Finding Constants for a Differential Equation

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The discussion focuses on finding constants A, B, and C for the function y = Ax^2 + Bx + C to satisfy the differential equation y'' + y' - 2y = x^2. Participants analyze the derivatives of y, leading to the equation 2A + 2Ax + B - 2(Ax^2 + Bx + C) = x^2. They suggest simplifying this to identify relationships between the coefficients, specifically noting that -2A must equal 1 to satisfy the right-hand side of the equation. After further discussion, the values A = -1/2, B = -1/2, and C = -3/4 are determined as the solution. The conversation highlights the importance of equating coefficients to solve differential equations effectively.
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Find constants A,B and C such that the function:
y = Ax^2 + Bx + C

satisfies the differential equation:
y'' + y' - 2y = x^2


\frac{d}{dx} (y) = \frac{d}{dx} (Ax^2 + Bx + C) = 2Ax + B
y' = 2Ax + B

\frac{d}{dx} (y') = \frac{d}{dx} (2Ax + B) = 2A
y'' = 2A

2A + 2Ax + B - 2y = x^2

I have been assigned a problem that is not yet covered for another 7 chapters.

I do not understand the question...

Any suggestions?

 
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Why didn't you continue with the substitution:
2A+2Ax+B-2Ax^{2}-2Bx-2C=x^{2}
 
arildno is right. Here is a hint.

2A+2Ax+B-2Ax^{2}-2Bx-2C=x^{2} can be cleaned up to

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

Do you see any terms x terms on the RHS of the equation? What does this tells you about 2A-2B? More over what should -2A equal to, so it can satisfy the RHS of the equation? Apply the same idea for the terms 2(A-C)+B
 
You should regard what Arildno wrote as what it should really be,viz. an identity

(2A-2C+B)+(2A-2B)x-2Ax^{2}\equiv x^{2}

Daniel.
 
A = -\frac{1}{2}

B = -\frac{1}{2}

C = -\frac{3}{4}
 
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