I was reading my notes and at some point I write that in solving a first order linear DE using the integrating factor the final solution should have only one arvitrary constant arising from the integration of $Q(x)e^{\int P(x)dx}$ and so $e^{\int P(x)dx}$ does not contain an arbitrary constant.
Essentially what this means is that if $$\frac{dy}{dx}+P(x)\cdot y=Q(x) \Rightarrow y=e^{-\int P(x)dx}\int Q(x)e^{\int P(x)dx}dx,$$the only constant that would determine a particular solution would be the one arising from the second integral and not the one in the exponent. Why is this?
I understand that since this is a firstorder DE is should only involve one constant, but what about what I ask above?
If I define $\int P(x)dx=f(x) + C$ and substitute above, this would yield $$y=e^{-f(x)+c}\int Q(x)e^{-f(x)+c}dx=e^{-f(x)}e^{2c}\int Q(x)e^{-f(x)}dx$$and now also define $\int Q(x)e^{-f(x)}dx=g(x)+k$ we have$$y=e^{-f(x)}e^{2c}\left(g(x)+k\right)=e^{2c}\cdot e^{-f(x)}g(x)+ke^{2c}\cdot e^{-f(x)}$$ so wouldn't the constant of integration $c$ also affect the final solution?
Thanks in advance!
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$\begingroup$The integrating factor of a differential equation is unique up to a nonzero multiplicative constant.
E.g., the integrating factor of the differential equation $$\dfrac{\mathrm{d}y}{\mathrm{d}x}+g(x)\,y=h(x)$$ is in general $Me^{\int g(x)\,\mathrm{d}x}\,\left(M\neq0\right),$ which attains the solution independently of $M$:\begin{equation} \begin{aligned} \left(Me^{\int g(x)\,\mathrm{d}x}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x}+\left(Me^{\int g(x)\,\mathrm{d}x}\right)g(x)\,y&=\left(Me^{\int g(x)\,\mathrm{d}x}\right)h(x)\\ e^{\int g(x)\,\mathrm{d}x}\dfrac{\mathrm{d}y}{\mathrm{d}x}+e^{\int g(x)\,\mathrm{d}x}g(x)\,y&=e^{\int g(x)\,\mathrm{d}x}h(x)\\ \dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{\int g(x)\,\mathrm{d}x}\,y\right)&=e^{\int g(x)\,\mathrm{d}x}h(x)\\ y&=\frac1{e^{\int g(x)\,\mathrm{d}x}}\int e^{\int g(x)\,\mathrm{d}x}h(x)\,\mathrm{d}x. \end{aligned} \end{equation}
Dropping the constant of integration (and any external modulus sign) from the integrating factor is equivalent to setting $M=1,$ which is the simplest choice of $M$.
$\endgroup$ $\begingroup$Take a close look at your result. Everything is divisible by $e^{2c}$, leaving $k$ as the only relevant constant.
In other words, if you do it without $c$, you can still get to your solution by just multiplying both sides by $e^2c$, which you are allowed to do anyway since it is a constant.
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