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