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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?

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


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

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    $\begingroup$ 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. $\endgroup$ – Lutz Lehmann Apr 4 '20 at 5:20
  • $\begingroup$ Yes that's true @LutzLehmann I just multiplied the equation by $F(x)$ Then I integrated the DE. $\endgroup$ – Aryadeva Apr 4 '20 at 5:26

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