# 2nd order DE inhomogeneous help!

• May 24th 2008, 06:37 AM
fuzzylogic25
2nd order DE inhomogeneous help!
hi, i am havin alot of trouble gettin the general solution to the ODE:

y'' - 7y' + 6y = 2(e^(3x)) + x(e^x)

i tried using the function for the particular solution as : C(e^3x) + (Ax+B)e^x but it doesnt work, like some of the constants cancel out and i cant get a proper solution!

how would i go about doin this question?
• May 24th 2008, 06:49 AM
mr fantastic
Quote:

Originally Posted by fuzzylogic25
hi, i am havin alot of trouble gettin the general solution to the ODE:

y'' - 7y' + 6y = 2(e^(3x)) + x(e^x)

i tried using the function for the particular solution as : C(e^3x) + (Ax+B)e^x but it doesnt work, like some of the constants cancel out and i cant get a proper solution!

how would i go about doin this question?

It should work. Eg. C = -1/3. Post your working.

Note: The same DE was posted by a different member at this thread: http://www.mathhelpforum.com/math-he...equations.html
• May 24th 2008, 07:42 AM
fuzzylogic25
yes i got c = -1/3 and also got A = -t/5 but i cant get a value for B.

i ended up with -5Ae^t = te^t

i had another expression (At+B)e^t but they all canceled out.

so now i have for y = (At+B)e^t = ((-t^2)/5 + B)e^t

what happens to the B? does it jus equal zero or something? im not sure?
• May 24th 2008, 07:54 AM
mr fantastic
Quote:

Originally Posted by fuzzylogic25
yes i got c = -1/3 and also got A = -t/5
[snip]

Do you mean A = -1/5?

Quote:

Originally Posted by fuzzylogic25
[snip] but i cant get a value for B.

i ended up with -5Ae^t = te^t

i had another expression (At+B)e^t but they all canceled out.

so now i have for y = (At+B)e^t = ((-t^2)/5 + B)e^t

what happens to the B? does it jus equal zero or something? im not sure?

Please don't just say what you did and what you ended up with. Show all your working (set out clearly and logically so that it's easy to follow).
• May 24th 2008, 08:11 AM
fuzzylogic25
ok sorry. i did mean A = -t/5 sorry i ended up using t's instead of x's coz of that post u linked me to.

well for y'' - 7y' + 6y = t(e^t) (i have taken out 2(e^(3t)) since i have no problem with that) - now im jus goin with t's

what i did was set y = (Ax+B)e^t
and then y' = (Ax+B)e^t + Ae^t
and then y'' = (Ax+B)e^t + 2Ae^t

so we have for y'' -7y' +6y :
(Ax+B)e^t + 2Ae^t - 7[(Ax+B)e^t + Ae^t] + 6[(Ax+B)e^t]

which when expanded is:
(Ax+B)e^t + 2Ae^t - 7(Ax+B)e^t - 7Ae^t + 6(Ax+B)e^t
which simplifies to:
-5Ae^t

so i equated the term to t(e^t) to get:
-5Ae^t = te^t
-5A = t
A = -t/5

and now i put that into y = (At+B)e^t = ((-t^2)/5 + B)e^t

so question is, what do i do with the B?

also i am sorry for writing it out with t's when i wrote the initial question in x's....
• May 24th 2008, 09:28 AM
galactus
Do you HAVE TO use undetermined coefficients?. Variation of Parameters is OK.

$y''-7y'+6y=2e^{3x}+xe^{x}$

By using the auxiliary equation $m^{2}-7m+6=(m-6)(m-1)$

We get $y_{c}=C_{1}e^{6x}+C_{2}e^{x}$

Compute the Wronskian:

$W=\begin{vmatrix}e^{6x}&e^{x}\\6e^{6x}&e^{x}\end{v matrix}=-5e^{7x}$

$W_{1}=\begin{vmatrix}0&e^{x}\\2e^{3x}+xe^{x}&e^{x} \end{vmatrix}=-e^{2x}\left(2e^{2x}+x\right)$

$W_{2}=\begin{vmatrix}e^{6x}&0\\6e^{6x}&2e^{3x}+xe^ {x}\end{vmatrix}=\left(2e^{2x}+x\right)e^{7x}$

$u'_{1}=\frac{W_{1}}{W}=\frac{e^{-5x}(2e^{2x}+x)}{5}$

$u'_{2}=\frac{W_{2}}{W}=\frac{-(2e^{2x}+x)}{5}$

Integrate The last two to get:

$u_{1}=\frac{-e^{-5x}(50e^{2x}+15x+3)}{375}$

$u_{2}=\frac{-(2e^{2x}+x^{2})}{10}$

$y_{p}=u_{1}e^{6x}+u_{2}e_{x}$

Now, after all that we can put it together.

$\boxed{y=y_{c}+y_{p}=C_{1}e^{6x}+C_{2}e^{x}-\frac{1}{3}e^{3x}-\frac{1}{10}x^{2}e^{x}-\frac{1}{25}xe^{x}-\frac{1}{125}e^{x}}$
• May 24th 2008, 02:39 PM
mr fantastic
Quote:

Originally Posted by fuzzylogic25
ok sorry. i did mean A = -t/5 sorry i ended up using t's instead of x's coz of that post u linked me to.

well for y'' - 7y' + 6y = t(e^t) (i have taken out 2(e^(3t)) since i have no problem with that) - now im jus goin with t's

what i did was set y = (At+B)e^t
and then y' = (At+B)e^t + Ae^t
and then y'' = (At+B)e^t + 2Ae^t

so we have for y'' -7y' +6y :
(At+B)e^t + 2Ae^t - 7[(At+B)e^t + Ae^t] + 6[(At+B)e^t]

which when expanded is:
(At+B)e^t + 2Ae^t - 7(At+B)e^t - 7Ae^t + 6(At+B)e^t
which simplifies to:
-5Ae^t

so i equated the term to t(e^t) to get:
-5Ae^t = te^t
-5A = t
A = -t/5

and now i put that into y = (At+B)e^t = ((-t^2)/5 + B)e^t

so question is, what do i do with the B?

also i am sorry for writing it out with t's when i wrote the initial question in x's....

My mistake. Clearly y = (At + B) e^t = At e^t + B e^t is redundant since B e^t is part of the homogenous solution.

So you try y = A t e^t. This doesn't work because you end up with -5 A e^t on the left and t e^t on the right. By definition A is a constant and so things aren't right.

So try y = (A t^2 + Bt) e^t as a particular solution .....
• May 24th 2008, 07:00 PM
fuzzylogic25
ahh ive tried y = (A t^2 + Bt) e^t as well and i jus get it down to:

2A-10At - 5B = t

I also tried y = (At^2 + Bt + C) e^t but get the same result as above. I tried this one too because from the answer galactus got there seems to be 3 terms for e^t.

I dont think iam doing it right (i did it same method as i showed u previously)...could someone show me how it is done using undetermined coefficients, i would really appreciate it coz im pretty lost now lol
• May 24th 2008, 07:08 PM
fuzzylogic25
ok wait sorry i jus realised what i did.

from 2A-10At - 5B = t
we equate -10At = t to get A = -1/10
and then 2A - 5B = 0
and then B = -1/25

so now we have for y= (At^2 + Bt)e^t = (-1/10t^2 -1/25t)e^t

so it works BUT why is my answer different to galactus???
i dont end up with a (-1/125e^t) term??
• May 24th 2008, 09:10 PM
mr fantastic
Quote:

Originally Posted by fuzzylogic25
ok wait sorry i jus realised what i did.

from 2A-10At - 5B = t
we equate -10At = t to get A = -1/10
and then 2A - 5B = 0
and then B = -1/25

so now we have for y= (At^2 + Bt)e^t = (-1/10t^2 -1/25t)e^t

so it works BUT why is my answer different to galactus???
i dont end up with a (-1/125e^t) term??

The $-\frac{1}{125} e^t$ term that galactus has can be put into the complementary solution:

$y_{c}= C_{1}e^{6t}+C_{2}e^{t} - \frac{1}{125} e^t = C_{1}e^{6t}+ \left( C_{2} - \frac{1}{125}\right) e^{t} = C_{1}e^{6t}+ C_{3} e^{t}$

where $C_3 = C_2 - \frac{1}{125}$ is just as arbitrary as $C_2$ .....