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Frequently Asked Questions (FAQ)
What is the solution for 16y^{''}-8y^'+y=0,y(0)=40,y^'(0)=-10 ?
- The solution for 16y^{''}-8y^'+y=0,y(0)=40,y^'(0)=-10 is y=40e^{t/4}-20e^{t/4}t