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Frequently Asked Questions (FAQ)
What is the solution for 25y^{''}-40y^'+16y=0,y(0)=2,y^'(0)=-3 ?
- The solution for 25y^{''}-40y^'+16y=0,y(0)=2,y^'(0)=-3 is y=2e^{(4t)/5}-(23e^{(4t)/5}t)/(5)