Constant of Integration of the Integrating Factor

Question

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!

Study the derivation of solving it once again. Or you are just done with the function you gave,it satisfy the lenier DE. — ABC, Mar 31 '13 at 13:57
I understand how to solve a linear DE using the IF. I just don't get why there is only one constant involved. — Zugzwang14, Mar 31 '13 at 14:07
Ok , do you have a reason why $C$ is not added in the Integration by parts[$\int fx.gx$] second integral (while we integrate g): $-\int( f'x\int (gx) .dx)dx$ .No, These are definations got by derivating a particular function to get the function whose integral is needed. In linear DE we multiply with $ e^{-P(x).dx}$ To get a function which is anti derivative of required one.It's a way , a defination to solve it.If we use a $C$ we would end up getting a wrong function. — ABC, Mar 31 '13 at 14:13
It's just typos. You should get $$y=e^{-(f(x)+C)}\int{Q(x)e^{f(x)+C} dx}=e^{-f(x)}\int{Q(x)e^{f(x)} dx}$$ — Ivan Loh, Mar 31 '13 at 14:36
Possible duplicate of Why is the constant of integration of the integrating factor irrelevant when solving first-order linear differential equations? — gen-ℤ ready to perish, May 04 '18 at 18:19

ryang · Answer 1 · 2023-02-27T19:00:10.407

The integrating factor of a differential equation is unique up to a nonzero multiplicative constant. For example, 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 from the integrating factor is equivalent to setting $M=1,$ which is the simplest choice of $M$.

score 0 · Answer 2 · answered Nov 30 '20 at 03:21

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.

Constant of Integration of the Integrating Factor

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