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Frequently Asked Questions (FAQ)
What is the solution for y^{''}+2y^'+26y=0,y(0)=4,y^'(0)=-2 ?
- The solution for y^{''}+2y^'+26y=0,y(0)=4,y^'(0)=-2 is y=e^{-t}(4cos(5t)+2/5 sin(5t))