Linear ordinary differential equations

A linear ODE is an ODE of the form: $y^{'} = a (x) \cdot y + b (x)$ where $a (x)$ and $b (x)$ are continuous functions.

Example

An example of linear ODE would be: $y^{'} = 5 x^{2} \cdot y + 2 x^{2}$ although, sometimes, we will find a linear ODE after doing some conversion. For example, we might have an ODE of the following form: $\frac{y^{'}}{5 x^{2}} = y + \frac{2}{5}$

and we can rewrite it (multiplying by $x^{2}$ ) to obtain the previous form.

In the particular case in which $b (x) = 0$ , we will say that the equation is homogeneous.

The resolution of this type of equations is divided into two steps.

Solve the homogeneous part.

We solve the equation: $y_{h}^{'} = a (x) \cdot y_{h}$ . This is a separable ODE, therefore, easy to solve. This solution is: $y_{h} (x) = k \cdot e^{\int a (x) d x}$

Example

In our example, we have: $y_{h}^{'} = 5 x^{2} y_{h} \Rightarrow y_{h} (x) = k \cdot e^{\int 5 x^{2} d x} = k \cdot e^{\frac{5}{3} x^{3}}$

Find a particular solution to the non homogeneous ODE.

We are going to use the method change of constants. Calling $y_{1} (x) = e^{\int a (x) d x}$ we look for a particular solution of the type $y_{p} (x) = u (x) \cdot y_{1} (x)$ .

Let's designate that as a solution: $\begin{matrix} y_{p}^{'} = u^{'} \cdot y_{1} + u \cdot y_{1}^{'} = u^{'} \cdot y_{1} + u \cdot a (x) \cdot y_{1} \\ y_{p}^{'} = a (x) \cdot y_{p} + b (x) = a (x) \cdot u \cdot y_{1} + b (x) \end{matrix}} \Rightarrow u^{'} = \frac{b (x)}{y_{1}}$

We solve the latter equation (it is enough to integrate both sides) and we already have a particular solution.

Example

In our example, we take $y_{1} (x) = e^{\frac{5}{3} x^{3}}$ Then we look for a particular solution of the form $y_{p} (x) = u (x) \cdot y_{1} (x)$ . We have seen that $u (x)$ has the following derivative $u^{'} = \frac{b (x)}{y_{1} (x)} = \frac{2 x^{2}}{e^{\frac{5}{3} x^{3}}} = 2 x^{2} \cdot e^{- \frac{5}{3} x^{3}}$

and, integrating, we obtain: $u (x) = \int 2 x^{2} \cdot e^{- \frac{5}{3} x^{3}} \cdot d x = - \frac{2}{5} e^{- \frac{5}{3} x^{3}}$

Thus, $y_{p} (x) = - \frac{2}{5} \cdot e^{- \frac{5}{3} x^{3}} \cdot e^{\frac{5}{3} x^{3}} = - \frac{2}{5}$

Finally, the solution of the linear equation is $y (x) = y_{h} (x) + y_{p} (x)$ .

Notice that the constant appears in the homogeneous solution (it does not make sense to put integration constants in the homogenous part).

Therefore, in our example, the solution of the ODE is: $y (x) = k \cdot e^{\frac{5}{3} x^{3}} - \frac{2}{5}$