I'm working on DEs and I've come across some speedbumps. I'm not very familiar with linear algebra and that might be the problem. However, I'm sure that there's a non-linear algebra explanation that I could understand.

Here's what I know right now:

There's a thing called the Principle of Superposition, which says that if $y_0$ and $y_1$ are linearly independent solutions to a given linear DE, then so is $C_0y_0+C_1y_1$. (The DE might have to be homogeneous, I'm not sure.) I don't understand why this is true.

If each solution is not a multiple of the other, then each is linearly independent from the other. Beyond that explanation, I don't know what it means.

Then, I want to learn how to show that this sum is a general solution. This is where my book introduces Cramer's Rule. Is there an intuitive explanation that will explain to me what is happening here?

I'm learning from online sources and trying to mash together an explanation but they all tend to skip over the details here, or assume some form of precogniscience. (You know, where an explanation of something low on the 'math tree' is given in terms of higher order mathematics.)

When I work with a 1st-order DE, I understand that a general solution is a 'family' of solutions, given the unknown constant. They're a one-dimensional family as it were. I can represent this family as a directional field.

With a 2nd-order DE, I guess that the general solution is a 2-dimensional family. I don't know how I would represent that geometrically. Maybe parametrically with a directional field? I'm just guessing.

Any explanations of any of these things would be very much appreciated.

  • 3
    $\begingroup$ Derivatives are linear operators, so sums of solutions are always solutions (e.g. $\frac{d}{dx}(f+g)=\frac{d}{dx}f+\frac{d}{dx}g$). The real trick is whether you can construct all solutions from some elementary ones. $\endgroup$
    – Andrew
    Commented Nov 28, 2012 at 17:32
  • $\begingroup$ Intuitively if there are solutions for which the equation equals to zero, when you add them up it makes 0 + 0 = 0. Multiply with two different constants: a*0 + b*0 = 0, is still 0. Just in words what is formally proven in the below answers. $\endgroup$ Commented Aug 1, 2021 at 17:19

3 Answers 3


This is true for a homogeneous linear differential equation. The reason is that you are looking for a solution to $$(\sum_{n=0}^k a_nD^n)y = 0$$ where D is the derivative operator. Now observe if $y_1$ and $y_2$ are solutions then observe that $$(\sum_{n=0}^k a_nD^n)y_1 = 0$$ $$(\sum_{n=0}^k a_nD^n)y_2 = 0$$ Now take any two real numbers $c_1$ and $c_2$ and observe that $$(\sum_{n=0}^k a_nD^n)(c_1y_1+c_2y_2) = c_1(\sum_{n=0}^k a_nD^n)y_1 + c_2(\sum_{n=0}^k a_nD^n)y_2 = 0$$

Thus we showed that a linear combination of solutions is also a solution.

  • $\begingroup$ Is it safe to say that even though $y$ is a function here, it is also a vector? If so, of what vector space it is? Also, what about if I have a system of linear equations? Is it plausible to say, for example, that in the $n$-dimensional ODE with constant coefficients $\mathbf{x}’ = A\mathbf{x}$, $A$ is acting as $a_0$ in $\frac{dy}{dx} = a_0{y}$? So the homogeneous equation is actually $\frac{d\mathbf{x}}{dt} - A\mathbf{x}= \mathbf{0}$, making $A$ the “constant coefficient” term. If so, is the solution just $e^A$? (!!!) What would a $n$-dimensional 2nd or 3rd ODE look like in matrix form? $\endgroup$ Commented Feb 13 at 20:27

Do you recall from Calculus that for constants $c$, differentiable functions $f(x)$ and $g(x)$, and integers $n\geq 0$ we have the following two facts? $$\frac{d^n}{dx^n}\bigl[c\cdot f(x)\bigr]=c\cdot\frac{d^n}{dx^n}\bigl[f(x)\bigr]\tag{1}$$ and $$\frac{d^n}{dx^n}\bigl[f(x)+g(x)\bigr]=\frac{d^n}{dx^n}\bigl[f(x)\bigr]+\frac{d^n}{dx^n}\bigl[g(x)\bigr]\tag{2}$$

The same idea is in play, here, with linear homogeneous DEs (they do need to be homogeneous).


The second question is still not answered ('how to show that this sum is a general solution'). I was looking for this myself, then came across this post. I think I understand it now. Thought I might as well post it here even though this is a very old question. If anyone ever reads this : let me know if there's something wrong with the explanation below...

Assume we have two different solutions $x(t)=\alpha(t)$ and $x(t)=\beta(t)$ to an ODE of the form $\frac{dx}{dt}+a(t)x(t)=0$.

For the two curves to cross each other we have to have :

$\alpha(t_0)=\beta(t_0)$ and $\frac{d}{dt}\alpha(t)|_{t=t_0} \neq \frac{d}{dt}\beta(t)|_{t=t_0}$ for some $t_0$ .

With $\frac{d \alpha}{dt}+a(t) \alpha(t)=0$ and $\frac{d \beta}{dt}+a(t) \beta(t)=0$ we see that this is impossible. Because the curves cannot cross they must be uniquely determined by their value for some $t=t_0$.

Now in a second order ODE like $\frac{d^2x}{dt^2}+a(t)\frac{dx}{dt} + b(t)x=0$ , two solutions can cross each other. But when we assume two solutions with : $\alpha(t_0)=\beta(t_0)$ and $\frac{d}{dt}\alpha(t)|_{t=t_0} = \frac{d}{dt}\beta(t)|_{t=t_0}$ and $\frac{d^2}{dt^2}\alpha(t)|_{t=t_0} \neq \frac{d^2}{dt^2}\beta(t)|_{t=t_0}$ for some $t_0$ we again see that this cannot be done.

So for two solutions $\alpha(t)$ and $\beta(t)$ with : $\alpha(t_0)=\beta(t_0)$ we see that the curves of $\frac{d\alpha}{dt}$ and $\frac{d\beta}{dt}$ can never cross each other because when $\frac{d\alpha}{dt}$ and $\frac{d\beta}{dt}$ are equal in $t_0$ then their derivatives $\frac{d}{dt}(\frac{d\alpha}{dt})$ and $\frac{d}{dt}(\frac{d\beta}{dt})$ must also be equal.

Thus curves of $\frac{dx}{dt}$ are uniquely determined by the choice of two boundary conditions : $x(t_0)$ and $\frac{dx}{dt}|_{t=t_0}$ . By integration of $\frac{dx}{dt}$ keeping the boundary conditions fixed, this in turn means that $x(t)$ must be completely determined by this choice of boundary conditions.

Same goes for higher order linear homogeneous ODE's of course.

The only thing we have to check now for the solution to be the complete solution is that we must be able to reach all possible function values with the constant coefficients we use in our solution. This can be done using the Wronskian determinant.


It is also possible for curves to cross each other while having the same derivative e.g.: $x = \sin t $ and $ x=t $ when $ t=0 $. But that would mean that the second derivative of at least one of the curves must be zero. They can also touch each other without crossing, but in both cases their derivatives must be equal in the point where they coincide ( at $t_0$ ). This means if we take the derivatives in the first order example above:

$\frac{d^2 \alpha}{dt^2}+\frac{d }{dt}a(t) \alpha(t) + a(t)\frac{d }{dt} \alpha(t)=0$ and $\frac{d^2 \beta}{dt^2}+\frac{d }{dt}a(t) \beta(t) +a(t) \frac{d }{dt}\beta(t)=0$

So at $t=t_0$ we must have : $\frac{d^2 \alpha}{dt^2}|_{t=t_0}=\frac{d^2 \beta}{dt^2}|_{t=t_0} $ . We can now continue this process for all higher derivatives and conclude that all higher derivatives of $\alpha$ and $\beta$ must be equal at $t=t_0$ .

This of course means that $\alpha$ and $\beta$ must be the same curves.

This last argument must be used to replace the first of course ! Whenever two curves have equal boundary conditions at some point, the two curves must be entirely the same because all their derivatives are the same at $t_0$ !

That implies they must be uniquely determined by their boundary conditions.

Wronskian tells us whether all possible boundary conditions can be satisfied. If that is the case we know we have the general solution.

  • $\begingroup$ Your "of course" statement about inferring that you have the same function by virtue of having all the same derivatives at one point is actually not true. For example, $f(x)=0$ and $f(x)=\begin{cases} e^{-1/x} & x>0 \\ 0 & x=0 \end{cases}$ have the same derivatives at $x=0$ but are not the same function. Instead to ensure that the general solution is actually general, you need to resort to Picard-Lindelof, which is straightforward to apply for a linear equation with leading coefficient $1$ and other coefficients bounded. $\endgroup$
    – Ian
    Commented Mar 9, 2023 at 18:02

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