Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I understand that if $y_1$ and $y_2$ are solutions, then $Ay_1+By_2$ is also a solution, and I understand by $A$ and $B$ can take any number. But why are there only 2 things (here $y_1$ and $y_2$, I encountered the word 'degrees of freedom' alluding to the same thing in Feynman's lectures), and we can be sure that there is not a $y_3$, that is not a multiple of $y_2$ of $y_2$, hiding away?

I assume that this problem generalises to having n independent solutions for an nth order linear homogeneous equation.

If it helps in the writing of your answers, am very much a beginner in differential equations.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

What is important here is that $y_1$ and $y_2$ are independent of each other. By independence it is meant that one cannot be equal to a constant multiple of the other. It can be proved that this condition (together with $y_1, y_2$ being solutions) is equivalent to showing that the Wronskian $y_1y_2'-y_2y_1'$ is non zero identically in the domain under consideration.

Once this is guaranteed it is easy to prove that any other solution $y$ can be formed by combining $y_1$ and $y_2$. To prove this, let $x_0$ be a point in the domain of $y$ and consider the system of equations in the variables $A$ and $B$:

$Ay_1(x_0) + By_2(x_0)=y(x_0)$

$Ay_1'(x_0) + By_2'(x_0)=y'(x_0)$

This system has a nontrivial solution provided $y_1y_2'-y_2y_1'$ is non zero which it is and so we can find a $A$ and B satisfying the given equations. Further since $y$ and $Ay_1+By_2$ are both solutions behaving identically at $x_0$ (and likewise their derivatives) and by a uniqueness theorem there can be only one, so $y=Ay_1+By_2$. In this way all other solutions are combinations of $y_1$ and $y_2$.

(For the proofs on the result on the Wronskian and on uniqueness I suggest you consult your text book.)

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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