Two Solutions to an ODE

(New Question of the Week)

This recent question involves an ordinary differential equation (ODE) and the relation between different solutions. It illustrates common difficulties in interpreting what a problem is asking for, as well as some communication problems involving language and notation.

Juares (whose native language is Portuguese) wrote to us a few weeks ago; I have cleaned up some of the translations:

The problem is:

Consider an ODE of the first order, dy/dx + t² y. It is known that y₁(t) and y₂(t) are solutions of this equation. The function y(t) = y₁(t) + y₂(t) is also a solution of this equation?

( ) true   ( ) false

Well, by making dy/y = -t²dx and integrating, solution becomes y = e^{-t3/3 + C₁}. I understand that there is only one solution, that one. And that there is no second solution as a function of t, but with a function of C.

For example y₁ would be y₁ =  e^{-t3/3 + C₁} one solution, the other solutions stem from varying C to C_{2, 3…}

And t varies in range, but which one? Infinite? I do not know more.

y₂ = 1/e^{t3/3 +C₁}, thus y(t) = e^{-t3/3 +C₁} + 1/e^{t3/3 +C₁}.

And since it is ODE of the first degree, it is not possible to say that it has two solutions, only in some cases, in this no. Answer (x) False (?)

Juares correctly solved the equation, but seems to misunderstand what it means to have more than one solution. There is one family of solutions, in which the constant C can vary; all solutions therefore look similar. The question does not imply that there are two unrelated solutions. Probably the main problem here (whether due to language issues or unfamiliarity with the concepts) is in seeing what the question is asking.

Doctor Fenton replied, first correcting some common typing errors. We become so accustomed to calling an independent variable x that we can write that even when there is no x involved; and it is common to omit an equal sign. It was easy enough to see what was intended:

Hi Juares,

You say that this is an ODE,

Consider ODE of the first order dy/dx + t²y.

but you have three variables, and this isn’t an equation, since there is no = sign.  I assume that the ODE is

dy/dt + t²y = 0 ,

so that t is the independent variable, and there is no x in the equation: y is a function of the single variable t.

The question was

Consider ODE of the first order dy/dx + t²y. It is known that y₁(t) and y₂(t) are solutions of this equation. The function y(t) = y₁(t) + y₂(t) is also a solution of this equation?

It isn’t necessary to solve the ODE to answer this question.

Just let y(t) = y₁(t) + y₂(t).  Then what is dy/dt?

By the addition property of differentiation,

dy/dt = d/dt(y₁ + y₂) = dy₁/dt + dy₂/dt .

Since y₁ is a solution of dy/dt + t²y = 0, then dy₁/dt + t²y₁ = 0, or dy₁/dt = -t²y₁.  What do you get if you replace dy₁/dt by -t²y₁ in the equation above, and also do the same thing for y₂, since it is also a solution of the ODE?

In your solution above, you say

Well, by making dy/y = -t²dx and integrating, solution becomes y = e^{-t3/3 + C₁} . I understand that there is only one solution, that one.

Actually, the formula  e^{-t3/3 + C₁} represents infinitely many solutions: each different value of C₁; gives a different solution to the ODE.  You can also write

e^{-t^3/3 + C₁} = e^-t3/3 * e^C₁ = (e^C₁)(e^-t3/3)

and since e^C₁  is just a constant C₂,

= C₂ e^-t3/3,

so that all the solutions of the DE have this same form.  However, for different values of C₂, they are different solutions: y₁ = e^-t3/3 , y₂ = 2 e^-t3/3, and y₃ = 20e^-t3/3 , and so on, are all different solutions to the ODE. If you add two solutions, such as

y₁ + y₂ = e^-t3/3 + 2e^-t3/3 = 3e^-t3/3, 

the sum is another function of the same form, so it is also a solution of the ODE.

To get a unique solution to the ODE, you must add an additional requirement: an initial value.  You specify a value t, t₀, and you require that y(t₀) = y₀ .  That requirement determines the value of C₂, so that only one of the functions of the given form passes through the point specified.

Does this help?

The quick answer to the question does not require solving the equation (and thus can be extended to any equation of the same type, teaching a general principle and perhaps serving as an example for a theorem); but when the actual solution is properly understood, it illustrates the general concept clearly.

What Doctor Fenton has done here is a good model for how students can use textbook exercises to deepen their understanding of a topic: Don’t stop with a quick answer to a problem, but explore what it means in detail. What Juares did bypassed the quick method, but practiced other important skills, and provided a way to illustrate the concepts.

Juares understood; he replied,

You said,

“Since y₁ is a solution of dy/dt + t²y = 0, then dy₁/dt + t^2y₁ = 0, or dy₁/dt = -t^2y₁.  What do you get if you replace dy₁/dt by -t^2y₁ in the equation above, and also do the same thing for y₂, since it is also a solution of the ODE?”

Could this be

(-t² y₁ – t² y₂) + t² (y₁ + y₂) = 0 ,

so

0 = 0,

so it’s true?

He was right:

You are correct that all terms cancel, so (y₁ + y₂) is also a solution of the ODE.

The sum of any two solutions of this equation is also a solution, because it still satisfies the differential equation. Next, you might want to think about what properties of the equation led to this result; when might it not work?