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## 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}= c

_{1}e^-x +c

_{2}+ xc

_{3}

then we multiply both sides by the annihilator

D^3<D^2(D+1)> = 8x^2<D^3>

we get y= c

_{1}e^-x +c

_{2}+ xc

_{3}+ x^2c

_{4}+ x^3c

_{5}

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}=c

_{1}e^3x + c

_{2}e^-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}=c

_{1}e^3x + c

_{2}e^-x + c

_{3}e^x

how do I go from there?

thank you!