Differential Equations: Stuck on one step of H.L.D.E problem

[Jeff12341234]

I need to find a,b,c,d, and e. I know how to do these problems the normal way but now he's giving us the answer and asking us to work backwards. I'm stumped. I think I night need to use some sord of system of equations but I'm not sure what it would look like..

 

[rude man]

You need to perform y', y'', y''' and y^iv from the given solution, substitute into the given diff. eq. then equate coefficients of like terms (in x) and solve for a, b, c and d.

 

[Jeff12341234]

I did that and now I have too many variables. I have a,b,c,d,e,x,c1,c2,c3, & c4

 

[rude man]

For y', don't collect terms in e^8x.
y' = (8c₂ + c₄)e^8x - 8c₃e^-8x + 8c₄xe^8x
all of which is multiplied by d.
Etc. for y, y'', etc.

You equate coefficients of every separate term in x. So for example x*exp(x) and exp(x) are two separate terms with their own separate coefficients. Coefficients can have only the constants a, b,c,d, and e. No x's.

 

[Jeff12341234]

How's that going to work though when you get down to creating a system of equations? The c1,c2,c3,c4 constants make the system unsolvable don't they?

 

[rude man]

How's that going to work though when you get down to creating a system of equations? The c1,c2,c3,c4 constants make the system unsolvable don't they?

Why? They are given constants.

 

[HallsofIvy]

The fact that y satisfyies the equation for any values lf c_1, c_2, etc. means that a, b, c, d, and e must be such that they cancel. Write the equation with some expression in a, b, c, d, and e times each constant and set those expressions equal to 0.

 

[Jeff12341234]

So is this right?

 

[Jeff12341234]

Additionally, I can distribute the a,b,c,d,&e to all of the terms and match them all up based on their x function but that only gives me 4 equations (one for xe^8x, one for e^8x, e^-8x, and Ec₁=0). I need 5 to find the 5 variables a,b,c,d,&e. What am I missing?

 

[Jeff12341234]

The system of equations I got is:

E+8D+64C+512B+4096A=0 <--- after dividing out xe^8xc₄
E+8D+64C+512B+4096A=0 <--- after dividing e^8xc₂
E-8D+64C-512B+4096A=0 <--- after dividing e^-8xc₃
E=0 <--- after dividing c₁
D+16c+192B+2048A=0 <--- after dividing e^8xc₄

but it still doesn't solve completely because the first two equations are the same..

you get: A=c/512, B=-c/64, C=-c/8, D=c, E=0
If you just guess that the constant c = 1 then it's right but that's not a good system..

 

[Mark44]

A better approach is to recognize that the characteristic equation is r(r - 8)²(r + 8) = 0. Expand this and you should be able to quickly determine the constants of your differential equation.

 

[rude man]

They can't be the same.

You have 5 independent equations: y,y',y'',y''' and y''''. And 5 unknowns a,b,c,d and e.

It also looks like you have not segregated the exp(x) and the xexp(x) terms as I pointed out you should. And how did you wind up with an exp(16x) in your 6th line from the bottom on post 8? Not likely in a linear diff. eq.!

Let's take a simpler example:
y = c1 + c2exp(x)+ c3xexp(x) is the solution to
ay' + by + c = 0.
Then y' = c2exp(x) + c3exp(x) + c3xexp(x)

So substitute ino the diff eq:
a(c2 + c3) exp(x) + ac3 xexp(x)] + bc1 + bc2 exp(x) + bc3 xexp(x)] + c = 0

from which we group coefficients of like terms in x including x⁰:

a(c2 + c3) + bc2 = 0 groups the exp(x) coeff.
ac3 + bc3 = 0 groups the x exp(x) coeff.
bc1 + c = 0 groups the exp(0) coeff.

& solve for a, b and c.
See now why you should not have included an x in your coefficients?

 

[Jeff12341234]

A better approach is to recognize that the characteristic equation is r(r - 8)²(r + 8) = 0. Expand this and you should be able to quickly determine the constants of your differential equation.

When you do that you get:
r⁴-8r³-64r²+512x
What would the systematic next step be because the real answer seems to be A=1/512, B=-1/64, C=-1/8, D=1, E=0

 

[Mark44]

When you do that you get:
r⁴-8r³-64r²+512x

You get r⁴ - 8r³ - 64r² + 512r = 0, so the DE would be y⁽⁴⁾ - 8y''' - 64y'' + 512y' + 0y = 0.

[STRIKE]Now that the coefficients of the DE are known, you can find the coefficients of the solution.[/STRIKE]

What would the systematic next step be because the real answer seems to be A=1/512, B=-1/64, C=-1/8, D=1, E=0

 

[Jeff12341234]

They can't be the same.

You have 5 independent equations: y,y',y'',y''' and y''''. And 5 unknowns a,b,c,d and e.

It also looks like you have not segregated the exp(x) and the xexp(x) terms as I pointed out you should. And how did you wind up with an exp(16x) in your 6th line from the bottom on post 8? Not likely in a linear diff. eq.!

Let's take a simpler example:
y = c1 + c2exp(x)+ c3xexp(x) is the solution to
ay' + by + c = 0.
Then y' = c2exp(x) + c3exp(x) + c3xexp(x)

So substitute ino the diff eq:
a(c2 + c3) exp(x) + ac3 xexp(x)] + bc1 + bc2 exp(x) + bc3 xexp(x)] + c = 0

from which we group coefficients of like terms in x including x⁰:

a(c2 + c3) + bc2 = 0 groups the exp(x) coeff.
ac3 + bc3 = 0 groups the x exp(x) coeff.
bc1 + c = 0 groups the exp(0) coeff.

& solve for a, b and c.
See now why you should not have included an x in your coefficients?

I've gone over it several times now. I'm following the same steps as you can see in the work below but I don't see *specifically* where I went wrong. I ultimately get:
factoring out c₄xe^8x --> 4096A+512B+64C+8D+E=0
factoring out c₄e^8x --> 2048A+192B+16C+D=0
factoring out c₂e^8x --> 4096A+512B+64C+8D+E=0
factoring out c₃e^-8x --> 4096A-512B+64C-8D+E=0
factoring out c₁ --> E=0

 

[Jeff12341234]

You get r⁴ - 8r³ - 64r² + 512r = 0, so the DE would be y⁽⁴⁾ - 8y''' - 64y'' + 512y' + 0y = 0.

Now that the coefficients of the DE are known, you can find the coefficients of the solution.

Wait, I don't follow. The question asks for a,b,c,d,&e of the DE. If you're saying that A=1,B=-8,C=-64, D=512, and E=0 then I'm done right? If that's true, the other solutions I found for a,b,c,d,&e also work so are they both right?

Additionally, I was under the impression that the constants for the solution can not be found without boundary values yet you're implying that they can be.

 

[Mark44]

Wait, I don't follow. The question asks for a,b,c,d,&e of the DE. If you're saying that A=1,B=-8,C=-64, D=512, and E=0 then I'm done right?

Yes - that's all they're asking for.

If that's true, the other solutions I found for a,b,c,d,&e also work so are they both right?

I think so. I haven't checked closely, but your constants seem to be a multiple of the ones I found.

Edit: Yes, my numbers are 512 times yours.

Additionally, I was under the impression that the constants for the solution can not be found without boundary values yet you're implying that they can be.

The constants of the solution (c₁, etc.) can't be found without initial conditions. I erred in saying that in my previous post, so I will take it out.