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Frequently Asked Questions (FAQ)
What is the solution for y^{''}+4y^'+5y=-10x+3e^{-x} ?
- The solution for y^{''}+4y^'+5y=-10x+3e^{-x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-2x+8/5+3/2 e^{-x}