Question

Find the general solution of the given differential equation

dr / d theta + rsec theta = cos theta

Answer 1

Question: $\frac {dr}{d\theta}+rsec(\theta)=cos(\theta)$

Solution:

Substituting $r=uv$ , $r'=u'v+uv'$ : (1)

$u'v+uv'+sec(\theta)uv=cos(\theta)$

$u'v+u(v'+sec(\theta)v)=cos(\theta)$ (2)

Setting the part inside () to 0 and separating variables:

$v'+sec(\theta)v=0$

$\frac {dv}{d\theta}=-sec(\theta)v$

$\int \frac {dv}{d\theta}=-\int sec(\theta)$

$ln(v)=-ln(sec(\theta)+tan(\theta)) + ln(k)$

$ln(v)=ln(\frac k {sec(\theta)+tan(\theta)})$

$v=\frac k {sec(\theta)+tan(\theta)}$

Putting $v$  from (3) into equation (2):

$u'\frac k {sec(\theta)+tan(\theta)}=cos(\theta)$

Using trigonometric identities $sec(\theta)=\frac 1 {cos(\theta)}$ , $tan(\theta)=\frac {sin(\theta)}{cos(\theta)}$ :

$du=\frac 1k cos(\theta)(sec(\theta)+tan(\theta))d\theta=\frac 1k (1+sin(\theta))d\theta$

$u=\frac 1k \int (1+sin(\theta))d\theta = \frac 1k (\theta-cos(\theta)+C)$

Finding r from (1):

$r=uv=\frac {\theta-cos(\theta)+C}{sec(\theta)+tan(\theta)}$

Answer: $r=\frac {\theta-cos(\theta)+C}{sec(\theta)+tan(\theta)}$