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\[y_P(x)=Ax^2+Bx+C\] \[y'_P=2Ax+B\] \[y''_P=2A\] \[ay''+by'+cy=Gx\] \[2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2\] I understand all of the above except the \[...=x^2\] part. Why is G(x)=x^2?

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You're just posting the particular solution to the ODE. Can you post the ODE itself so that we can give it a better look? I mean, you obtained some strange coefficients I didn't see until you decided to operate on the derivatives linearly.
oh I'm sorry I left one part out \[y''+y'-2y=x^2\] soo sorry....I guess I'm just tired, cant read my calc book =P
I have another question, which I'll probably figure out as soon as I post it :) I understand that: 2A=1 but why is: 2Ax+B=0 and Ax^2+Bx+C=0? should't they equal 1 as well?

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You're looking at the solution process incorrectly. Let \(y''+y'-2y=x^2\). Since this is an inhomogeneous ODE with source term \(x^2\), the particular solution is assumed to have the form \(y_p=Ax^2+Bx+C\). Differentiating as you did\[\begin{align}y_p'&=2Ax+B\\y_p''&=2A\end{align}\]and combining linearly yields\[\begin{align}(2A)+(2Ax+B)-2(Ax^2+Bx+C)&=x^2\\2A+2Ax+B-2Ax^2-2Bx-2C&=x^2\\(-2A)x^2+(2A-2B)x+(2A+B-2C)&=(1)x^2+(0)x+(0)\end{align}\]Observe that the \(-2A\) term on the LHS corresponds to the \(x^2\) on the RHS, while the \(2A-2B\) and \(2A+B-2C\) terms both correspond to \(0\). Hence, this implies that \(-2A=1\). I'm sure you can wrap the rest up from here.

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