find largest interval over which the general solution is defined
$$\begin{align}
\frac{dy}{dx}+\frac{sin(x)}{cos(x)}y=\frac{1}{cos(x)}
\end{align}
$$
Integrating factor will be
$$\begin{align}\rightarrow I= e^{\int \frac{sin(x)}{cos(x)}dx}=e^{-ln|cos(x)|}=\frac{1}{cos(x)}\end{align}$$
Multiply I to each term of the equation
$$\begin{align}\rightarrow \frac{1}{cos(x)}\frac{dy}{dx}+\frac{1}{cos(x)}\frac{sin(x)}{cos(x)}y=\frac{1}{cos(x)}\frac{1}{cos(x)}\end{align}$$
Integrate both sides:
$$\begin{align}\rightarrow \frac{d}{dx}(\frac{1}{cos(x)}y)=\int \frac{1}{cos^2(x)}\end{align}$$ (left side of the resulting equation is automatically the derivative of the integrating factor and y)
$$\begin{align}\rightarrow \frac{sin(x)}{cos^2(x)}y=tan(x)+C\end{align}$$
Explicit solution
$$\begin{align}\rightarrow y=\frac{cos^2(x)}{sin(x)}(tan(x)+C)\end{align}$$
$$\begin{align}\rightarrow y=\frac{tan(x)}{sec(x)}+\frac{C}{sec(x)}\end{align}$$
if the steps were right this is so far I can go, having problems to find the largest interval from the solution and transient term, any help?
Best Answer
This is a linear differential equation. Therefore it is defined on the largest intervals for which the functions used in the differential equation are all defined.
Hence solutions are defined on intervals for which $\cos x$ doesn't vanish. If you want a solution defined at $0$, the interval is $(-\frac{\pi}{2},\frac{\pi}{2})$