}'=\pmatrix{5r -3s + 2t \\ 3r' \\ 2r'+x }$$ Integrate the last two lines: $$\pmatrix {r' \\ s \\ t}=\pmatrix{5r -3s + 2t \\ 3r +c_2\\ 2r+\frac 12 x^2+c_3 }$$ $$\pmatrix {r' \\ s \\ t}=\pmatrix{x^2+2c_3-3c_2 \\ 3r +c_2\\ 2r+\frac 12 x^2+c_3 }$$ Integrating first line: $$\pmatrix {r \\ s \\ t}=\pmatrix{\frac 1 3 x^3+x(2c_3-3c_2)+c_1 \\ 3r +c_2\\ 2r+\frac 12 x^2+c_3 }$$ $$y(x)=\pmatrix{\frac 1 3 x^3+x(2c_3-3c_2)+c_1 \\ x^3+3x(2c_3-3c_2)+3c_1 +c_2\\ \frac 2 3 x^3+2x(2c_3-3c_2)+2c_1+\frac 12 x^2+c_3 }$$ You have to apply the initial conditions and find the three constants. Note that the method of eigenvectors and eigenvalues works fine too. You have $\lambda=0$ as eigenvalue with multiplicity $3$. $$(c_1,c_2,c_3)=(1,-2,0)$$ $$ \implies y(x)=\pmatrix{\frac 1 3 x^3+6x+1 \\ x^3+18x+1\\ \frac 2 3 x^3+\frac 12 x^2 +12x+2}$$