YAMT: A quick math problem

[VTboy]

y"' - 3y" +3y' -y = e^x -x -16

Solve the DE.

I get

y=c1e^x + c2(xe^x) + c3(x^2e^x) + 1/6(x^3e^x) -13 +X

Does this look right you.

My yc was c1e^x + c2(xe^x) + c3(x^2e^x)

yp was Ax^3e^x + B +Cx

 

[BlueWeasel]

I haven't done a DE since I took the class in 1998......I hated it with a passion.

 

[VTboy]

Bump

 

[her209]

The answer is 3.

 

[bonkers325]

how did you find Yp? i just learned this stuff in diff.eq and i never did a practice problem yet. my Yc comes out the same as yours.

-bonk's lil bro

 

[VTboy]

To find yp I used the anihalator method.

y??? ? 3y? + 3y? ? y = e^x + - x +16
L=D^3 -3D^2 + 3D -1
L=(D-1)^3

Auxiliary Equation: (m-1)^3 =0

m=1 (multiplicity 3)

yc= c1e^x + c2xe^x + c3x^2e^x

L1=(D-1)D^2
L1L(y) = (D-1)^4D^2(y)

Auxiliary Equation: (m-1)^4m^2 =0
m=1 (multiplicity 4)
m=0 (multiplicity 2)

y= c1ex + c2xe^x + c3x^2e^x + c4x^3e^x + c5 + c6x

yp = c4x^3e^x + c5 + c6x

comes from the y= part with the yc part removed.

 

[bonkers325]

auxilary equation?

-edit

nvm i figured it out lol

 

[VTboy]

Another why to get yp is this

If g(x) is in the form
e^(ax) then yp=Ae^(ax)
k (constant) yp=A
Ax+b then yp= Ax+B

So our g(x) was e^x -x -16

For the e^x part we have Yp=Ae^x
For the -x part we have Yp=Bx+C ( I used B and C instead of A because I already used A in the e^x part)
For the -16 we can say Yp=D

We combine them by adding the terms together to get

yp=Ae^x + Bx + C + D since C and D are both constants we can just combine them to get yp=Ae^x + Bx + E

Since we already have constant times e^x in our yc we must multiply that by x untill we no longer have it in our yc. So we multiply it by x^3 since both a costant times xe^x and x^2e^x is in our yc.

So your final Yp=Ax^3e^x+Bx+E

 

[bonkers325]

in my notes my instructor said that we should at most use x^2. is that just for second order differentials? we're using 3rd order differential equation, so we can use x^3?

 

[VTboy]

Originally posted by: bonkers325
in my notes my instructor said that we should at most use x^2. is that just for second order differentials? we're using 3rd order differential equation, so we can use x^3?

I dont know my book and teacher make no mention of x^2 being the most you should use. You dont want a value in yp to also be in yc though.

 

[bonkers325]

Originally posted by: VTboy
Another why to get yp is this

If g(x) is in the form
e^(ax) then yp=Ae^(ax)
k (constant) yp=A
Ax+b then yp= Ax+B

So our g(x) was e^x -x -16

For the e^x part we have Yp=Ae^x
For the -x part we have Yp=Bx+C ( I used B and C instead of A because I already used A in the e^x part)
For the -16 we can say Yp=D

We combine them by adding the terms together to get

yp=Ae^x + Bx + C + D since C and D are both constants we can just combine them to get yp=Ae^x + Bx + E

Since we already have constant times e^x in our yc we must multiply that by x untill we no longer have it in our yc. So we multiply it by x^3 since both a costant times xe^x and x^2e^x is in our yc.

So your final Yp=Ax^3e^x+Bx+E

wouldnt Yp = Ax^3e^x + Bx^4 + Ex^3 ??
since you're multiplying x^3 into the equation

edit-

ohh you should break up Yp into several equations so u can use the summation of Yp... right?

 

[Pepsei]

Please stop posting.

 

[VTboy]

You only multiply the term that is in yc by x^3 not all of the terms.

 

[VTboy]

Originally posted by: Pepsei
Please stop posting.

I see plenty of people post math threads in this forum. I can post one if I want.

 

[bonkers325]

how did u solve for the coefficients

are you supposed to plug in Yp, Yp', Yp'', Yp''' into the original equation? im looking at the example from my notes, and this is how he solves for the coefficients

 

[VTboy]

To find the coefficients I find yp', yp", and yp"' and solve for
yp"'-3y"+3y'-y=e^x-x-16