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!
Hi AlanC877,
$\displaystyle \frac{dy}{dx}+ysecx\;=\;secx$
$\displaystyle \frac{dy}{dx}\;=\;\left(1-y\right)secx$
$\displaystyle \frac{dy}{\left(1-y\right)}\;=\;secxdx$
$\displaystyle \frac{dy}{\left(y-1\right)}+secxdx\;=\;0$
From now on, it can be solved as seperable differential equation. If you still have problem, write again.
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!
$\displaystyle \int sec x dx = \int sec x \frac{sec x + tan x}{sec x + tan x}dx$
set, $\displaystyle u=sec x + tan x $
then we find, $\displaystyle du=(sec x\; tan x + sec^2x) dx $
substitute $\displaystyle du = (sec x\; tan x + sec^2x) dx, u = sec x + tan x $
After substitution we get;
$\displaystyle
\int secx\frac{sec x + tan x}{sec x + tan x}dx\;=\;\int\frac{\left(sec^2x+secx\;tanx\right)d x}{secx+tanx}\;=\;\int\frac{du}{u}$
Solve integral, we get; $\displaystyle \ln|u|+C$
Then substitute back $\displaystyle u=tanx+secx$, and we get;
$\displaystyle \int secx\;=\;\ln|secx+tanx| + C$