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Let the first order differential equation be

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say $v(x)$ then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor $v(x)$ from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as $\large e^{\int P(x)dx}=v(x)$

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$v$ can not change its sign, because $\ln|v(x)|$ always has a finite value. For the integrating factor, you just need one of the solutions of the homogeneous equation, so choosing the symbolically minimal $v(x)=e^{\int_0^x P(s)ds}$ is sufficient.

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I don't believe the absolute value matters here. Don't forget that integral is also going to have a constant of integration, which when rewritten, will leave you with some positive constant multiplying your integrating factor. For example

$$e^{\int xdx}=e^{x+C}=e^Ce^x$$

That $e^C$ is just some positive constant. It is also fine to say if $|v(x)|=e^Ce^x$, then $v(x)$ can equal $-e^Ce^x$. Multiplying the integrating factor by any constant (well, may want to avoid zero...) yields the same result.

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  • $\begingroup$ That constant of integration cancels in main solution and I am asking whether choosing negative value of v(x) matters for main solution.I don't understand the last statement u said may be you are contradicting yourself or may be I am wrong $\endgroup$ – Vishvajeet Patil May 13 '14 at 11:46
  • $\begingroup$ @VishvajeetPatil The problem is you're assuming we're neglecting the absolute value, but we don't have to. We can have $v(x)=\pm e^{P(x)dx}$. The choice of integrating factor doesn't matter much, but there is a choice. We often just go with a constant of integration of $0$ because it's the simplest. Just because we make one choice does not mean we neglect the others. $\endgroup$ – Mike May 13 '14 at 22:26

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