Differential Equation: Is it appropriate?

[Saladsamurai]

Homework Statement

y''' + 2y'' - 3y' = 0


I solved the above using the following, where D is the differential operator:

D(D - 1)(D + 3)y = 0

Let u(x) = (D + 3)y

Hence,

(D² - D)[u(x)] = 0. This 2nd Order in u(x), linear, homogenous with constant coefficients. Hence the solution given by characteristic equation is (from lambda = {0,1}):

u(x) = C1 + C2*exp(x)

Plugging this result back into the factorized DE:

(D + 3)y = C1 + C2*exp(x)

y' + 3y = C1 + C2*exp(x)

which can be solved by integration factor:

y(x) = (1/3)*C1 + (1/4)C2*exp(x) + C3*exp(-3x)

Checking with Wolfraalpha gives:

y(x) = (-1/3)*C1*exp(-3 x)+C2*exp(x)+C3

which looks similar to mine, but is a little off. Namely the negative sign and the fact that the factor of (1/3) is associated with an exponential and my factor of (1/3) is associated with a constant only.

Any thoughts on this?

Also: Would it be appropriate to simply write (1/3)*C1 as just plain old C1? that is, can't I just "absorb" all of the numerical coefficients into the parametric coefficients?

thanks!

 

[rock.freak667]

If wikipedia solved it the same way, with the D operator, then something went wrong. You can however, rewrite constants such as 2c as C₁ instead to make things easier to read and write.

 

[Saladsamurai]

If wikipedia solved it the same way, with the D operator, then something went wrong. You can however, rewrite constants such as 2c as C₁ instead to make things easier to read and write.

I'm sorry, but could you clarify? I have no idea what algorithm Wolfram uses to solve it's equations. Are you thinking that I screwed up along the way? I think that's what you are saying.

 

[rock.freak667]

I'm sorry, but could you clarify? I have no idea what algorithm Wolfram uses to solve it's equations. Are you thinking that I screwed up along the way? I think that's what you are saying.

No no, what I am saying is that you said Wolfram's answer had (1/3)C₁e^-3x and your answer has (1/3)C₁e^-3x

w.r.t Wolfram's answer, you can just write -C₁ as another constant, say 'A' giving (1/3)Ae^-3x.

Which is in the same form as yours.

 

[jambaugh]

You can rename your arbitrary constants and absorb constant multipliers into them.
A exp(-3x) + B exp(x) + C

All are equivalent but you may get different forms with those extra multipliers depending on your order of integration. The software packages are not always smart enough to absorb multiplier of arbitrary constants. Once they pick the constant its just another variable.

 

[Mark44]

You really shouldn't have those fractional constants together with the subscripted constants C₁, C₂, and C₃. They're not wrong, but they don't add anything useful.

Your solution should look something like this: y = C₁ + C₂e^t + C₃e^-3t.

Also, you can go directly from the characteristic equation (here in factored form), r(r - 1)(r + 3) = 0, to your linearly independent solutions.
y₁ = e^0t = 1
y₂ = e^t
y₃ = e^-3t

The general solution of your differential equation is the set of all linear combinations of your three linearly independent solutions; namely y = C₁ + C₂e^t + C₃e^-3t.

If you have initial conditions, you can solve for the three constants to get a unique solution.

 

[Saladsamurai]

<snip> ... and the fact that the factor of (1/3) is associated with an exponential and my factor of (1/3) is associated with a constant only.

No no, what I am saying is that you said Wolfram's answer had (1/3)C₁e^-3x and your answer has (1/3)C₁e^-3x

w.r.t Wolfram's answer, you can just write -C₁ as another constant, say 'A' giving (1/3)Ae^-3x.

Which is in the same form as yours.

Hi rock.freak Actually, the problem was that my (1/3)C1 is not attached to an exponential at all. However, Wolfram's is.

Maybe I should just stop checking my answers with Wolfram and do it the old fashioned way. I thought it would be a way to speed things up a bit, but it seems to be raising more questions than it answers.

Thanks again!

 

[jambaugh]

Hi rock.freak Actually, the problem was that my (1/3)C1 is not attached to an exponential at all. However, Wolfram's is. [...]

That's only because you and Wolfram chose different names, permute Wolfram's C1, C2, and C3 names.