# Math Help - Missing Constants in Derivative Equation

1. ## Missing Constants in Derivative Equation

Find the constants A, B, and C such that the function $y = Ax^2 + Bx + C$ satisfies the differential equation $y'' + y' -2y = x^2$.

Here is what I have done so far:

$y' = 2Ax + B$

$y'' = 2A$

$2A + 2Ax + B - 2y = x^2$
$2A(1 + x) = x^2 + 2y - B$

But now I don't know where to go next. Can anybody help, please? Thanks.

2. Originally Posted by joatmon
Find the constants A, B, and C such that the function $y = Ax^2 + Bx + C$ satisfies the differential equation $y'' + y' -2y = x^2$.

Here is what I have done so far:

$y' = 2Ax + B$

$y'' = 2A$

$2A + 2Ax + B - 2y = x^2$
Do you remember that $y = Ax^2 + Bx + C$? Replace the "y" in that formula by that.

$2A(1 + x) = x^2 + 2y - B$

But now I don't know where to go next. Can anybody help, please? Thanks.

3. Yes, but when I attempt to set these as an equation, I can never get the variables to eliminate. I always have three variables, but only 2 equations. I am totally spinning my wheels and need more help. Thanks.

4. Plugging into the ODE gives

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

Collecting all of the powers of x gives

$-2Ax^2+(2A-2B)x+(2A+B-2C)=1x^2+0x+0$

For this equation to be true the coefficients on the polynomials must be equal this gives the system of equations

$\begin{array}{c c c c} -2A & & &=1 \\ 2A & +2B & &=0 \\ 2A&+B&-2C&=0
\end{array}$

5. Whenever you have and equation in "x", you have an infinite number of equations!

While setting coeffients in $-2Ax^2+(2A-2B)x+(2A+B-2C)=1x^2+0x+0$ is probablt simples, another way would be to subtract $x^2$ from both sides of the equation to get $(-2A- 1)x^2+(2A-2B)x+(2A+B-2C)= 0$ and use the fact that the only way a polynomial can be identically 0 (0 for all x) is if the coefficients are all 0, giving the same equations.

The most basic way, and the reason I said "an infnite number of equations", is to just put whatever values of x you want into the equation:
With $2A(1 + x) = x^2 + 2(Ax^2 + Bx + C) - B$ if x= 0, you have $2A= 2C- B$, with x= 1, $4= 1+ 2(A+ B+ C)- B$, and with x= -1, $0= 1+ 2(A- B+ C)- B$. That also gives you three equations to solve for A, B, and C.