# General solution to ODE

I'm pretty comfortable solving most ODE's you'd find in a differential equations class at this point, but just realized that there is a conceptual gap in my knowledge, particularly, how does one know when a solution is the general solution?

My understanding of a general solution is that all particular solutions can be derived from the general solution, which is a powerful restriction. I'm not sure how to be certain this is the case. For a linear ODE, one can apply the superposition principle to multiple particular solutions to obtain a new one. I understand how this works, but as I understand, for an $n^{th}$ degree linear ODE, a linear combination of $n$ independent solutions will yield a general solution. Is this true, and if so, why? It makes some intuitive sense in a linear algebra perspective, but I'm not sure how to begin formalizing it.

Thanks!

## 2 Answers

So let us define the following $n^{th}$ degree linear ODE as the following:

\begin{align} x^{(n)} + \sum_{i=0}^{n-1} a_i x^{(i)} = \Psi(t) \end{align}

where, to keep things simple, $a_i \; \forall i$ are constants. We can then define a new set of variables where $y_{i} = x^{(i)} \;\forall i \in \lbrace 0, 1, \cdots, n-1 \rbrace$. Using these variables, we can create a system of $n$ first order ODEs with the following form:

\begin{align} \dot{\boldsymbol{y}} &= A \boldsymbol{y} + D \Psi(t) \end{align}

where $D = [0, 0, \cdots, 0, 1]^{T}$ and $A$ can be defined as the following:

\begin{align} A = \begin{bmatrix} \boldsymbol{0} & I \\ -\boldsymbol{a}^{T} & \end{bmatrix} \end{align}

where $\boldsymbol{a}^{T} = [ a_0, a_1, \cdots, a_{n-2}, a_{n-1} ]$ (makes up entire bottom row), $I$ is a $(n-1) \times (n-1)$ identity matrix, $\boldsymbol{0}$ is a $(n-1) \times 1$ matrix of zeros. We can then assume we know the eigendecomposition of $A$ such that $A = Q \Lambda Q^{-1}$. Making this substitution into the system of first order equations and simplifying gives us the following:

\begin{align} Q^{-1} \dot{\boldsymbol{y}} &= \Lambda Q^{-1} \boldsymbol{y} + Q^{-1} D \Psi(t) \end{align}

Since $A$ is a constant matrix, $Q$, $\Lambda$, and $Q^{-1}$ are constant matrices as well. This means we can define $\boldsymbol{z} = Q^{-1}\boldsymbol{y}$, which means $\dot{\boldsymbol{z}} = Q^{-1} \dot{\boldsymbol{y}}$. Substituting all this and just defining $\hat{\boldsymbol{\Psi}}(t) = Q^{-1} D \Psi(t)$, we get the following:

\begin{align} \dot{\boldsymbol{z}} &= \Lambda \boldsymbol{z} + \hat{\boldsymbol{\Psi}}(t) \\ &\text{or}\\ \dot{z}_{k} &= \lambda_{k} z_{k} + \hat{\Psi}_{k}(t) \;\;\; \forall k \end{align}

Using the latter form gives you $n$ independent ODEs you can solve. Once you solve them for the solutions $z_{k}(t) \; \forall k$, you can get the solution of $\boldsymbol{y}(t)$ by doing:

\begin{align} \boldsymbol{y}(t) &= Q \boldsymbol{z}(t) \\ \boldsymbol{y}(t) &= \sum_{k=1}^{n} \boldsymbol{q}_{k} z_{k}(t) \end{align}

where $\boldsymbol{q}_{k}$ is the $k^{th}$ column of $Q$. As you can see, $\boldsymbol{y}(t)$ becomes a linear combination of the $n$ solutions $z_{k}(t)$ found in the diagonalized system.

From memory, I think for an $n$th degree ODE, a linear combination of $n$ independent solutions also is a solution.

The general solution should account for all particular solutions.

• The general solution to the ODE will generally only be a linear combination of particular solutions if it's a linear equation. Jan 23, 2017 at 19:17
• You more or less just restated what I wrote in the question. My question is why? Jan 23, 2017 at 19:19