Homogeneous linear equations of order n with constant coefficients

As happen with the linear systems of order 1, one ODE of order $n$ has $n$ linearly independent solutions in such a way that any solution to an homogeneous ODE will be a linear combination of these linearly independent solutions. Therefore, to solve the ODE will consist in finding these $n$ functions.

The ODE we consider is $$a_n \cdot y^{(n)} (x)+a_{n-1} \cdot y^{(n-1)}+ \ldots +a_1 \cdot y'+ a_0 \cdot y=0$$ where $a_i$ are constants.

An example of a honogenous ODE of order $n$ would be:$$y''+y=0$$

Then we define the characteristical polynomial of the ODE as: $$a_n\cdot \lambda^n+a_{n-1} \cdot \lambda^{n-1}+ \ldots + a_1 \cdot \lambda+a_0=0$$ and we look at its $n$ roots.

The characteristical polynomial is easy to write, it is just necessary to change $y$ for $\lambda$ and to raise it to the corresponding order of the derivative.

For example, in the ODE that we have given, the associated characteristic polynomial is: $\lambda ^2+1=0$.

This polynomial has two combined complex roots: $\lambda_1=i, \ \lambda_2=-i$

Then

• If $\lambda$ are real and simple the solution is of the form: $e^{\lambda x}$
• If $\lambda$ are real of multiplicity m the solutions are of the form:$e^{\lambda x}, x\cdot e^{\lambda x}, x^2\cdot e^{\lambda x}, \ldots, x^{m-1} \cdot e^{\lambda x}$
• If $\lambda=a+bi$ are complex and simple, the two solutions are of the form: $e^{ax}\cos (bx), e^{ax}\sin (bx)$ (there are two solutions because whenever a complex root exists the conjugate also appears)
• If $\lambda=a+bi$ are complex of multiplicity m, the two solutions are of the form: $$e^{ax}\cos (bx),x \cdot e^{ax}\cos (bx), \ldots, x^{m-1}e^{ax}\cos (bx) \\ e^{ax}\sin (bx), x\cdot e^{ax}\sin (bx), \ldots, x^{m-1} e^{ax}\sin (bx)$$

Then, once we find these $n$ solutions, the general solution of the ODE will be a linear combination of these $n$ solutions.

Let's come back to the example at the beginning. As our polynomial took two (simple) combined complex roots we are in case 3. Therefore the solution is: $$y(x)=c_1 \cdot \cos x+ c_2 \cdot \sin x$$ where the constants will be determined by the initial conditions (in the event that we would know them).