What is the steady state response of a system of two exponentials to sinusoidal excitation?
The system with impulse response:
$$I(t)=I_1e^{-t/\lambda_1}+I_2e^{-t/\lambda_2}$$

with Laplace Transform:
$$I(s)=I_1\frac{1}{s+1/\lambda_1}+I_2\frac{1}{s+1/\lambda_2}$$

and the input:
$$L(t)=a+bsin(\omega)$$

with Laplace Transform:
$$L(s)=\frac{a}{s}+\frac{bw}{s^2+w^2}$$

will obtain the following output, in Laplace Domain:
$$IL(s)=I(s)L(s)=(I_1\frac{1}{s+1/\lambda_1}+I_2\frac{1}{s+1/\lambda_2}) (\frac{a}{s}+\frac{bw}{s^2+w^2})$$

with this very simple Inverse Laplace Transform, as the final solution:
$$IL(t)=
-\frac{I_1 \lambda_1 e^{-t/\lambda_1} (a \lambda_1^2 w^2 + a - b \lambda_1 w)}{\lambda_1^2 w^2 + 1} 
- \frac{I_2 \lambda_2 e^{-t/\lambda_2} (a \lambda_2^2 w^2 + a - b \lambda_2 w)}{\lambda_2^2 w^2 + 1} 
+ a I_1 \lambda_1 + a I_2 \lambda_2 
+ \frac{\{b (sin(t w) (I_1 (\lambda_1 \lambda_2^2 w^2 + \lambda_1) + I_2 \lambda_2 (\lambda_1^2 w^2 + 1))
- w cos(t w) (I_1 \lambda_1^2 (\lambda_2^2 w^2 + 1) + I_2 \lambda_2^2 (\lambda_1^2 w^2 + 1)))\}}{(\lambda_1^2 w^2 + 1) (\lambda_2^2 w^2 + 1)}$$

As you can see, this kind of symbolisms can be readily solved in Wolfram Alpha through this:

inverselaplacetransform((c/(s+1/m)+d/(s+1/n))*(a/s+bw/(s^2+w^2)))