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

Printable View

- Apr 9th 2011, 11:56 PMsalohcinseahhave solution but i cannot understand
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 - Apr 10th 2011, 12:02 AMProve It
If $\displaystyle \displaystyle y=Ax^2 + Bx + C$ and $\displaystyle \displaystyle y'' = 2A$, then what's $\displaystyle \displaystyle y'' + 9y$?

- Apr 10th 2011, 12:10 AMsalohcinseah
2A+9Ax^2 +9Bx+9C=x^2

- Apr 10th 2011, 01:01 AMProve It
- Apr 10th 2011, 01:09 AMsalohcinseah
yes it make more sense now thank you so much

- Apr 10th 2011, 01:23 AMtonio

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