# Thread: First Order Differential Equation Help

1. ## First Order Differential Equation Help

Solving this differential question is supposed to only be worth 6 marks in a past exam question, but my working leads me to a very difficult integral. I was wondering if I have missed something in my solution?

$cos(x)y' + y = sec(x) = tan(x) \newline y' + sec(x)y = sec(x)(sec(x) + tan(x)) \newline \mu = exp(\int sec(x)dx) = sec(x) + tan(x) \newline \mu y' + \mu sec(x)y = \mu sec(x)(sec(x) + tan(x)) \newline \frac{\mathrm{d}}{\mathrm{d} x}y(sec(x)+tan(x)) = sec(x)(sec(x) + tan(x))^2 \newline y(sec(x) + tan(x)) = \int sec(x)(sec(x) + tan(x))^2 dx$

2. Originally Posted by StaryNight
Solving this differential question is supposed to only be worth 6 marks, but my working leads me to a very difficult integral. I was wondering if I have missed something in my solution?

$cos(x)y' + y = sec(x) = tan(x) \newline y' + sec(x)y = sec(x)(sec(x) + tan(x)) \newline \mu = exp(\int sec(x)dx) = sec(x) + tan(x) \newline \mu y' + \mu sec(x)y = \mu sec(x)(sec(x) + tan(x)) \newline \frac{\mathrm{d}}{\mathrm{d} x}y(sec(x)+tan(x)) = sec(x)(sec(x) + tan(x))^2 \newline y(sec(x) + tan(x)) = \int sec(x)(sec(x) + tan(x))^2 dx$
No it looks good so far

for the last integral

$\displaystyle \int \sec(x)(\sec(x)+\tan(x))^2dx$

let $u=\sec(x)+\tan(x) \implies du=[\sec(x)\tan(x)+\sec^2(x)]dx \iff du=u\sec(x)dx \iff \frac{du}{u}=\sec(x)dx$

putting this is gives the integral

$\displaystyle \int u^2\frac{du}{u}=\int udu$

3. Of course! I should have seen this.
Originally Posted by TheEmptySet
No it looks good so far

for the last integral

$\displaystyle \int \sec(x)(\sec(x)+\tan(x))^2dx$

let $u=\sec(x)+\tan(x) \implies du=[\sec(x)\tan(x)+\sec^2(x)]dx \iff du=u\sec(x)dx \iff \frac{du}{u}=\sec(x)dx$

putting this is gives the integral

$\displaystyle \int u^2\frac{du}{u}=\int udu$