Linear first order ODE with exponential coefficient

Question

I have a problem. I have to solve a linear first order, homogeneous ODE with an exponential coefficient.The equation is as follows; $\frac{dP(x)}{dt}-e^{-ax}P(x)=0$. Clearly, its impossible to find an IF here, because that would result in an exponential of an exponential.Are there any alternate methods to handle? Thank you in advance.

Answer 1

$$\frac{dP}{dx}-e^{ax} P=0 \implies \int \frac{dp}{P}=\int e^{ax} dx \implies \log P= \frac{e^{ax}}{a}+C \implies P(x)=C_1 e^{\frac{e^{ax}}{a}}$$

Answer 2

Exponential of exponential do not make a problem. If you prefer, let $P(x)=e^{Q(x)}$ $$\frac{dP(x)}{dx}=e^{-ax}P(x)\implies \frac{dQ(x)}{dx}=e^{-ax}$$