Suppose that $T$, $X$ and $C$ are random variables with $X\sim \mathrm{bin}(1,r)$ and $T\mid X=x\sim \exp(\lambda_x)$ for $x=0,1$ and $C\sim\exp(\lambda_C)$ such that $C$ is independent of $T$ and $X$. Let $p,q\in (0,1)$ such that $p+q\in (0,1)$ be given probabilities and define $t=-\log(1-p)$. Assume furthermore that $ P(T\leq t\mid X=x)=p+qx,\quad x=0,1. $ Then the parameter $\lambda_x$ is given by $ \lambda_x=-\frac{\log(1-p-qx)}{t} $ and in particular $\lambda_0=1$. In the above setting everything but $\lambda_C$ are known parameters.
My problem is the following: Given a particular value of $P(T\leq C\wedge t)$, how do I find the value of $\lambda_C>0$ that realizes this probability (and for what values of the probability is it possible to find such $\lambda_C$?)
My attempt at this problem, is to note that $ P(T\leq C\wedge t)=(1-r)P(T\leq C\wedge t\mid X=0)+rP(T\leq C\wedge t\mid X=1) $ and so we only have to find an expression of $P(T\leq C\wedge t\mid X=x)$ for $x=0,1$. This can be done in the following way: $ \begin{align*} P(T\leq &C\wedge t\mid X=x)=\int_0^\infty\left(1-\exp(-\lambda_x\cdot y\wedge t)\right)\,\lambda_C\,e^{-\lambda_c y}\;\mathrm d y\\ &=1-\int_0^t \lambda_C \exp(-(\lambda_x+\lambda_C)y)\,\mathrm dy-\int_t^\infty\lambda_C\exp(-\lambda_xt)\exp(-\lambda_c y)\,\mathrm dy\\ &=1+\frac{\lambda_C}{\lambda_x+\lambda_C}\exp(-(\lambda_C+\lambda_x)t)-\frac{\lambda_C}{\lambda_x+\lambda_C}-\exp(-(\lambda_C+\lambda_x)t). \end{align*} $ However, this is where I am stuck. I tried using that $\exp(-\lambda_x t)=(1-p-qx)$ but without success. Does anyone have an idea on how to move on from here? Also I should probably mention that I do not know if the problem is possible to solve analytical.