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Frequently Asked Questions (FAQ)
What is the solution for (1/4)y^{''}+y^'+y=x^2-2x ?
- The solution for (1/4)y^{''}+y^'+y=x^2-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-4x+7/2