What is the differential equation that relates input and output voltage of a circuit?

Question

I have the transfer function of this circuit, which is

I am being asked to find the differential equation that represents the relationship between input and output voltage. Since the transfer function H(s)=Vo(s)/Vi(s), I thought that I would be able to get this equation by taking the inverse laplace transform of H(s). But I can't figure out the answer, and when I put it into MATLAB I get this ridiculously long answer full output: 5*symsum((exp(root(z^3 + 30*z^2 + 200*z + 75, z, k)*t)*root(z^3 + 30*z^2 + 200*z + 75, z, k))/(60*root(z^3 + 30*z^2 + 200*z + 75, z, k) + 3*root(z^3 + 30*z^2 + 200*z + 75, z, k)^2 + 200), k, 1, 3) + 75*symsum(exp(t*root(z^3 + 30*z^2 + 200*z + 75, z, k))/(3*root(z^3 + 30*z^2 + 200*z + 75, z, k)^2 + 60*root(z^3 + 30*z^2 + 200*z + 75, z, k) + 200), k, 1, 3)

Am I going about this incorrectly? Is the differential equation I'm looking for not the inverse laplace transform? Any help is appreciated, I am quite confused.

Answer 1

If you take the inverse Laplace Transform of the transfer function, you get the impulse response in time (\$h(t)\$). This is not yout intention here. In order to get the corresponding linear differential equation, you can work as below:

$$ H(s)= \frac{75(s+1)}{s(s+5)(s+25)+75(s+1)} $$

Considering \$V_i(s)\$ and \$V_o(s)\$ as the input and output of the system, respectively:

$$ H(s)= \frac{V_o(s)}{V_i(s)} = \frac{75s+75}{s^3+30s^2+200s+75} $$

$$ (s^3+30s^2+200s+75)V_o(s) = (75s+75)V_i(s) $$

We know that transfer function assumes zero initial conditions. So, when converting from the \$s\$ domain to time domain, is enough to ensure that each \$s^n\$ factor is translated to an operation \$\frac{d^n.(t)}{dt^n}\$ (or n-derivative):

$$ \frac{d^3v_o(t)}{dt^3}+30\frac{d^2v_o(t)}{dt^2}+200\frac{dv_o(t)}{dt}+75v_o(t)=75\frac{dv_i(t)}{dt}+75v_i(t) $$