# Thread: have solution but i cannot understand

1. ## have 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?

2. If $\displaystyle y=Ax^2 + Bx + C$ and $\displaystyle y'' = 2A$, then what's $\displaystyle y'' + 9y$?

3. 2A+9Ax^2 +9Bx+9C=x^2

4. Originally Posted by salohcinseah
2A+9Ax^2 +9Bx+9C=x^2
Yes, now rewrite the LHS and RHS as

$\displaystyle 9Ax^2 + 9Bx + 2A + 9C = x^2 + 0x + 0$.

Does it make sense that $\displaystyle 9A = 1, 9B = 0$ and $\displaystyle 2A + 9C = 0$?

5. yes it make more sense now thank you so much

6. Originally Posted by salohcinseah
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?

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 $9Ax^2+9Bx+(9C+2A)=x^2$ , 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, $9A=1\iff A=1/9$

-- the lineal coef. in both sides must be equal, $9B=0\iff B=0$

-- free coef's must be equal, $9C+2A=0\iff 9C+2/9=0\iff C=-2/81$

Tonio