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
    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.


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