Solving Second order non homogeneous DE using variation of parameters?

Question

The process I was taught in class was as follows:

1. Transform the second order non homogeneous DE into a system of first order DE.

2. Solve the homogeneous part.

3. Assume the solution to the original equation is of the form X=ψ(t)*C(t)

4. Use ψ(t)*C'(t)=G(t) to solve for C'(t)

5. Use C'(t) to solve for C(t)

6. Plug C(t) into equation X=ψ(t)*C(t)

7. Get y= from the top row of the matrix X.

 

So following these steps:

Problem: y''-2y'-3y=-3t*e^(-t)

 

Changed into matrix form:

X'=|0 1|X+|0      |

    |3 2|     |-3te^-t|

Or written out:

x'₁=x₂

x'₂=3x₁+2x₂-3te^-t

 

Solving for the homogeneous part only I get

x₁=C₁*e^3t+C₂*e^-t

x₂=C₁*3e^3t-C₂*e^-t

 

Assuming that the solution is of the form X=ψ(t)*C(t) will just change every C into a C(t)

 

Then using ψ(t)*C'(t)=G(t)

|e^3t    e^-t| |C'₁(t)| = |0      |

|3e^3t -e^-t| |C'₂(t)| = |-3te^-t|

 

I used addition to eliminate C₂

 

e3t*C'₁(t)+e^-t*C'₂(t)=0

+3e^3t*C'₁(t)-e^-t*C'₂(t)=-3te^-t

 

4e^3t*C'₁(t)=-3te^-t

C'₁(t)=-3/4te^-4t

 

Plugging this back into equation one to get C'₂(t)

e^3t*-3/4*te^-4t+e^-tC'₂(t)=0

C'₂(t)=3/4t

 

Solving for C₁(t) and C₂(t):

C₁(t)=∫C'₁(t) dt = -3/4∫te^-4t

u=t            v=-1/4e^-4t

du=dt        dv=e^-4t dt

C₁(t)=-3/4[-1/4*te^-4t-∫-1/4*e^-4tdt

C₁(t)=3/16*te^-4t+3/64*e^-4t+k₁

 

C₂(t)=∫C'₁(t) dt = 3/4∫ t dt

C₂(t)=3/8 * t² + k₂

 

Plugging these back into X=ψ(t)*C(t)

X= |e^3t    e^-t|  |3/16*te^-4t+3/64*e^-4t+k₁|

     |3e^3t -e^-t|  |3/8*t²+k₂                              |

X= |3/16*te^-t+3/64*e^-t+e^3tk₁+3/8*t²e^-t+e^-tk₂|

     |9/16*te^-t+9/64*e^-t+3e^3tk₁-3/8*t²e^-t-e^-tk₂ |

 

But this gives me a final solution of y=k₁*e^3t+k₂e^-t+3/8*t²e^-t+3/16*te^-t+3/64e^-t

My book is telling me the correct answer should be y=k₁*e^3t+k₂e^-t+3/16*te^-t+3/8*t²e^-t

 

 

I'm having this same problem on several other problems. A couple are coming out with the correct answer. So I'm just not sure where I'm going wrong. 

Answer 1

Hello Allison,

 

Your solution is perfect.

I have gone through all steps start to end.

 

Your answer is CORRECT

 

Because ---

See last line of your solution----

 

k2*e^(-t)+ 3/64*e^(-t)= some other constant(book say - it is k2)-- ok no problem.

 

So your answer and book answer are perfectly matching -- no difference.

 

Surendra