How to calculate the probability of x is bigger than y while they have the same distribution but different mean values?

Question

I am reading an engineering book. The authors mentioned that variables $x,y$ are exponentially distributed with a mean $a^2+N_o$ and $N_o$, respectively. We can calculate the probability of error by direction integration, that is: $$p_e=\mathbb{P}\{y>x\}=[2+\frac{a^2}{N_o}]^{-1}$$ Where the probability density function of the r.v. is $f(u)=\frac{1}{\mu}\cdot\exp^{\frac{-u}{\mu}}$ with $\mu$ is the mean. My question is how to calculate the error probability since $x$ and $y$ have different mean values? To my knowledge, I think we can calculate $\mathbb{P}\{y>0\}$ with $\int_{0}^{\infty} \frac{1}{N_o}\cdot\exp^{\frac{-u}{N_o}}du$ (please correct me if I am wrong). Thus, how do I extend it to two r.v. with different mean values?