Polynomial higher order DE in D notation form.

[Hercuflea]

Homework Statement

My teacher likes to teach us the D-notation methods for higher order DE's. I am having a hard time with this one and I can't seem to find the formula for the general solution

Find a fundamental set of the equation (D-1)$^{2}$(D$^{2}$-6D+13)$^{3}$y = 0

Homework Equations

Trying to figure this out. Split the formula into two equations, the fundamental set of each of these will be part of the fundamental set of the entire equation.

The Attempt at a Solution

Case 1: (D-1)$^{3}$y = 0

Let y = e$^{t}$z

(D-1)$^{3}$e$^{t}$z = 0

e$^{t}$(D-1+1)$^{3}$z = 0 Divide by e^t and add 1 inside parentheses.

D$^{3}$z = 0

z = c1+c2t + c3t$^{2}$

y1 = [e$^{t}$, te$^{t}$, t$^{2}$e$^{t}$]



Case 2:

(D$^{2}$-6D+13)$^{3}$y = 0

(D$^{2}$-6D+9+4)$^{3}$y = 0

((D-3)$^{2}$-4i$^{2}$)$^{3}$y = 0

((D+(-3-2i))$^{3}$(D+(-3+2i))$^{3}$y = 0

Now what? I have my roots in the form alpha + beta(i), but I don't know the equation to find the fundamental set.

 

[Mark44]

Homework Statement

My teacher likes to teach us the D-notation methods for higher order DE's. I am having a hard time with this one and I can't seem to find the formula for the general solution

Find a fundamental set of the equation (D-1)$^{2}$(D$^{2}$-6D+13)$^{3}$y = 0

Homework Equations

Trying to figure this out. Split the formula into two equations, the fundamental set of each of these will be part of the fundamental set of the entire equation.

The Attempt at a Solution

Case 1: (D-1)$^{3}$y = 0

Let y = e$^{t}$z

(D-1)$^{3}$e$^{t}$z = 0

e$^{t}$(D-1+1)$^{3}$z = 0 Divide by e^t and add 1 inside parentheses.

No, you can't do this (add 1 inside the parentheses).

I'm not sure you understand what the notation means. For example, (D -1)y means dy/dt - y. D means differentiation with respect to the independent variable (d/dt) and 1 means the identity operation.

(D - 1)³z is the same as D³z - 3D²z + 3Dz - z.

D$^{3}$z = 0

z = c1+c2t + c3t$^{2}$

y1 = [e$^{t}$, te$^{t}$, t$^{2}$e$^{t}$]



Case 2:

(D$^{2}$-6D+13)$^{3}$y = 0

(D$^{2}$-6D+9+4)$^{3}$y = 0

((D-3)$^{2}$-4i$^{2}$)$^{3}$y = 0

((D+(-3-2i))$^{3}$(D+(-3+2i))$^{3}$y = 0

Now what? I have my roots in the form alpha + beta(i), but I don't know the equation to find the fundamental set.

 

[Hercuflea]

No, you can't do this (add 1 inside the parentheses).

I'm not sure you understand what the notation means. For example, (D -1)y means dy/dt - y. D means differentiation with respect to the independent variable (d/dt) and 1 means the identity operation.

(D - 1)³z is the same as D³z - 3D²z + 3Dz - z.

My teacher taught us that (unless i am misunderstanding) when we have (D-r)$^{k}$e$^{rt}$y = 0, you can put the e$^{rt}$ on the left side, and by doing this you also have to add (or subtract if the original r has a positive sign) r inside the parentheses to isolate D.

 

[HallsofIvy]

Homework Statement

My teacher likes to teach us the D-notation methods for higher order DE's. I am having a hard time with this one and I can't seem to find the formula for the general solution

Find a fundamental set of the equation (D-1)$^{2}$(D$^{2}$-6D+13)$^{3}$y = 0

Homework Equations

Trying to figure this out. Split the formula into two equations, the fundamental set of each of these will be part of the fundamental set of the entire equation.

I wouldn't consider it necessary to do that. Just note that the characteristic equation is
$(r- 1)^3(r^2- 6r+ 13)= 0$, just the equation you have written differently. Some people would just use "D" instead of "r" as a parameter rather than as a differential operator.
Now, of course, you use the fact that if the product of two numbers is 0 at least one must be 0 to split the equation as you have.

The Attempt at a Solution

Case 1: (D-1)$^{3}$y = 0

Let y = e$^{t}$z

(D-1)$^{3}$e$^{t}$z = 0

e$^{t}$(D-1+1)$^{3}$z = 0 Divide by e^t and add 1 inside parentheses.

D$^{3}$z = 0

z = c1+c2t + c3t$^{2}$

y1 = [e$^{t}$, te$^{t}$, t$^{2}$e$^{t}$

With a little practice, you can determine that solution immediately without going through all of that.

Case 2:

(D$^{2}$-6D+13)$^{3}$y = 0

(D$^{2}$-6D+9+4)$^{3}$y = 0

((D-3)$^{2}$-4i$^{2}$)$^{3}$y = 0

Yes, this factor of the characteristic equation has the complex conjuate solutions 3- 2i and 3+ 2i.

((D+(-3-2i))$^{3}$(D+(-3+2i))$^{3}$y = 0

Now what? I have my roots in the form alpha + beta(i), but I don't know the equation to find the fundamental set.

If r is a solution to the characteristic equation then $e^{rt}$ is a solution to the differential equation. And, of course, $e^{(a+ bi)t}= e^{at}e^{ibt}$ and now you can use the fact that $e^{ibt}= cos(bt) + i sin(bt)$. What that reduces to is that if $a\pm bi$ are solutions to the characteristic equation, then $e^{at}cos(bt)$ and $e^{at}sin(bt)$ are independent solutions to the differential equation. Here, the general solution is $Ae^t+ Bte^t+ Ct^2e^t+ De^{3t}cos(2t)+ Ee^{3t}sin(2t)$ or, equivalently, $Ae^t+ Bte^ti+ Ct^2e^t+ e^{3t}(Dcos(2t)+ Esin(2t)$.

 

[Hercuflea]

If r is a solution to the characteristic equation then $e^{rt}$ is a solution to the differential equation. And, of course, $e^{(a+ bi)t}= e^{at}e^{ibt}$ and now you can use the fact that $e^{ibt}= cos(bt) + i sin(bt)$. What that reduces to is that if $a\pm bi$ are solutions to the characteristic equation, then $e^{at}cos(bt)$ and $e^{at}sin(bt)$ are independent solutions to the differential equation. Here, the general solution is $Ae^t+ Bte^t+ Ct^2e^t+ De^{3t}cos(2t)+ Ee^{3t}sin(2t)$ or, equivalently, $Ae^t+ Bte^ti+ Ct^2e^t+ e^{3t}(Dcos(2t)+ Esin(2t)$.

Ok, so my professor was saying that when you have conjugate solutions to the characteristic equation, you only need to include one of them in the general solution. I guess you applied this identity here?

 

[Mark44]

No, that's not what HallsOfIvy was saying. He said that if a ± ib were solutions to the characteristic equation, then e^at(cos(bt) + sin(bt)) would be solutions to the differential equation. There wasn't anything about discarding one of the conjugates.

 

[HallsofIvy]

It is true that since the conjugate of a+ bi is a- bi, all the information about both (the values of a and b) is contained in either one. However, you need both $e^{(a+bi)t}= e^{at}e^{bti}= e^{at}(cos(bt)+ i sin(bt))$ and $e^{(a- bi)t}= a^{at}(cos(bt)- isin(bt))$ to be able to eliminate the "i" and write the general solution as $e^{at}(Ccos(bt)+ Dsin(bt))$ with both C and D real constants.

(Of course, if your differential equation had non-real coefficients, then so would the characteristic equation so its solutions would NOT necessarily come in conjugate pairs and, in that case, it would be best to leave the solutions in the form $e^{(a+bi)t}$.

 

[Hercuflea]

Ok thanks guys that really cleared some things up!