## MathSofiya Group Title $y_P(x)=Ax^2+Bx+C$ $y'_P=2Ax+B$ $y''_P=2A$ $ay''+by'+cy=Gx$ $2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2$ I understand all of the above except the $...=x^2$ part. Why is G(x)=x^2? 2 years ago 2 years ago

1. across

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.

2. MathSofiya

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

3. MathSofiya

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?

4. MathSofiya

@across

5. across

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.