Find the general solution in chemistry?

[Success]

Find the general solution of y"'-y"-y'+y=2e^-t+3.

Here's the work:
r³-r²-r+1=r²(r-1)-(r-1)=(r-1)(r²-1)=(r²-1)²(r+1)
r=1, -1
y=c₁e^t+c₂te^t+c₃e^-t

The answer in the textbook is y=c₁e^t+c₂te^t+c₃e^-t+(1/2)te^-t+3 but I don't know how to get the last 2 terms. Help me...

 

[STEMucator]

Find the general solution of y"'-y"-y'+y=2e^-t+3.

Here's the work:
r³-r²-r+1=r²(r-1)-(r-1)=(r-1)(r²-1)=(r²-1)²(r+1)
r=1, -1
y=c₁e^t+c₂te^t+c₃e^-t

The answer in the textbook is y=c₁e^t+c₂te^t+c₃e^-t+(1/2)te^-t+3 but I don't know how to get the last 2 terms. Help me...

I believe what they have there is the particular solution $y_p$.

 

[Success]

Let me try to find the particular solution and see if it makes sense then.

 

[Boorglar]

Yes, what you've found is the solution to the homogeneous equation (where the right hand side is 0). You need to add the particular solution.

 

[HallsofIvy]

There are two standard ways of finding a "specific solution" to a non-homogeneous linear equation to be added to the general solution to the associated homogeneous equation:
1) "Variation of Parameters". Given independent solutions y1 and y2 to the associated homogeneous equation, look for a specific solution to the entire equation of the form uy1+ vy2, solving for u and v.

2) "Undetermined Coefficients". Guess the correct form for the specific solution up to coefficients that have to be determined.

The second only works if you can guess the correct form and that will be only when the "right hand side" (or "non-homogeous part") of the equation is itself one of the various kinds of functions one expects to get as solutions to a linear homogeneous equation with constant coefficients. The first works for any "right hand side" but is much more difficult computationally.

The " kinds of functions one expects to get as solutions to a linear homogeneous equation with constant coefficients" are (1) exponentials, (2) polynomials, (3) sines and cosines, and (4) combinations of those.

Here, the "right hand side" is 2e^{-t}+ 3 which is of that kind so we ca use "undetermined coefficients". Yes, the general solution to the "associated homogeneous equation" is C_1e^t+ C_2te^{t}+ C_3e^{-t}. For a "right hand side" involving e^{-t} we would normally try Ae^{-t} but that is already a solution of the associated homogeneous equation so we try, instead, Ate^{-t} (just as the fact that "1" is a double root of the characteristic equation gives e^t and te^t as solutions to the associated homogeneous equation).

That is, set y(t)= Ate^{-t}+ B in the equation and solve for A and B.

 

[Success]

HallsofIvy, how do I do that? Can you show complete work?

 

[STEMucator]

HallsofIvy, how do I do that? Can you show complete work?

He's already done most of the work for you by guessing the correct form of the particular solution. All you would have to do at this point is plug it in and solve for the coefficients.

Undetermined coefficients relies on the fact that you can guess the form of the particular solution.

 

[Redbelly98]

That is, set y(t)= Ate^{-t}+ B in the equation and solve for A and B.

HallsofIvy, how do I do that? Can you show complete work?

No, he can't. Or he shouldn't -- since showing the complete work would be against our forum guidelines. To proceed, you should:

1. Substitute the expression HallsofIvy wrote into the differential equation you have.
2. Take derivatives, as necessary, of HallsofIvy's expression.
3. Combine like terms together, and examine them to figure out what A and B are.

If you get stuck, post all the work you were able to do up until the point where you got stuck.

 

[Mark44]

Find the general solution of y"'-y"-y'+y=2e^-t+3.

Here's the work:
r³-r²-r+1=r²(r-1)-(r-1)=(r-1)(r²-1)=(r²-1)²(r+1)
r=1, -1

You have a mistake here that I don't think anyone noticed. The final factorization should be (r - 1)²(r + 1), NOT (r²[/color] - 1)²(r + 1).

y=c₁e^t+c₂te^t+c₃e^-t


The answer in the textbook is y=c₁e^t+c₂te^t+c₃e^-t+(1/2)te^-t+3 but I don't know how to get the last 2 terms. Help me...

 

[Success]

Mark44, yeah, you're right.

 

[Success]

Okay, I've took the derivatives for all of them and I got y'=Ce^(-t)(-t+1), y"=Ce^(-t)(t-2) and y'''=Ce^(-t)(-t+3). I combined like terms and got Ce^(-t)(-t+3-t+2+t-1+t)=2e^-t, C=1/2, so yeah, the answer makes sense. But can anyone tell me how to get t in (-t+3-t+2+t-1+t)? The last term.