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Frequently Asked Questions (FAQ)
What is the solution for y^{''}+6y+8=0,y(0)=0,y^'(0)=12 ?
- The solution for y^{''}+6y+8=0,y(0)=0,y^'(0)=12 is y= 4/3 cos(sqrt(6)t)+2sqrt(6)sin(sqrt(6)t)-4/3