Integrating Factor Greetings: What is it & Why Use It?

[amaresh92]

greetings
what does a integrating factor tells about a differential equation?
in order to find the solution for a exact equation we multiply the equation by integrating factor(I.F).
as intergrating factor=e^integration(p)dx
i.e given by I.F=e^gx where gx is integration of p
now as we have integrated the p we did not add any constant to the integration which is gx.
may i know the reason why we did not add any constant over there.
advanced thanks

 

[HallsofIvy]

You appear to be mixing different concepts. Any time we have a first order differential equation, $dy/dx= f(x,y)$, rewritten as $dy- f(x,y)dx= 0$ then there exist an "integrating factor", $\mu(x)$ such that $\mu(x)dy- \mu(x)f(x,y)dx$ is an "exact" differential. That is, that there exist a function F(x) such that $dF= \mu(x)dy- \mu(x)f(x,y)dx$. Then the differential equation becomes $dF= 0$ so that F(x, y)= C is a solution to the equation.

We don NOT "multiply the equation by integrating factor" "in order to find the solution for a exact equation"- exact equations are easy by themselves. We multiply non-exact equations by an integrating factor to make the equation exact.

However, it is only in the case that the differential equation is linear, that is of the form $dy/dx+ p(x)y= f(x)$ that we have a specific formula for that integrating factor
$$e^{\int p(x) dx}$$

To see why we do not need to add the constant of integration, look at what happens if we do:
$$e^{\int p(x)dx+ C}= e^{\int p(x)dx}e^{C}$$
Now multiplying the entire equation by that makes it "exact":
$$e^{\int p(x)dx}e^C dy/dx+ e^{\int p(x)dx}e^Cp(x)y= e^{\int p(x)dx}e^Cf(x)$$
$$\frac{d\left(\int p(x)e^Cy\right)}{dx}= e^{\int p(x)dx}e^Cf(x)$$

Now, because $e^C$ is a constant we can take it out of the derivative on the left, divide both sides by it, and get rid of it. Since any constant of integration gives the same thing we don't need it.