General Solution of 2nd Order Nonhomogeneous using undetermined coefficients.

[rhm95]

I understand that the general solution is the addition of the homogeneous solution to a particular solution of the non-homogeneous equation. My textbook threw me a curve-ball on a problem that is both unexplained and has no examples.

The given problem is of the form y''+y'+y=g(t): y''-2y'-3y=-5t*e^(-t). I am unable to find the particular solution of the non-homogeneous solution.

I assume g(t) is composed of e^-t and the linear equation At+B. so y=e^-t (At+B)
y'=Ae^-t -e^-t(At+B) = e^-t (A-At-B)
y''=e^-t(-A) - e^-t(A-At-B)= e^-t(At-2A+B)

plug back into og equation

e^-t(At-2A+B) - 2e^-t(A-At-B) - 3e^-t(At+B) = -5te^-t

e^-t cancels from equation

(At-2A+B) - 2(A-At-B) - 3(At+B) = -5t

t(A+2A-3A) + (-2A+B-2A+2B-3B) = -5t

-4A = -5t

A=1.25t

This is what threw me off. Firstly B disappeared so I though it was unable to be solved for. I also have never gotten a function as the solution for a undetermined coefficient. I rolled with it though, and tried plugging back into original equation and deriving again to solve for B.

y=1.25t²e^-t + Be^-t = e^-t(1.25t²+B)

y'=2.5te^-t - e^-t(1.25t²+B)=e^-t(2.5t-1.25t²-B)

y''= (2.5-2.5t)e^-t - e^-t(2.5t-1.25t²-B)=e^-t(2.5-2.5t-2.5t+1.25t²+B)=e^-t(1.25t²-5t+2.5+B)

plug back into original equation

e^-t(1.25t²-5t+2.5+B) - 2e^-t(2.5t-1.25t²-B) - 3e^-t(1.25t²+B) =-5te^-t

e^-t divides out

(1.25t²-5t+2.5+B) - 2(2.5t-1.25t²-B) - 3(1.25t²+B) = -5t

t²(1.25+2.5-3.75)+t(-5-5)+(2.5+B+2B-3B)=-5t

The B's cancel out and I can't solve the equation. I have a feeling I am using the wrong equation for g(t). Can anyone give me some suggestions?

 

[rhm95]

2nd Order Non-homogeneous equation solution using undetermined coefficients

I understand that the general solution is the addition of the homogeneous solution to a particular solution of the non-homogeneous equation. My textbook threw me a curve-ball on a problem that is both unexplained and has no examples.

The given problem is of the form y''+y'+y=g(t): y''-2y'-3y=-5t*e^(-t). I am unable to find the particular solution of the non-homogeneous solution.

I assume g(t) is composed of e^-t and the linear equation At+B. so y=e^-t (At+B)
y'=Ae^-t -e^-t(At+B) = e^-t (A-At-B)
y''=e^-t(-A) - e^-t(A-At-B)= e^-t(At-2A+B)

plug back into og equation

e^-t(At-2A+B) - 2e^-t(A-At-B) - 3e^-t(At+B) = -5te^-t

e^-t cancels from equation

(At-2A+B) - 2(A-At-B) - 3(At+B) = -5t

t(A+2A-3A) + (-2A+B-2A+2B-3B) = -5t

-4A = -5t

A=1.25t

This is what threw me off. Firstly B disappeared so I though it was unable to be solved for. I also have never gotten a function as the solution for a undetermined coefficient. I rolled with it though, and tried plugging back into original equation and deriving again to solve for B.

y=1.25t²e^-t + Be^-t = e^-t(1.25t²+B)

y'=2.5te^-t - e^-t(1.25t²+B)=e^-t(2.5t-1.25t²-B)

y''= (2.5-2.5t)e^-t - e^-t(2.5t-1.25t²-B)=e^-t(2.5-2.5t-2.5t+1.25t²+B)=e^-t(1.25t²-5t+2.5+B)

plug back into original equation

e^-t(1.25t²-5t+2.5+B) - 2e^-t(2.5t-1.25t²-B) - 3e^-t(1.25t²+B) =-5te^-t

e^-t divides out

(1.25t²-5t+2.5+B) - 2(2.5t-1.25t²-B) - 3(1.25t²+B) = -5t

t²(1.25+2.5-3.75)+t(-5-5)+(2.5+B+2B-3B)=-5t

The B's cancel out and I can't solve the equation. I have a feeling I using the wrong equation for g(t). Can anyone give me some suggestions?

 

[rhm95]

I also tried y=(At^2+Bt+C)e^-t, and found values for A and B, but C cancels out and becomes non-solvable as well.

 

[HallsofIvy]

I understand that the general solution is the addition of the homogeneous solution to a particular solution of the non-homogeneous equation. My textbook threw me a curve-ball on a problem that is both unexplained and has no examples.

The given problem is of the form y''+y'+y=g(t): y''-2y'-3y=-5t*e^(-t). I am unable to find the particular solution of the non-homogeneous solution.

Well, no, the equation "y''- 2y'- 3y= -5t e^(-t)" is not "of the form y''+ y'+ y= g(t)"! Perhaps you meant "of the form y''+ ay'+ by= g(t)" for constants a and b (here a= -2 and b= -3.

I assume g(t) is composed of e^-t and the linear equation At+B. so y=e^-t (At+B)

No. You did not first solve the associated homogeneous equation, y''- 2y'- 3y= 0. That has characteristic equation $\displaystyle r^2- 2r- 3= (r+ 1)(r- 3)= 0$ and characteristic roots r= -1 and r= 3. The general solution to the associated homogeneous equation is $\displaystyle y= Ce^{-t}+ De^{3t}$. Since $\displaystyle e^{-t}$ is already a solution to the associated homogeneous equation, you need to try $\displaystyle y= e^{-t}(At^2+ Bt)$, multiplying what you have by t.

y'=Ae^-t -e^-t(At+B) = e^-t (A-At-B)
y''=e^-t(-A) - e^-t(A-At-B)= e^-t(At-2A+B)

plug back into og equation

e^-t(At-2A+B) - 2e^-t(A-At-B) - 3e^-t(At+B) = -5te^-t

e^-t cancels from equation

(At-2A+B) - 2(A-At-B) - 3(At+B) = -5t

t(A+2A-3A) + (-2A+B-2A+2B-3B) = -5t

-4A = -5t

A=1.25t

This is what threw me off. Firstly B disappeared so I though it was unable to be solved for. I also have never gotten a function as the solution for a undetermined coefficient. I rolled with it though, and tried plugging back into original equation and deriving again to solve for B.

y=1.25t²e^-t + Be^-t = e^-t(1.25t²+B)

y'=2.5te^-t - e^-t(1.25t²+B)=e^-t(2.5t-1.25t²-B)

y''= (2.5-2.5t)e^-t - e^-t(2.5t-1.25t²-B)=e^-t(2.5-2.5t-2.5t+1.25t²+B)=e^-t(1.25t²-5t+2.5+B)

plug back into original equation

e^-t(1.25t²-5t+2.5+B) - 2e^-t(2.5t-1.25t²-B) - 3e^-t(1.25t²+B) =-5te^-t

e^-t divides out

(1.25t²-5t+2.5+B) - 2(2.5t-1.25t²-B) - 3(1.25t²+B) = -5t

t²(1.25+2.5-3.75)+t(-5-5)+(2.5+B+2B-3B)=-5t

The B's cancel out and I can't solve the equation. I have a feeling I am using the wrong equation for g(t). Can anyone give me some suggestions?