I was reading about Second Order Linear DE's and I became confused. According to the existence and uniqueness theorem, an initial value problem has only one unique solution pg(146). However, the text later says that a given differential equation has more than one fundamental set of solutions (top of pg 153). I thought that the fundamental set of solutions was the unique solution. Could someone please clarify this for me. Thanks

My reference text is Elementary Differential Equations and Boundary Value Problems(Boyce & DiPrima) 9th edition

  • $\begingroup$ I have that book in my library. Could you please give a more exact point in the text where this statement is. I think I (and others) will have a better chance to give a satisfying answer that way. $\endgroup$
    – mickep
    Commented Apr 12, 2017 at 7:14

2 Answers 2


Consider $y' = y$.

$y_1(t) = e^t$ is certainly a solution. So is $y(t) = 2 e^t$ or $y(t) = 15 e^t$. In fact, $y(t) = c e^t$ is a solution for any value of $c$. Once you add an initial condition to the problem, then there is a unique solution. But the differential equation by itself admits an infinite family of solutions.

Since all of the solutions I wrote down are the same up to a scalar factor (i.e. linearly dependent), we only need to specify one fundamental solution $y_1(t) \not \equiv 0$, giving the fundamental solution set $\{y_1(t) \}$. Then we can recover all possible solutions by taking linear combinations: $$y(t) = c_1 y_1(t).$$

Moving to a second order differential equation, consider $$y'' - y =0.$$ The following are all solutions (i.e., satisfy the differential equation):

  • $y(t) = e^t$
  • $y(t) = e^{-t}$
  • $y(t) = \cosh(t)$
  • $y(t) = \sinh(t)$
  • $y(t) = e^{t} - e^{-t} + 2 \cosh(t) + 3 \sinh(t)$
  • Infinitely many more possibilities omitted for brevity.

Since we want to write all possible solutions of the differential equation, we note that we can express $\sinh$ and $\cosh$ in terms of exponentials: $$\cosh(t) = \frac{e^{t} + e^{-t}}{2} \qquad \sinh(t) = \frac{e^t - e^{-t}}{2}.$$

Thus we can express any answer involving $\sinh$ and $\cosh$ strictly in terms of exponentials: $$y(t) = c_1 e^t + c_2 e^{-t}.$$ However, we could have also just used $\sinh$ and $\cosh$ instead, as we can write $e^t$ and $e^{-t}$ as linear combinations of $\sinh(t)$ and $\cosh(t)$. Thus we could have written that all solutions take the form of $$y(t) = c_1 \sinh(t) + c_2 \cosh (t).$$

Thus we have many different fundamental solution sets:

  • $\{ y_1(t) = e^t, ~y_2(t) = e^{-t} \}$
  • $\{ y_1(t) = \sinh(t), ~ y_2(t) = \cosh(t) \}$
  • $\{ y_1(t) = 14 e^t, ~y_2(t) = -6 \cosh(t) \}$
  • Many many more. All of these will generate all possible solutions to the differential equation upon taking linear combinations.

However, the following is NOT a fundamental solution set: $$\{ y_1(t) = e^t, ~ y_2(t) = e^{-t}, ~ y_3(t) = \sin(t), ~y_4(t) = \cosh(t) \}$$ as the functions given form a linearly dependent set. However, any pair of these functions are linearly independent, so we can form a fundamental solution set by taking exactly two of the functions from this set.

If you've had some exposure to linear algebra, these notions precisely correspond to finding a basis for a vector space. Recall that a basis must satisfy these two properties:

  1. Spanning: anything in the vector (sub-)space can be written as a linear combination of the basis vectors.
  2. The basis vectors are linearly independent: $\sum a_i \mathbf{b_i} = 0 \iff a_i = 0$ for all $i$.

However, we now have a vector space of all functions that are second-differentiable and we want a basis for the subspace of all solutions to the linear differential equation.


For a first order ODE, an initial value problem consists of the ODE and one initial condition y(x0) = y0. Once the arbitrary constant of the general solution is determined, you have a unique solution. This solution is called a "particular solution". For the second order equations, the idea is similar.

So we have a second order ODE: (1) y'' + p(x)y' + q(x)y = 0

We have a general solution, (2) y = c1y1 + c2y2.

Then we have the theorem: A general solution of an ODE (1) on an open interval I is a solution (2) in which y1 and y2 are solutions of (1) on I that are not proportional, and c1 and c2 are arbitrary constants. These y1, y2 are called a basis of solutions of (1). A particular solution of (1) on I is obtained if we assign specific values to c1 and c2 in (2).

c1 and c2 sometimes must be restricted on some interval to avoid complex expressions.

More, a basis of solutions of (1) on an open interval I is a pair of linearly independent solutions of (1) on I.

To be linearly independent, k1y1(x) + k2y2(x) can only equal zero if both k1 = 0 and k2 = 0.

I think what confuses you is the difference between "the" unique solution and "a" unique solution.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .