$\vec Y'=\begin{pmatrix}-4 & 2 \\ 2 & -1\end{pmatrix} \vec Y+ \begin{pmatrix}x^{-1} \\2x^{-1}+4\end{pmatrix}$

$\vec Y'=\begin{pmatrix}-4 & 2 \\ 2 & -1\end{pmatrix} \vec Y+ \begin{pmatrix}x^{-1} \\2x^{-1}+4\end{pmatrix}$

The eigenvalues are $0,-5$, the eigenvectors are $\begin{pmatrix}1 \\2\end{pmatrix} ,\begin{pmatrix}-2 \\1\end{pmatrix} $ then the solutions of the homogeneous ode are $\begin{pmatrix}1 \\2\end{pmatrix} ,e^{-5x}\begin{pmatrix}-2 \\1\end{pmatrix} $

How am I supposed to find the solution to the non-homogeneous ode ?

Thanks !

The answer is $$\boxed{{\bf Y}(x)=c_{1}\begin{bmatrix}-2\\ 1 \end{bmatrix}e^{-5x}+c_{2}\begin{bmatrix} 1\\ 2 \end{bmatrix}e^{0\cdot x}+\frac{1}{5}\begin{bmatrix} -\frac{8}{5}+8x+\log x\\ \frac{4}{5}+16x+2\log x \end{bmatrix}}$$

The Nonhomogeneous system is given by \begin{align*} \underbrace{\frac{{\rm d}}{{\rm d}x}\begin{bmatrix} y_{1}(x)\\ y_{2}(x)\end{bmatrix}}_{{\bf Y}'(x)}=\underbrace{\begin{bmatrix}-4 & 2 \\ 2 & -1 \end{bmatrix}}_{\bf A}\underbrace{\begin{bmatrix} y_{1}(x) \\ y_{2}(x)\end{bmatrix}}_{{\bf Y}(x)}+\underbrace{\begin{bmatrix} \frac{1}{x}\\ \frac{2}{x}+4\end{bmatrix}}_{{\bf F}(x)} \end{align*}

• Complementary solution: $${\bf Y}'(x)={\bf A}{\bf Y}(x)$$ The eigenvalues are give by $$\lambda_{1}=-5,\quad \lambda_{2}=0$$ and the eigenvectors are given by $${\bf v}_{1}=\begin{bmatrix} -2 \\ 1\end{bmatrix},\quad {\bf v}_{2}=\begin{bmatrix} 1 \\ 2\end{bmatrix}$$ Since $\lambda_{1}\not=\lambda_{2}$ so the complementary solution is of the form $${\bf Y}_{c}(x)=c_{1}{\bf v}_{1}e^{\lambda_{1}x}+c_{2}{\bf v}_{2}e^{\lambda_{2}x}$$ Hence $${\bf Y}_{c}(x)=c_{1}\begin{bmatrix}-2\\ 1 \end{bmatrix}e^{-5x}+c_{2}\begin{bmatrix} 1\\ 2 \end{bmatrix}e^{0\cdot x}$$

• Particular solution: $${\bf Y}'(x)={\bf A}{\bf Y}(x)+{\bf F}(x)$$ Using variation of parameters:
  The fundamental matrix is of the form $$\Phi(x)=\begin{bmatrix} -2e^{-5x} & 1 \\ e^{-5x} & 2 \end{bmatrix},\quad \Phi^{-1}(x)=\begin{bmatrix} -\frac{2}{5}e^{5x} & \frac{1}{5} e^{5x}\\ \frac{1}{5} & \frac{2}{5} \end{bmatrix}$$The particular solution is of the form $${\bf Y}_{p}(x)=\Phi(x)\int \Phi^{-1}(x){\bf F}(x){\rm d}x$$ Hence, we have \begin{align*} {\bf Y}_{p}(x)&=\Phi(x)\int \Phi^{-1}(x){\bf F}(x){\rm d}x\\ &=\begin{bmatrix} -2e^{-5x} & 1 \\ e^{-5x} & 2 \end{bmatrix} \int \begin{bmatrix} -\frac{2}{5}e^{5x} & \frac{1}{5} e^{5x}\\ \frac{1}{5} & \frac{2}{5} \end{bmatrix} \begin{bmatrix} \frac{1}{x}\\ \frac{2}{x}+4\end{bmatrix}{\rm d}x\\ &=\frac{1}{5}\begin{bmatrix} -2e^{-5x} & 1 \\ e^{-5x} & 2 \end{bmatrix}\int \begin{bmatrix} -2e^{5x} & e^{5x} \\ 1 & 2 \end{bmatrix} \begin{bmatrix} \frac{1}{x} \\ \frac{2}{x}+4\end{bmatrix}{\rm d}x\\ &=\frac{1}{5}\begin{bmatrix} -2e^{-5x} & 1 \\ e^{-5x} & 2 \end{bmatrix}\int \begin{bmatrix}-\frac{2}{x}e^{5x}+\left(\frac{2}{x}+4\right)e^{5x}\\ \frac{1}{x}+2\left( \frac{2}{x}+4\right) \end{bmatrix}{\rm d}x\\ &=\frac{1}{5}\begin{bmatrix} -2e^{-5x} & 1 \\ e^{-5x} & 2 \end{bmatrix} \begin{bmatrix} \frac{4}{5}e^{5x}\\ 8x+5\log x\end{bmatrix}\\ &=\frac{1}{5}\begin{bmatrix} -\frac{8}{5}+8x+\log x\\ \frac{4}{5}+16x+2\log x \end{bmatrix} \end{align*}

• General solution: $${\bf Y}(x)={\bf Y}_{c}(x)+{\bf Y}_{p}(x)$$ Therefore the answer is $$\boxed{{\bf Y}(x)=c_{1}\begin{bmatrix}-2\\ 1 \end{bmatrix}e^{-5x}+c_{2}\begin{bmatrix} 1\\ 2 \end{bmatrix}e^{0\cdot x}+\frac{1}{5}\begin{bmatrix} -\frac{8}{5}+8x+\log x\\ \frac{4}{5}+16x+2\log x \end{bmatrix}}$$