How Do Multiple Solutions to a Differential Equation Work?

[oneamp]

Something has been bothering me. I see a phrase like this:

Plugging our two roots into the general form of the solution gives the following solutions to the differential equation.

y_1 = ae^xt and y_2 = be^vt

So here we have 2 solutions, I assume, since the text calls them solutions.

It goes on to say, "superposition: c_1 y_1 + c_2 y_2 is also a solution, for any constants"

So not it seems we have 3, or actually infinity if any constant can be used, solutions.

Further, my teacher says that certain DEs can only have one solution. Then he goes on to use these techniques, which define solutions in these different forms!

Please, clear this up for me!

Thank you

 

[SteamKing]

You post is a bit hazy in certain areas.

For ODEs, there are an infinite number of solutions unless a sufficient number of initial conditions are specified. Remember, when we integrate something, there is a constant of integration which is also produced. The constant can be any number. Applying the initial conditions to the solution allows us to determine a specific value of the constant of integration which satisfies the differential equation AND the initial condition.

For linear differential equations only, if you have more than one solution to an equation, then any linear combination of these solutions also satisfies the original DE. If you want to determine the particular values of the constants, then you must apply a sufficient number of initial conditions.

 

[HallsofIvy]

Something has been bothering me. I see a phrase like this:

Plugging our two roots into the general form of the solution gives the following solutions to the differential equation.

y_1 = ae^xt and y_2 = be^vt

So here we have 2 solutions, I assume, since the text calls them solutions.

Your text calls them solutions? Did you not trying to put them and their derivatives into the equation to verify that they are, in fact, solutions?

It goes on to say, "superposition: c_1 y_1 + c_2 y_2 is also a solution, for any constants"

So not it seems we have 3, or actually infinity if any constant can be used, solutions.

Further, my teacher says that certain DEs can only have one solution. Then he goes on to use these techniques, which define solutions in these different forms!

Please, clear this up for me!

Thank you

You should have learned that "the set of all solutions to an n^th order, linear differential equation form an n dimensional vector space". That means, in particular, that if f and g are both solutions so is Cf+ Dg for any constants C and D. And that means that any n^th order, liner differential equation has an infinite number of solutions.

No, your teacher did NOT tell you that "certain DEs can only have one solution". He told you that certain differential equation problems have a unique solution- a "differential equation problem" being a differential equation with additional conditions. An n^th order linear differential has a unique solution satisfying conditions such as the value of the function and it first n-1 derivatives at a given value of x (an "initial value problem") or values of the function at n different values of x. The first is the "Fundamental Existence and Uniqueness Theorem for initial value problems" which is proved in any differential equations textbook.

 

[mesa]

Something has been bothering me. I see a phrase like this:

Plugging our two roots into the general form of the solution gives the following solutions to the differential equation.

y_1 = ae^xt and y_2 = be^vt

So here we have 2 solutions, I assume, since the text calls them solutions.

It goes on to say, "superposition: c_1 y_1 + c_2 y_2 is also a solution, for any constants"

So not it seems we have 3, or actually infinity if any constant can be used, solutions.

Further, my teacher says that certain DEs can only have one solution. Then he goes on to use these techniques, which define solutions in these different forms!

Please, clear this up for me!

Thank you

Superposition gave me a bit of trouble too, perhaps it would be best just to start from the beginning and run through each step and I bet that will help with your confusion (it certainly did for me!)

Your post starts off with the ‘roots’. This part is pretty straight forward, basically what we do is assume that y is some exponential in hopes of making an equation of the form ay’’+by’+cy=0 true so we say,

y=e^(rt) where ‘r’ is an unknown value to make this equation true. From here we can now write out y’ and y’’ as,
y’=re^(rt)
y’’=r^2e^(rt)

Substitiute these back into the differential equation,
ar^2e^(rt)+bre^(rt)+e^(rt)=0

Our e^(rt) can be taken out leaving,
ar^2+br+c=0
Which is just a simple quadratic to solve for our ‘r’ values. If we get two roots we will have two different values for y from each like so,

y_1=e^(r_1t)
and
y_2=e^(r_2t)

Now using the general solution,
y(t)=c_1e^(r_1t)+c_2e^(r_2t)
we take the first two derivatives and plug everything back into our differential equation, after simplifying we get,

e^(r_1t)c_1(ar_1^2+br+c)+e^(r_2t)c_2(ar_2^2+br_2+c)=0

Since we know (ar_1^2+br+c) and (ar_2^2+br_2+c) are both equal to zero and that e^(r_1t) and e^(r_2t) will never be zero that means our c_1 and c_2 can be any value and still satisfy this equation.

Hope this helps!