What is the solution for 16y^{''}-8y^'+145y=0,y(0)=1 ?

The solution for 16y^{''}-8y^'+145y=0,y(0)=1 is y=e^{t/4}(cos(3t)+c_{2}sin(3t))

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16y^{''}-8y^'+145y=0,y(0)=1

16 y^{^{''}} - 8 y^{^{'}} + 145 y = 0, y (0) = 1

y = e^{\frac{t}{4}} (cos (3 t) + c_{2} sin (3 t))

Solution steps

16 y^{''} (t) - 8 y^{'} (t) + 145 y = 0

Solve linear ODE:

y = e^{\frac{t}{4}} (cos (3 t) + c_{2} sin (3 t))

y = e^{\frac{t}{4}} (cos (3 t) + c_{2} sin (3 t))

16 \cdot y^{^{''}} + 24 \cdot y^{^{'}} + 73 \cdot y = 0, y (0) = [4]

64 y^{^{''}} - 81 y = 0

y^{^{''}} + 4 y = 0, y (3 π) = - 3, y (0) = - 3

y^{^{''}} - y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for 16y^{''}-8y^'+145y=0,y(0)=1 ?
The solution for 16y^{''}-8y^'+145y=0,y(0)=1 is y=e^{t/4}(cos(3t)+c_{2}sin(3t))