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Frequently Asked Questions (FAQ)
What is the solution for y^{''}+6y^'+25y=100,y(0)=0,y^'(0)=0 ?
- The solution for y^{''}+6y^'+25y=100,y(0)=0,y^'(0)=0 is y=e^{-3t}(-4cos(4t)-3sin(4t))+4