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\[\large y_p=Ax^2+2Bx+C\]\[\large y_p^{'}=2Ax+2B\]

u need any help........

u know whats L[dy/dx] =... ?

Can't we use integrating factor to solve it?

what Callisto said probabely is the best bet :)

\[y'+p(x)y=q(x)\]
\[\alpha = exp[\int p(x)dx]\]\[y=\frac{1}{\alpha}\int \alpha [q(x)] dx\]

in your case p(x)=1, q(x) = x^2+2x

\(y=e^{-x}\int e^x(x^2+2x)dx , after \:\: simplification, y=x^2\)

ohh.....yes!