Separable ordinary differential equations

Let's consider, $y^{'} = f (x, y)$ with $y$ , an ODE of first order . We will say that the EDO is separable if we can rewrite it as $h (y) \cdot y^{'} = g (x)$ , that is, if we can move everything that depends on $y$ to one side of the equality and everything that depends on $x$ to the other.

Example

An example of separable ODE would be $y^{'} = 2 x y$ , since we can put everything that depends on the varible $y$ to one side of the equality and everything that depends on $x$ to other by dividing the entire equationby $y$ : $y^{'} = 2 x y ⟹ \frac{1}{y} y^{'} = 2 x$ In our case, then $h (y) = \frac{1}{y}, g (x) = 2 x$

Then we integrate both sides of the equality and we obtain the solution: $h (y) \cdot y^{'} = g (x) ⟹ h (y) \cdot \frac{d y}{d x} = g (x) ⟹ h (y) d y = g (x) d x ⟹$ $\int h (y) d y = \int g (x) d x + C$ Let's note that we have added a constant, since, when integrating a function, we do not know if there was a constant. Now we try to isolate $y$ in terms of $x$ and obtain the solution.

Example

For example, in the case shown previously: $\frac{1}{y} y^{'} = 2 x \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x} = 2 x \Rightarrow \frac{d y}{y} = 2 x \cdot d x \Rightarrow \int \frac{d y}{y} = \int 2 x \cdot d x + C \Rightarrow \Rightarrow \ln | y | = x^{2} + C \Rightarrow | y | = e^{x^{2} + C} = e^{x^{2}} \cdot e^{C} = K \cdot e^{x^{2}}, k > 0 \Rightarrow$ $\Rightarrow y (x) = k \cdot e^{x^{2}}, k > 0$

Something that is worth noting is that when we operated to get all the variables in one side, we may be losing some of the solutions. In a way in order to put all the $y$ on one side we assumed that $y \neq 0$ . However, if we look at the ODE we can realize that $y = 0$ is actually a solution of the ODE, as far as $k$ is also zero.

As we have already said, sometimes, we will have to solve a PVI. In the example we found all the solutions of the ODE. To find the solution of a PVI it is enough to impose the initial conditions and find the concrete constant that assures that the initial condition is satisfied.

Example

Let's consider, for example the PVI: ${\begin{matrix} y^{'} = 2 x y \\ y (0) = 1 \end{matrix}$ From the previous example we know that the solutions are: $y (x) = k \cdot e^{x^{2}} k \in R$ .

Let's look, then, tat he value of $k$ so that we have $y (0) = 1$ : $y (0) = 1 \Rightarrow 1 = y (0) = k \cdot e^{0} \Rightarrow k = 1$ Therefore, the solution to our PVI is: $y (x) = e^{x^{2}}$ .

We are going to show some more examples:

Example

Solve the ODE: $y^{'} = 4 x e^{-} y$ This is a separable ODE since we can put everything that depends on $x$ to one side and everything that depends on $y$ to other.

In effect: $y^{'} \cdot e^{y} = 4 x$ Now we proceed as we said: $y^{'} = \cdot e^{y} = 4 x \Rightarrow \frac{d y}{d x} e^{y} = 4 x \Rightarrow e^{y} \cdot d y = 4 x \cdot d x \Rightarrow \int e^{y} \cdot d y = \int 4 x \cdot d x \Rightarrow$ $\Rightarrow e^{y} = 2 x^{2} + C \Rightarrow y (x) = \ln (2 x^{2} + C)$ where $C$ is the constant that would be determined by the initial conditions.

Example

Solve the ODE: $2 x + 5 = y^{'} \cdot \sin y$ We observe that in this case we already have separated variables.

So let's proceed to the integrals: $2 x + 5 = y^{'} \sin y \Rightarrow 2 x + 5 = \sin y \cdot \frac{d y}{d x} \Rightarrow \int (2 x + 5) d x = \int \sin y d y \Rightarrow$ $\Rightarrow x^{2} + 5 x + C = - \cos y \Rightarrow y (x) = \arccos (- x^{2} - 5 x - C)$

Example

Solve the ODE: $y^{'} = x \cdot (y^{2} + 1)$ It is again a separable ODE when we divide all the terms by $y^{2} + 1$ . So its solution is obtained in the following way:

$\frac{y^{'}}{y^{2} + 1} = x \Rightarrow \frac{d y}{d x} \cdot \frac{1}{y^{2} + 1} = x \Rightarrow \frac{d y}{y^{2} + 1} = x \cdot d x \Rightarrow \int \frac{d y}{y^{2} + 1} = \int x \cdot d x \Rightarrow$ $\Rightarrow \arctan y = x + C \Rightarrow y (x) = \tan (x + C)$