1. ## Differential Equation

Hi, I'm stuck on the following question:

Find the general solution to the following differential equation:

(dy/dx)+y.secx=secx

Any help available would be greatly appreciated!

2. Hi AlanC877,

$\frac{dy}{dx}+ysecx\;=\;secx$

$\frac{dy}{dx}\;=\;\left(1-y\right)secx$

$\frac{dy}{\left(1-y\right)}\;=\;secxdx$

$\frac{dy}{\left(y-1\right)}+secxdx\;=\;0$

From now on, it can be solved as seperable differential equation. If you still have problem, write again.

3. is it correct
dy/dx=secx(1-y) ?
If so - seperate variables.

4. That does make sense but I have been asked to find it by finding the integrating factor first.

i.e. P(x) would be set to equal secx, then the integral of this (secx) would be found. The integrating factor would be e^(the integral).
You then multiply both sides of the equation by the integrating factor, integrate both sides and rearrange to get an equation for y. Do you know this method?!

Really what I'm stuck on is integrating secx is suppose!

5. I know that type of solution but -i think- this is much easier.

And $\int{secx} = \ln{|secx+tanx| + C}$

6. Yep, I would agree but its the stupid way that we have to for the syllabus.

How did you get that integral of secx? I think I'm just bein stupid but I can't remember the method.

Thanks for your help so far.

7. $\int sec x dx = \int sec x \frac{sec x + tan x}{sec x + tan x}dx$

set, $u=sec x + tan x$

then we find, $du=(sec x\; tan x + sec^2x) dx$

substitute $du = (sec x\; tan x + sec^2x) dx, u = sec x + tan x$

After substitution we get;
$
\int secx\frac{sec x + tan x}{sec x + tan x}dx\;=\;\int\frac{\left(sec^2x+secx\;tanx\right)d x}{secx+tanx}\;=\;\int\frac{du}{u}$

Solve integral, we get; $\ln|u|+C$

Then substitute back $u=tanx+secx$, and we get;

$\int secx\;=\;\ln|secx+tanx| + C$