Since this is alinear equation- and especially since it is a linear equation with constant coefficients, there is a simpler approach to it. First, find the general solution to the "associated homogeneous equation". That is the equation we get by dropping "g(t)", the only part that has no "f(t)", the unknown function in it.
That gives \(\displaystyle f'= -A f\) where "A" is your \(\displaystyle \lambda_N+\lambda_{1C}+ \lambda_{1I}\).
We can write that as \(\displaystyle \frac{df}{f}= -A dt\) and integrate both sides to get \(\displaystyle ln(f)= -\int A dt\). Since A is a constant here, that is \(\displaystyle ln(f)= -At+ C\) for C an arbitrary constant. Then, taking the exponential of both sides, \(\displaystyle f(t)= e^{-At+ C}= C' e^{-At}\) where \(\displaystyle C'= e^C\).
Now we look for a single solution to the entire equation using a method called "variation" of parameters. We look for a solution to the entire equation of the for \(\displaystyle f(t)= u(t)e^{-At}\) for some function u(t) replacing the constant C'. (That is the "parameter" we are "varying".)
With \(\displaystyle f(t)= u(t)e^{-At}\), we use the product rule to differentiate: \(\displaystyle f'(t)= u'(t)e^{-At}- Au(t)e^{-At}\). Putting those into the differential equation, \(\displaystyle u'(t)e^{-At}- Au(t)e^{-At}= -Au(t)e^{-At}+ g(t). The two "\(\displaystyle -Aue^{-At}\)" cancel as a consequence of \(\displaystyle e^{-At}\) being a solution to the associated homogeneous equation. That leaves \(\displaystyle u'e^{-At}= g(t)\) so that \(\displaystyle u'= g(t)e^{At}\) and \(\displaystyle u(t)= \int g(t)e^{At}dt\).
Because, in your problem, \(\displaystyle g(t)= \frac{V_{irr}}{d}C_w(t)\) that can be writtenas \(\displaystyle u(t)= \frac{V_{irr}}{d}\int_0^t C_w(s)e^{As}ds+ D\). Notice that I have changed the "dummy variable" in the integral to s, to avoid confusion with the "real" variable, t, and have written the "indefinite" integral as a definite integral from 0 to t with the added constant, D, to allow for other possible starting values than "0".
Since \(\displaystyle f(t)= u(t)e^{-At}\), we now have \(\displaystyle f(t)= \left(\frac{V_{irr}}{d}e^{-At}\right)\int_0^t C_w(s)e^{As}ds+ De^{-At}\) or,
since your \(\displaystyle A= \lambda_N+\lambda_{1C}+ \lambda_{1I}\), \(\displaystyle f(t)= \left(\frac{V_{irr}}{d}e^{-(\lambda_N+\lambda_{1C}+ \lambda_{1I})t}\right)\int_0^t C_w(s)e^{(\lambda_N+\lambda_{1C}+ \lambda_{1I})s}ds+ De^{-(\lambda_N+\lambda_{1C}+ \lambda_{1I})t}\)\)