Inhomogeneous system of differential equations where $\det A(t)=0$ 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?
 A: $$y'=\begin{bmatrix}5 & -3 & 2 \\ 15 & -9 & 6 \\ 10 & -6 & 4\end{bmatrix}y+\begin{pmatrix}0\\0\\x\end{pmatrix}, y(0) = (1,1,0)$$
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 $3$.
$$(c_1,c_2,c_3)=(1,-2,0)$$
$$ \implies y(x)=\pmatrix{\frac 1 3 x^3+6x+1 \\  x^3+18x+1\\ \frac 2 3 x^3+\frac 12 x^2 +12x+2}$$
A: Hint.
As we can verify
$$
y'_2=y'_1+y'_3-x
$$
or
$$
y_2=y_1+y_3-\frac 12 x^2+C
$$
so substituting $y_2$ into the DEs
$$
y'_1=g_1(y_1,y_2,y_3)\\
y'_3=g_3(y_1,y_2,y_3)+x
$$
we have the reduced system
$$
\cases{y'_1=g_1(y_1,y_1+y_3-\frac 12 x^2+C,y_3)\\
y'_3=g_3(y_1,y_1+y_3-\frac 12 x^2+C,y_3)+x
}
$$
