How do I solve for the particular solution using the annihilator method?

[Gogeta007]

Homework Statement

1. y''' + y'' = 8x^2
2. y'' - 2y' - 3y = 4e^x - 9

Homework Equations

Annihilator for any polinomial D^n+1

annihilator for e functions = e^ax = (D-a)^n

The Attempt at a Solution

1. getting the complementary eq.
values of m = 0, 0, -1
y_c= c₁e^-x +c₂ + xc₃

then we multiply both sides by the annihilator
D^3<D^2(D+1)> = 8x^2<D^3>

we get y= c₁e^-x +c₂ + xc₃ + x^2c₄ + x^3c₅

and from the original eq. the x^2 term has a coefficient of 8
therefore c4 = 8
sice there is no x^3 term then c5 should be zero ?
and what about c4?

the answer on the back says the answer is:
c1+c2x+c3e^-x + 2/3 x^4 + 8/3x^3 + 8x^2

===========

2.values of m=3, 2

y_c=c₁e^3x + c₂e^-x
if we multiply by the annihilator alpha = 1 in this case therefore annihilator would be (D-1)
<D-1><D^2-2D-3>=4e^x-9 <D-1>
so we have
y_c=c₁e^3x + c₂e^-x + c₃e^x

how do I go from there?

thank you!

 

[Dustinsfl]

Use the wronskian and cramer's rule.

 

[vela]

Homework Statement

1. y''' + y'' = 8x^2
2. y'' - 2y' - 3y = 4e^x - 9

Homework Equations

Annihilator for any polinomial D^n+1

annihilator for e functions = e^ax = (D-a)^n

The Attempt at a Solution

1. getting the complementary eq.
values of m = 0, 0, -1
y_c= c₁e^-x +c₂ + xc₃

then we multiply both sides by the annihilator
D^3<D^2(D+1)> = 8x^2<D^3>

we get y= c₁e^-x +c₂ + xc₃ + x^2c₄ + x^3c₅

You should have one more power of x.

and from the original eq. the x^2 term has a coefficient of 8
therefore c4 = 8
sice there is no x^3 term then c5 should be zero ?
and what about c4?

the answer on the back says the answer is:
c1+c2x+c3e^-x + 2/3 x^4 + 8/3x^3 + 8x^2

You're thinking y=8x^2, but it's actually y'''+y''=8x^2. You need to plug your trial solution back into the differential equation and then solve for the coefficients.

 

[vela]

Homework Statement

2. y'' - 2y' - 3y = 4e^x - 9

2. values of m=3, 2

m=2?

y_c=c₁e^3x + c₂e^-x
if we multiply by the annihilator alpha = 1 in this case therefore annihilator would be (D-1)
<D-1><D^2-2D-3>=4e^x-9 <D-1>
so we have
y_c=c₁e^3x + c₂e^-x + c₃e^x

how do I go from there?!

You need a slightly different annihilator since

(D-1)(4 e^x-9) = 9

Again, once you get your trial solution, plug it back into the differential equation and then solve for the coefficients.

 

[Gogeta007]

Thank you all. I have one more

y'' + y' + y = xsinx

The annihilator for xsinx is (D²+1)² which would yield 4 answers that are imaginary (+i,-i,+i,-i)
when you do the particular solution, since the answers are repeating do you put an x before the constant?
so far i have that y_c=e^-x/2(c_1cos<(sqr3)/2> + c_2sin<(sqr3)/2>)
then the annihilator values for D
y_p= c_3cosx + c_4sinx + c_5cosx + c_6sinx
or is it
y_p= c_3cosx + c_4sinx + xc_5cosx + xc_6sinx

and if its the latter, to get the coefficients you have to take up to the third derivative. . .using product rules? (...im hoping not) or is there another "easier" way?thank you

 

[vela]

Your complementary solution is wrong. You have a third-order differential equation, so you should have three independent solutions. Yours only has two.

You need the second particular solution you wrote down because you need four linearly independent solutions. You multiply by powers of x to make the repeated solutions independent.

 

[Gogeta007]

my bad. . .it is double prime not triple.
*edited*
thanks for pointing that out

 

[vela]

my bad. . .it is double prime not triple. *edited*

Well, at least you only have to differentiate y_p twice.

The arguments of the cosine and sine in the complementary solution should have an x in it.