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Frequently Asked Questions (FAQ)
What is the solution for y^{''}+6y^'+25y=0,y(0)=0,y^'(0)=2 ?
- The solution for y^{''}+6y^'+25y=0,y(0)=0,y^'(0)=2 is y= 1/2 e^{-3t}sin(4t)