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I have the following equation that I wish to represent in the form of a control system block diagram:

$a \frac{d x(t)}{dt} + b(t)x(t) = u(t)$

The variable $b(t)$ and $u(t)$ are both inputs that will change with time.

If I take the Laplace transform I get: $B(s) + a s X(s) = U(s)$

I am interested in the transfer function that describes $\frac{X(s)}{U(s)}$ but I cannot separate variables to get this...

How can I describe the model above in either a transfer function or state space model? I'd like to simulate it as it is part of a control system I'm interested in.

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I assume that you define $B(s)$ as $\mathcal{L}\{b(t)\}(s)$. But that term does not occur by itself. It only occurs as a product with $x(t)$. Even if you would try to calculate the Laplace transform of the product of $x(t)$ and $b(t)$, then you still get $\mathcal{L}\{b(t)\,x(t)\}(s)\neq X(s)\,B(s)$.

This is because multiplication in the time domain results in convolution in the frequency domain (and vice versa).

Your system is not time invariant, because $b(t)$ is a function of time. You can only really define transfer function for linear time invariant (LTI) systems.

If $b(t)$ is periodic ($b(t)=b(t+T)$) then you could discretize your system with a time step of $T$. And for that model you could formulate a discrete transfer function in the z-domain. But in the current form you can't say much in general about its behavior in the frequency domain.

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    To paraphrase you: the key issue is that adding the term $b(t) x(t)$ breaks the LTI assumptions. I was wondering if I could take the log of the inputs to the system to allow me to sum them together, but I suppose this would break the 'linear' part of LTI assumptions. – 2017-01-27
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    @Hefaestion I am not sure if taking taking the log of something would help remove the time variant part. But you are correct that it then would be nonlinear and thus still not LTI. – 2017-01-27