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The equation

$$xy'(x)=A(x)y(x),$$

when $A$ is a scalar, is simple: its solution would be

$$y(x)=\exp\left(\int A(x)\frac{dx}x\right).$$

But what if $A$ is a matrix? Does this equation have a similarly closed-form solution in this case? My attempts at solving it have so far stuck at integration of $y'(x)y^{-1}(x)$ after separating the variables. The integral doesn't seem to be a logarithmic function of $y$, since $y$ and $y'$ don't necessarily commute.

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    My apologies for my now-deleted incorrect answer. My answer worked only when $A(x) = f(x)M$ for some scalar function $f(x)$ and constant $M$. I overlooked that for more general matrix functions, $\frac {de^A}{dx}$ need not be $A'e^A$. – 2017-02-19

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One cannot solve for arbitrary $n\times n$ matrix $A(x)$, as this problem is equivalent to finding the general solution of an arbitrary scalar $n$th order linear ODE. Already for $n=2$, this would mean the ability to solve Schroedinger equation in 1D for any potential.

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The solution can be written in terms of the Dyson series $$Y(x)=U(x)Y_{0}$$ $$U(x)=\sum_{n=0}^{\infty}(-1)^n\prod_{k=0}^{n}\int_{x_{0}}^{x_{k-1}}{dx_{k}}\frac{A(x_{k})}{x_{k}}$$ Simple exponentiation of the integral of matrix works only if $$\Big[A(x),\int{A(x)}\frac{dx}{x}\Big]=0$$

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    Could you elaborate what the $x_i$ integration limits mean? – 2017-02-26