2nd-order Nonhomogeneous Differential Equation

[Elmer Correa]

Homework Statement

Finding the general solution:
y”+4y’+4y=t*e^(-2t)

Homework Equations

The Attempt at a Solution

So I got the complementary solution pretty easily as y= c1*e^(-2t)+c2*te^(-2t)
I haven’t been able to find a particular solution using the method of undetermined coefficients. I don’t understand why this is since I was taught that any sum or product of exponential functions, sines, cosines, and polynomials for the nonzero term. Yet the form e^(-2t)*(At+B) doesn’t work in this case according to online calculator.
Any help explaining what I’m doing wrong would Ben much appreciated.

 

[Ray Vickson]

Homework Statement

Finding the general solution:
y”+4y’+4y=t*e^(-2t)

Homework Equations

The Attempt at a Solution

So I got the complementary solution pretty easily as y= c1*e^(-2t)+c2*te^(-2t)
I haven’t been able to find a particular solution using the method of undetermined coefficients. I don’t understand why this is since I was taught that any sum or product of exponential functions, sines, cosines, and polynomials for the nonzero term. Yet the form e^(-2t)*(At+B) doesn’t work in this case according to online calculator.
Any help explaining what I’m doing wrong would Ben much appreciated.

Nobody can help you figure out what you are doing wrong unless you show us what you actually tried.

Anyway, as an alternative to the method of undetermined coefficients you might try the method of variation of parameters. See, eg.,
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

 

[ehild]

Finding a particular solution with undetermined coefficient:

 

[Elmer Correa]

Nobody can help you figure out what you are doing wrong unless you show us what you actually tried.

Anyway, as an alternative to the method of undetermined coefficients you might try the method of variation of parameters. See, eg.,
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

Fair enough.
Starting with my assumed form of a particular equation:
y= e^(-2t)*(At+B)
y'=-2e^(-2t)(At+B)+A*e^(-2t)
y''=4e^(-2t)(At+B)-4Ae^(-2t)

Plugging this all back into the original differential equation:
e^(-2t)*(4At+4B-4A-8At-8B+4A+4At+4B)=e^(-2t)*t
4At+4B-4A-8At-8B+4A+4At+4B=t+0
The issue is that I can't equate like terms because everything cancels out, something I was told should never happen if the right side of a second order constant coefficient nonhomogeneous equation is in the form here. What went wrong? Am I making an arithmetic error or is one of my assumptions wrong?

PS: I got the correct answer using variation of parameters, what's frustrating me is that I shouldn't have to use it in this case and I don't understand why I seemingly do have to.

 

[ehild]

Fair enough.
Starting with my assumed form of a particular equation:
y= e^(-2t)*(At+B)
The issue is that I can't equate like terms because everything cancels out, something I was told should never happen if the right side of a second order constant coefficient nonhomogeneous equation is in the form here. What went wrong? Am I making an arithmetic error or are one of my assumptions wrong?

.

Your assumed particular solution is the general solution of the homogeneous equation, no wonder that it resulted in zero when substituted into the left hand side of the differential equation.
According to Post #3 (iii) use y=t²(At+B)e^-2t as particular solution.

 

[Elmer Correa]

Your assumed particular solution is the general solution of the homogeneous equation, no wonder that it resulted in zero when substituted into the left hand side of the differential equation.
According to Post #3 (iii) use y=t²(At+B)e^-2t as particular solution.

I'm not quite sure why I didn't look at your original post more closely, thanks.