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.



2 Answers 2


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.