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Thread: Missing Constants in Derivative Equation

  1. #1
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    Missing Constants in Derivative Equation

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

    Here is what I have done so far:

    $\displaystyle y' = 2Ax + B$

    $\displaystyle y'' = 2A$

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

    But now I don't know where to go next. Can anybody help, please? Thanks.
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  2. #2
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    Quote Originally Posted by joatmon View Post
    Find the constants A, B, and C such that the function $\displaystyle y = Ax^2 + Bx + C$ satisfies the differential equation $\displaystyle y'' + y' -2y = x^2$.

    Here is what I have done so far:

    $\displaystyle y' = 2Ax + B$

    $\displaystyle y'' = 2A$

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

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

    But now I don't know where to go next. Can anybody help, please? Thanks.
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    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.
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  4. #4
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    Plugging into the ODE gives

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

    Collecting all of the powers of x gives

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

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

    $\displaystyle \begin{array}{c c c c} -2A & & &=1 \\ 2A & +2B & &=0 \\ 2A&+B&-2C&=0
    \end{array}$
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  5. #5
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    Whenever you have and equation in "x", you have an infinite number of equations!

    While setting coeffients in $\displaystyle -2Ax^2+(2A-2B)x+(2A+B-2C)=1x^2+0x+0$ is probablt simples, another way would be to subtract $\displaystyle x^2$ from both sides of the equation to get $\displaystyle (-2A- 1)x^2+(2A-2B)x+(2A+B-2C)= 0$ 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 $\displaystyle 2A(1 + x) = x^2 + 2(Ax^2 + Bx + C) - B$ if x= 0, you have $\displaystyle 2A= 2C- B$, with x= 1, $\displaystyle 4= 1+ 2(A+ B+ C)- B$, and with x= -1, $\displaystyle 0= 1+ 2(A- B+ C)- B$. That also gives you three equations to solve for A, B, and C.
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