Derivative of operator with respect to parameters From Shankar's QM book pg. 56:
For an operator $\theta(\lambda)$ that depends on a parameter $\lambda$ defined by $$\theta(\lambda)=e^{\lambda\Omega}$$
where $\Omega$ is also a constant operator, we can show that $$\frac{d}{d\lambda}\theta(\lambda)= e^{\lambda\Omega}\Omega=\theta(\lambda)\Omega .\tag{1.9.7}$$
Hence if we are confronted with the above differential equation, its solution is given by
$$\theta(\lambda)=Ce^{\lambda\Omega}$$
where $C$ is a constant operator.
My question is why does the constant operator $C$ appear?
 A: For essentially the same reason that it appears in differential equations of functions. The differential equation
$$\frac{\text{d}\theta(t)}{\text{d}\lambda} = \theta(\lambda) \Omega$$
defines a family of operators, given by $$\theta(\lambda) = C e^{\lambda \Omega}.$$ Different choices of the constant operator $C$ lead to different operators $\theta(\lambda)$, all of which satisfy the same differential equation. In other words, the choice of $C=\mathbb{I}$ is just one of the possibilities. This mirrors the case when you're working with functions, the solution to a differential equation of the form $f'(t) = a\times f(t)$ is the family of functions $f_c(t) = c \exp(at)$, where $c$ is a constant that is set by the value of $f(t)$ at $t=0$.
Another way to see explicitly that any choice of the operator $C$ satisfies this equation is by explicitly writing out the operator in its power-series form, i.e.:
\begin{align}
\theta(\lambda) = C e^{\lambda \Omega} &= C + \lambda C \Omega + \frac{\lambda^2}{2!} C \Omega^2 + \frac{\lambda^3}{3!} C \Omega^3 + ... \\
\implies \frac{\text{d}\theta}{\text{d}\lambda} &= 0\,\,\mathbb{I} + C \Omega + \lambda C \Omega^2 + \frac{\lambda^2}{2!}  \lambda C \Omega^3 + ...\\
&= C \left( \mathbb{I} + \lambda \Omega + \frac{\lambda^2}{2!} \Omega^2 + ...\right) \Omega \\
&= C e^{\lambda\Omega} \Omega,\\
\text{i.e. }\quad \frac{\text{d}\theta}{\text{d}\lambda} &= \theta(\lambda) \Omega.
\end{align}
Note that since $C$ and $\Omega$ need not commute, so I pulled $C$ out to the left, and $\Omega$ out to the right. Thus, $\theta(\lambda) = C\exp{\lambda\Omega}$ satisfies the differential equation for any constant operator $C$.
