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Math Help - 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 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.
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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 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.
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  3. #3
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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

    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<br />
\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 -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.
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