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Frequently Asked Questions (FAQ)
What is the solution for x^{''}+4x^'+8x=e^{-2t},x(0)=1,x^'(0)=-2 ?
- The solution for x^{''}+4x^'+8x=e^{-2t},x(0)=1,x^'(0)=-2 is x= 3/4 e^{-2t}cos(2t)+1/4 e^{-2t}