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

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16y^{''}-8y^'+y=0,y(0)=40,y^'(0)=-10

16 y^{^{''}} - 8 y^{^{'}} + y = 0, y (0) = 40, y^{^{'}} (0) = - 10

y = 40 e^{\frac{t}{4}} - 20 e^{\frac{t}{4}} t

Solution steps

16 y^{''} (t) - 8 y^{'} (t) + y = 0

Solve linear ODE:

y = 40 e^{\frac{t}{4}} - 20 e^{\frac{t}{4}} t

y = 40 e^{\frac{t}{4}} - 20 e^{\frac{t}{4}} t

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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