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