Question

First Order Linear Differential Equation
Please help with step by step thank you :_)

Answer 1

\[\frac{ dr }{ d \theta }+ r \tan \theta=\sec \theta \]

Answer 2

\[r'+r\tan\theta=\sec\theta\]
\[r'+r\frac{\sin\theta}{\cos\theta}=\frac{1}{\cos\theta}\]
\[r'\cos\theta+r\sin\theta=1\]
from inspection and a knowledge that \(\sin^2\theta+cos^2\theta=1\) we can say that
\[r = \sin\theta\] is one solution, but also notice that
\[r = \sin\theta+A\cos\theta\] is also a solution because
\[r'\cos\theta+r\sin\theta = (\cos\theta-A\sin\theta)\cos\theta + (\sin\theta+A\cos\theta)\sin\theta\]
\[=\sin^2\theta+cos^2\theta + A\sin\theta\cos\theta-A\cos\theta\sin\theta = \sin^2\theta+cos^2\theta\]

Answer 3

i solved it with a friend and got the same answer, thanks!