# Issue Using Integrating Factor

Let $$F(x)$$ be an arbitrary continuously differentiable function, and consider the following differential equation

$$y'+ \frac{F'(x)}{F(x)}y= \frac{1}{F(x)}.$$

If I use the Integrating Factor which says that for an equation $$y'+ P(x)y= Q(x),$$

the general solution is

$$y=e^{-\int P(x)dx}\int Q(x)e^{\int P(x)dx}dx+Ce^{-\int P(x)dx}.$$

Using the fact that

$$\int \frac{F'(x)}{F(x)} dx = \ln(F(x)),$$

this simplifies to

$$y=(1+C)F(x)^{-1}.$$

However, if I substitute this solution into the initial differential equation I get

$$-\frac{(1+C)F'(x)}{F(x)^2} + \frac{F'(x)}{F(x)}\frac{(1+C)}{F(x)} = \frac{1}{F(x)},$$

or

$$0 = \frac{1}{F(x)}.$$

Where is my mistake?

• $\int 1 \neq 1$ – Ninad Munshi Apr 3 '20 at 20:18
• Rewrite the differential equation as $(y'F+F'y)=1$ and note that the LHS is $(yF)'$. – dan_fulea Apr 3 '20 at 20:25
• Your solution for y is not correct. That's why when you plug it in the DE it gives you something wrong. – Aryadeva Apr 3 '20 at 20:37

## 1 Answer

You don't really need an integrating factor: $$y'+ \frac{F'(x)}{F(x)}y= \frac{1}{F(x)}$$ $$F(x)y'+ F'(x)y= 1$$ $$(F(x)y)'= 1$$ $$F(x)y= x+c$$ $$\implies y(x)= \dfrac {\color{blue}{x+c}}{F(x)}$$ And not $$y(x)= \dfrac {\color{red}{1+c}}{F(x)}$$

$$y=e^{-\int P(x)dx}\int Q(x)e^{\int P(x)dx}dx+Ce^{-\int P(x)dx}.$$ $$y=\dfrac {1}{F(x)}\int dx+\dfrac C{F(x)}$$ $$y=\dfrac {x}{F(x)}+\dfrac C{F(x)}$$

• In the first step, you are using $F$ as integrating factor, you multiply the equation with it to get integrable expressions. You are just not using the formula for it. – Lutz Lehmann Apr 4 '20 at 5:20
• Yes that's true @LutzLehmann I just multiplied the equation by $F(x)$ Then I integrated the DE. – Aryadeva Apr 4 '20 at 5:26