Before moving on to numerical methods for the solution of ODEs we begin by revising basic analytical techniques for solving ODEs that you will of seen at undergraduate level.
If your equation is of the form
$$ \frac{dy}{dx}=\frac{f(x)}{g(y)} \,, $$
then it can be reformulated as
$$\int g(y)\,dy = \int f(x)\,dx + C \,, $$
where $C \in \Re$ is a constant of integration.
This can be solved by integration, either directly or by substitution, to give a relation between $y$ and $x$.
The following example will show you how to use to use the separation of variables to solve a first-order ODE.
We wish to solve
$$ \frac{dy}{dx} = \frac{x}{y} \,, $$
subject to $ y(0) = 1 $.
The equation can be reformulated as
$$ \int y\,dy = \int x\,dx + C \,, $$
which can be solved to give
$$ \frac{y^2}{2} = \frac{x^2}{2} + C \,,$$
which simplifies to
$$ y = \pm \sqrt{x^2 + A} \,, $$
where $ A = 2C $.
This is the general solution to the differential equation. We now need to apply the initial condition $y(0)=1$.
$$ 1 = y(0) = \pm\sqrt{A} \,, $$
Therefore $A=1$, and we take the positive root.
The solution is $y = \sqrt{x^2+1}$.
If your equation is of the form
$$ \frac{dy}{dx} + f(x)y = g(x) \, $$
then it can be reformulated by introducing an integrating factor.
The integrating factor for the above equation is defined as
$$ \varphi(x) = e^{\int f(x) \,dx} \,, $$
(Note: that this is not unique as any multiple would also be an integrating factor).
In order to solve the original equation we multiply through by the integrating factor to give
$$ \varphi(x)\frac{dy}{dx} + \varphi(x)f(x)y(x) = \varphi(x)g(x) \,, $$
by construction of the integrating factor
$$ \frac{d\varphi}{dx} = \varphi(x)f(x) \,. $$
The original equation can be written as
$$ \frac{d}{dx}\left(\varphi(x)y(x)\right) = \varphi(x)g(x) \,.$$
This can be solved by integration of the right hand side, either directly or by substitution.
The solution is
$$ y(x) = \frac{\int\varphi(x)g(x)\,dx}{\varphi(x)} $$
Note: You need to be careful to consider where the solutions exist, that is, you need $\varphi(x) \neq 0$.
Therefore no solutions exist for $x$ such that $\varphi(x) \neq 0$.
The following example will show you how to use integrating factors to solve a first-order ODE.
We wish to solve
$$ \frac{dy}{dx} + \frac{y}{x} = 1 \,, $$
subject to $y(1) = 0$.
The integrating factor for this equation is given by
$$ \varphi(x) = e^{\int\frac{1}{x}\,dx} = x \,. $$
Therefore the original equation may be written as
$$ \frac{d}{dx}\left(xy\right) = x \,. $$
Integrating the right-hand side gives
$$ y(x) = \frac{x}{2} + \frac{C}{x} \,, $$
where $C \in \Re$ is a constant of integration, and $x \neq 0$. This is the general solution to the differential equation. We now need to apply the initial condition $y(1) = 0$.
$$ 0 = \frac{1}{2} + \frac{C}{1} \,.$$
Therefore $ C = \frac{-1}{2} $, and the solution is $ y(x) = \frac{x}{2} - \frac{1}{2x}$, valid for $x>0.$