Differential Equation- Undetermined Coefficient

[jrsweet]

Homework Statement

Find a particular solution to the differential equation
2 y'' - 1 y' - 1 y = -1 t^2 + 2 t + 3 e^{4 t} .

Homework Equations

The Attempt at a Solution

I have attempted this problem many times. I think I am having trouble assuming what the general form is.

assume yp=(At+B)^2+Ce^(4t)=(A^2)(t^2)+2ABt+(B^2)+Ce^(4t)
yp'=2(A^2)t + 2AB + 4Ce^(4t)
yp''=2(A^2)+16Ce^(4t)

So, 2yp''-yp'-yp= 4(A^2)+32Ce^(4t)-2(A^2)t-2AB-4Ce^(4t)-(A^2)(t^2)-2ABt-(B^2)-Ce^(4t)

So, 27Ce^(4t)=3e^(4t) => C=(1/9)
-(A^2)(t^2)=-t^2 => A=1
t(-2(A^2)+2AB) = 2 => B=2
4(A^2)-2AB-(B^2)=0 ... but it doesn't work.



I have tried this problem multiple times in multiple different ways this being my last attempt. It's for an online homework and I can't seem to get the right answer, although I feel like I do it right every time...

 

[Mark44]

Homework Statement

Find a particular solution to the differential equation
2 y'' - 1 y' - 1 y = -1 t^2 + 2 t + 3 e^{4 t} .

Homework Equations

The Attempt at a Solution

I have attempted this problem many times. I think I am having trouble assuming what the general form is.

assume yp=(At+B)^2+Ce^(4t)=(A^2)(t^2)+2ABt+(B^2)+Ce^(4t)

Try this instead for your particular solution:
y_p = A + Bt + Ct² + De^4t

yp'=2(A^2)t + 2AB + 4Ce^(4t)
yp''=2(A^2)+16Ce^(4t)

So, 2yp''-yp'-yp= 4(A^2)+32Ce^(4t)-2(A^2)t-2AB-4Ce^(4t)-(A^2)(t^2)-2ABt-(B^2)-Ce^(4t)

So, 27Ce^(4t)=3e^(4t) => C=(1/9)
-(A^2)(t^2)=-t^2 => A=1
t(-2(A^2)+2AB) = 2 => B=2
4(A^2)-2AB-(B^2)=0 ... but it doesn't work.

I have tried this problem multiple times in multiple different ways this being my last attempt. It's for an online homework and I can't seem to get the right answer, although I feel like I do it right every time...

For your general solution, you'll need the particular solution + the solution to the homogeneous problem 2y'' - y' - y = 0. The solution to the homogeneous problem will be y_h = c₁e^r₁t + c₂e^r₂t for some constants r₁ and r₂ that you need to find.

 

[jrsweet]

I tried that as well:

yp=A+Bt+Ct^2+De^(4t)
yp'=B+2Ct+4De^(4t)
yp''=2C+16De^(4t)

2yp''-yp'-yp= 4C+32De^(4t)-B-2Ct-4De^(4t)-A-Bt-Ct^2-De^(4t)
= 27De^(4t)-Ct^2-2Ct-Bt-A-B-4C

So, 27De^(4t)=3e^(4t)
D= (1/9)

-Ct^2=-t^2
C=1

-2Ct-Bt=2t
-2(1)-B=2
B=-4

-A-B-4C=0
-A+4-4(1)=0
A=0

SOOoooo... yp=t^2-4t+(1/9)e^(3t)

What is wrong with thisss?! I don't get it. It's getting frustrating.

Someone shed some light if you could.

 

[HallsofIvy]

I tried that as well:

yp=A+Bt+Ct^2+De^(4t)
yp'=B+2Ct+4De^(4t)
yp''=2C+16De^(4t)

2yp''-yp'-yp= 4C+32De^(4t)-B-2Ct-4De^(4t)-A-Bt-Ct^2-De^(4t)
= 27De^(4t)-Ct^2-2Ct-Bt-A-B-4C

This should be "+ 4C".

So, 27De^(4t)=3e^(4t)
D= (1/9)

-Ct^2=-t^2
C=1

-2Ct-Bt=2t
-2(1)-B=2
B=-4

-A-B-4C=0
-A+4-4(1)=0
A=0

-A- B+ 4C= 0
-A+ 4+ 4(1)= 0
A= 8

SOOoooo... yp=t^2-4t+(1/9)e^(3t)

What is wrong with thisss?! I don't get it. It's getting frustrating.

Someone shed some light if you could.

 

[jrsweet]

I already tried that answer as well... It's a no go. There must be a glitch in the system or something...

 

[Mark44]

Then
1. Make sure the problem you worked is the same one that was given.
2. If it is, check your particular solution. It should be true that 2y_p'' - y_p' - y = -t² + t + 3e^4t. If this equation is an identity, then you have found what the problem asked for. If the system says different, then you have a good case to make that it is wrong. Assuming that the system gives an answer, you should then be able to show that its particular solution is NOT a solution of the given differential equation.