Question

Calc help i don't get the last part...

Answer 1

What is it that you do not understand?

In the last step, we are \(constructing\) the differential equation's solutions by inspection.

Answer 2

You have that \((-2A)x^2+(2A-2B)x+(2A+B-2C)=(1)x^2+(0)x+(0)\).

This implies that

\(-2A=1\),
\(2A-2B=0\), and
\(2A+B-2C=0\).

This is a system of three equations and three unknowns. Find them.

Answer 3

how??

Answer 4

That's an algebra 1 (or 2) problem:

\(-2A=1\implies A=-1/2\), etc.

Answer 5

ok.. and from there

Answer 6

that is kinda what i need explained

Answer 7

\( y'' + y' - y = x^2 \) is your given differential equation!!
and y is of the form \( y = Ax^2 + Bx + C \)

\( ax^2 + bx +c = 2x^2 + 3x + 1 \) Find the values of a,b,c ... <--- similar to it

Answer 8

ya?

Answer 9

As soon as you find \(A\), \(B\), and \(C\), you plug them into \(y=Ax^2+Bx+C\) to obtain your final result.

Answer 10

and once you plug them in, what do you solve for?

Answer 11

That's it; you're done.

Answer 12

but they get the value of A B and C on the very bottom

Answer 13

and What is that part (1)x^2 + (0)x + (0) about and how did they get that?

Answer 14

And?

Read the question. It says: "Find constants \(A\), \(B\), and \(C\) such that the function \(y=Ax^2+Bx+C\) satisfies this equation."

Answer 15

I would recommend you re-read the chapter.

Answer 16

yea ... first value of A,
then value of A+B ---> gives value of B
then value of A+B+C --> gives value of C

Answer 17

y = Ax^2 + Bx + C

y' = 2Ax + B

y'' = 2A

so putting this in the equation:

2A + 2Ax + B - 2(Ax^2 + Bx + C) = x^2


-2Ax^2 +2(A-B)x +2A + B - 2C = x^2

so now compare coefficients of x^2 , x and x^0 (1)

Answer 18

one last thing, why does A = B?

Answer 19

because comparing the coefficients of x

2(A-B)x = 0x

2(A-B) = 0