Definition. First-order differential equations are differential equations that only involve the first order derivatives of a specific variable with respect to another variable.
Examples:
$$ \frac{dy}{dx}=2x, \qquad y’= \arctan(x) + e^x, \qquad (2xe^x)\left(\frac{dy}{dx}\right)=\sin(\arctan(x)) $$
Definition. Degrees of freedom refers to the number of independent data that can successfully estimate a parameter.
This may feel a little vague. But think about this; you have computed the result of an indefinite integral, say
$$ \int 2x\ dx $$
and you get $x^2 + C$, where C is an arbitrary constant. How many pre-determined data, or observation, do you need to determine the value of $C$? The answer is one. Assume that you are given the fact that when you plug 2 into $x$, you get the result 4. Then, you say, with a little abuse of notation,
$$ \begin{gathered} (x=2)^2 + C = 4 \\ 4 + C = 4 \\ C=0 \end{gathered} $$
Therefore, we obtain the value of $C$, which means that one independent data is adequate to obtain the full solution, and so the degree of freedom is one.
In first-order differential equations, there only appears the first derivative of a variable, so, in the end, the degree of freedom of the differential equation is one. In general, it can be underlined that the degree of freedom of the solution of a differential equation is the same as its order.
It is better to start with an example.
$$ y’=2x^2y, \qquad y(0)=1. $$
Here, we know that $y’=2x^2y$ is an ordinary differential equation, furthermore, it is a first-order equation, whose degree of freedom is one. $y(0)=1$ is called the initial condition. Together, it is called an Initial Value Problem.
These type of problems ask us to solve the differential equation first, and then, plugging in the values $x=0$ and $y=1$ simultaneously, to compute the value of the constant, and finally, have the solution of the initial value problem.
Notation: IVP stands for Initial Value Problem and it is used in general.
Example(1) Solve the IVP given by
$$ y’=2x^2y, \qquad y(0)=1 $$
Solution: We first solve the equation and obtain a general solution as follows:
$$ \begin{aligned} \frac{dy}{dx} &= 2x^2y \\ \frac{1}{y}dy &= 2x^2dx \\ \int \frac{1}{y}\ dy &= \int 2x^2\ dx \\ \ln \lvert y \rvert &= \frac{2x^3}{3} + C \\ \lvert y \rvert &= \tilde{C} \cdot e^{\left(\frac{2x^3}{3}\right)} \end{aligned} $$
We emphasize the transformation from $e^C$ to $\tilde{C}$, since both are arbitrary constants. Now, we have obtained a general solution. It follows from plugging in the initial condition that
$$ \begin{aligned} \lvert 1 \rvert &= \tilde{C} \cdot e^0 \\ 1 &= \tilde{C} \cdot 1 \\ 1 &= \tilde{C} \end{aligned} $$
Therefore, the solution of the IVP is given by
$$ \lvert y \rvert = \tilde{C} \cdot e^{\frac{2x^3}{3}} $$
Example(2) Solve the IVP given by
$$ y’ = \frac{(1+y^2)(1+x^2)}{xy}, \qquad y(1)=1, \qquad x>0 $$
Solution: We can straightforwardly observe that we are also given the condition $x>0$. It is common to assume that, therefore, we may end up with an expression where $\ln(x)$ appears. However, this observation does not affect our method to solve the problem. Similarly, we first solve the equation and obtain a general solution, and then, by plugging in the initial condition, get the solution of the IVP.
$$ \begin{aligned} \frac{dy}{dx} &= \frac{(1+y^2)(1+x^2)}{xy} \\ \left(\frac{y}{1+y^2}\right)dy &= \left(\frac{1}{x} + x \right)dx \\ \int \left(\frac{y}{1+y^2}\right)\ dy &= \int \left(\frac{1}{x} + x \right)\ dx \\ \frac{\ln(1+y^2)}{2} &= \ln(x) + \frac{x^2}{2} + C \end{aligned} $$
Now, we have the general solution. We, then, consider the initial condition, by plugging in $y=1$ and $x=1$:
$$ \begin{aligned} \frac{\ln(1+1)}{2} &= \ln(1) + \frac{1^2}{2} + C\\ \frac{\ln(2)}{2} &= \frac{1}{2} + C\\ \frac{\ln(2)-1}{2} &= C \\ \ln\left(\sqrt{\frac{2}{e}}\right) &= C \end{aligned} $$
Therefore, we conclude that the solution of the IVP is given by the implicit equation
$$ \frac{\ln\left(1+y^2\right)}{2} = \ln(x) + \frac{x^2}{2} + \ln\left(\sqrt{\frac{2}{e}}\right). $$