Why is this the general solution of this DE?

Question

I am reading a device physics text (Sze Physics of Semiconductor Devices, 3e, Chapter 2.4.3) and the author makes the claim that the solution $y(x)$ to a simple linear DE of the form $$y' +P(x)y = Q(x)$$ along with the boundary condition at $x = 0$ which I will just call $y(0)$ is $$y(x) = \frac{\int_0^{x}Q\exp \left(\int_0^{x''} P \,dx' \right) \,dx'' + y(0)}{\exp \left(\int_0^x P \,dx' \right)}.$$ My differential equations knowledge is weak as it's been a long time. I have some hazy sense that I should be looking for an integrating factor or something of the sort, but I would greatly appreciate if someone could step through the argument leading to this solution. Thank you!

Answer 1

Hint Multiply both sides of the original equation by the integrating factor $$\mu(x) := \exp \int_0^x P(s) \,ds ,$$ so that $$\mu'(x) = P(x) \exp \int_0^x P(s) \,ds = P(x) \mu(x),$$ whence the differential equation becomes $$y'(x) \mu(x) + y(x) \mu'(x) = Q(x) \mu(x) ,$$ or $$\frac{d}{dx} \left[y(x) \mu(x) \right] = Q(x) \mu(x) .$$ Now, integrate both sides, apply the Fundamental Theorem of Calculus to the left-hand side, and rearrange to solve for $y(x)$.