# Thread: Second Order Non-Homogenous Differential Equations

1. ## Second Order Non-Homogenous Differential Equations

I have 3 non-homogeneous differential equations. The first i have done myself and believe to be correct, but can someone just confirm. The second and third i just need some guidance on what form the particular solution will take and the rest i will try to do myself:

1) y" - y' - 12y = e^6x here i used yp = ce^6x and got the general solution as yh = Ae^-3x + Be^4x + e^6x/18

2) y" - y' -6y = e^-x + 12x and yp = ???? y(0)=1, y'(0)=-2

3) y" + 2y' + 5y = cos3x and yp = ???? y(0)=1, y'(0)=0

2. ## Re: Second Order Non-Homogenous Differential Equations

2) I would try

$y_{p}=Ae^{-x}+Bx+C.$

3) Try

$y_{p}=A\cos(3x)+B\sin(3x).$

3. ## Re: Second Order Non-Homogenous Differential Equations

Originally Posted by NFS1
I have 3 non-homogeneous differential equations. The first i have done myself and believe to be correct, but can someone just confirm. The second and third i just need some guidance on what form the particular solution will take and the rest i will try to do myself:

1) y" - y' - 12y = e^6x here i used yp = ce^6x and got the general solution as yh = Ae^-3x + Be^4x + e^6x/18

2) y" - y' -6y = e^-x + 12x and yp = ???? y(0)=1, y'(0)=-2

3) y" + 2y' + 5y = cos3x and yp = ???? y(0)=1, y'(0)=0
In 1) you seem forgotten to add a constant. ({e^6}/{144}).

Give us your solutions for all the three, and we will try to help more.

4. ## Re: Second Order Non-Homogenous Differential Equations

surely is the constant not e^6x/18 as

>>yp=ce^6x so y'p=6e^6x and y"p=36e^6x
>>sub into the diff. equ. to get (36c - 6c - 12c)e^6x=e^6x
>> so 18c=1 hence c=1/18 therefore yh = Ae^-3x + Be^4x + e^6x/18

so how is it e^6x/144??

5. ## Re: Second Order Non-Homogenous Differential Equations

Your solution to the first problem is correct, I believe, though I would encourage you to use parentheses around the arguments of functions. That is, write it as y = Ae^-3x + Be^4x + e^(6x)/18, or better yet use LaTeX.

6. ## Re: Second Order Non-Homogenous Differential Equations

Originally Posted by Ackbeet
2) I would try

$y_{p}=Ae^{-x}+Bx+C.$

3) Try

$y_{p}=A\cos(3x)+B\sin(3x).$
2) so >> yp=Ae^-x +Bx + C
>>yp = ce^-x + 12Bx + c ???
>> y'p = -ce^-x + 12B and y"p= ce^-x

is the above correct???

7. ## Re: Second Order Non-Homogenous Differential Equations

You can take a look @ wolphram alpha if you want.

8. ## Re: Second Order Non-Homogenous Differential Equations

Originally Posted by NFS1
2) so >> yp=Ae^-x +Bx + C
>>yp = ce^-x + 12Bx + c ???
>> y'p = -ce^-x + 12B and y"p= ce^-x

is the above correct???
Not sure why you have the 12 in there, but your working is on the right track to getting the correct answer.

9. ## Re: Second Order Non-Homogenous Differential Equations

Originally Posted by Ackbeet
Not sure why you have the 12 in there, but your working is on the right track to getting the correct answer.
I thought as the rhs of the 2nd euation is e^-x + 12x so yp= ce^-x + 12b + c

10. ## Re: Second Order Non-Homogenous Differential Equations

Originally Posted by NFS1
I thought as the rhs of the 2nd euation is e^-x + 12x so yp= ce^-x + 12b + c
Perhaps, but it's going to be absorbed into the b anyway. Incidentally, mathematics is case-sensitive. B is not equal to b, so I would advise paying more attention to having a consistent case for variables throughout your work.

11. ## Re: Second Order Non-Homogenous Differential Equations

ok so im having difficulty on solving the 2nd, below is what i have done:

diff equa. >> y" - y' -6y = e^-x + 12x
particular equa.>> yp = Ae^-x + Bx + C so y'p= -Ae^-x + B and y"p= Ae^-x
sub into diff eq >> Ae^-x -(-Ae^-x + B) - 6(Ae^-x + Bx + C) = e^-x + 12x
simplify >> -4e^-x - 6Bx - 6C - B = e^-x + 12x

how would i solve from here??

12. ## Re: Second Order Non-Homogenous Differential Equations

Originally Posted by NFS1
ok so im having difficulty on solving the 2nd, below is what i have done:

diff equa. >> y" - y' -6y = e^-x + 12x
particular equa.>> yp = Ae^-x + Bx + C so y'p= -Ae^-x + B and y"p= Ae^-x
sub into diff eq >> Ae^-x -(-Ae^-x + B) - 6(Ae^-x + Bx + C) = e^-x + 12x
simplify >> -4Ae^-x - 6Bx - 6C - B = e^-x + 12x

how would i solve from here??
It should be: -4Ae^-x - 6Bx - 6C - B = e^-x + 12x

Get rid of stuff on the right side.

(-4A-1)e^-x +(-6B - 12)x + (-6C - B) = 0

So, (-4A-1) = 0, (-6B - 12) = 0, (-6C - B) = 0 .

Solve for A, B, C.

13. ## Re: Second Order Non-Homogenous Differential Equations

Originally Posted by SammyS
It should be: -4Ae^-x - 6Bx - 6C - B = e^-x + 12x

Get rid of stuff on the right side.

(-4A-1)e^-x +(-6B - 12)x + (-6C - B) = 0

So, (-4A-1) = 0, (-6B - 12) = 0, (-6C - B) = 0 .

Solve for A, B, C.
ok i have A=-1/4 B==-2 C=-1/3
subbed into to get general solution as >> yh= Ae^3x + Be^-2x - 1/4e^-x - 2x - 1/3

subbed in initial conditions to get A= 19/12 and B=0
so initial value problem is yh = 19/12e^3x + 0e^-2x - 1/4e^-x - 2x - 1/3
simplified >>yh = 19/12e^3x- 1/4e^-x - 2x - 1/3

can sum1 confirm if that is correct??