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Let $A(t) \in \mathbb R^{2\times 2}$ and $b(t) \in \mathbb R^2$ continuous functions in an open interval $I$. Consider the system $$(1) \space X'=A(t)X+b(t).$$

Let $X_1,X_2$ be linearly independent solutions of the associated homogeneous system. Show that it is always possible to find a particular solution $X_p(t)$ of $(1)$ of the form $$X_p(t)=c_1(t)X_1(t)+c_2(t)X_2(t),$$ with $c_1$ and $c_2$ $C^1$ functions in $I$.

I want to find a particular solution of (1) such that $$X_p=c_1(t)X_1(t)+c_2(t)X_2(t),$$ but then ${X_p}'=A(t)X_p+b(t)$, which means $$c_1{X_1}'+c_2{X_2}'+{c_1}'X_1+{c_2}'X_2=A(t)(c_1X_1+c_2X_2)+b(t)$$

Using the fact that $X_1$ and $X_2$ are homogeneous solutions, we are left with $${c_1}'X_1+{c_2}'X_2=b(t)$$

I know that as $X_1,X_2$ are linearly independent, then any solution of the homogeneous system can be expressed as a linear combination of those two but I don't know how to use this fact.

I don't know how to continue from this point so I would appreciate some help.

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The fundamental system is given by $\Phi(t)=(X_1(t),X_2(t))\in M(2\times 2)$. So every solution can be expressed of those two functions. The general solution of the homogenous system is of the form $\Phi(t)c(t)$ with a $C^1$-curve $c\colon I\to\mathbb{R}^2$ and $c(t)=(c_1(t),c_2(t))$. The System $X'=A(t)X+b(t)$ is now of the form $$ \Phi'(t)c(t)+\Phi(t)c'(t)=A(t)\Phi(t)c(t)+b(t) $$ which leads to $\Phi(t)c'(t)=b(t)$ or $c'(t)=\Phi(t)^{-1}b(t)$ with the inverse matrix $\Phi(t)^{-1}$. The curve is $$ c(t)=\int \Phi(t)^{-1}b(t)\mathrm{d}t $$ and the particular solution is given by $$ X_p(t)=\Phi(t)\int \Phi(t)^{-1}b(t)\mathrm{d}t $$ This procedure is called the variation of parameters or variation of constants.

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