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Frequently Asked Questions (FAQ)
What is the solution for 25y^{''}-20y^'+4y=0,y(5)=0,y^'(0)=-e^2 ?
- The solution for 25y^{''}-20y^'+4y=0,y(5)=0,y^'(0)=-e^2 is y=-5e^{2+(2t)/5}+e^{2+(2t)/5}t