hi
y"+9 y = x^2
let yp= Ax^2 +Bx + c
y'=2Ax+B
y"=2A
thus giving you 2A+9(Ax^2 +Bx+C)=x^2
here where i am confused
Equate X^2 : 9A
thus A = 1/9
Equate x :B
9B=0
B=0
someone mind explaining it to me in layman term?
Thanks in advance
hi
y"+9 y = x^2
let yp= Ax^2 +Bx + c
y'=2Ax+B
y"=2A
thus giving you 2A+9(Ax^2 +Bx+C)=x^2
here where i am confused
Equate X^2 : 9A
thus A = 1/9
Equate x :B
9B=0
B=0
someone mind explaining it to me in layman term?
Thanks in advance
I don't fully understand: the above is a differential equation so you must have some
track in mathematics already walked (at the very least some basic courses in calculus
and linear algebra), so how much layman terminology do you need?
Anyway, you've a polynomial identity $\displaystyle 9Ax^2+9Bx+(9C+2A)=x^2$ , and we must
know that two pol's (whose coefficients are taken from the same field) are equal iff
coefficients of the corresponding same powers of x are equal, thus:
-- quadratic coefficient in both sides of the above equality must be the same, $\displaystyle 9A=1\iff A=1/9$
-- the lineal coef. in both sides must be equal, $\displaystyle 9B=0\iff B=0$
-- free coef's must be equal, $\displaystyle 9C+2A=0\iff 9C+2/9=0\iff C=-2/81$
Tonio