What is the solution for 16y^{''}-8y^'+145y=0,y(0)=-2,y^'(0)=1 ?

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

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16y^{''}-8y^'+145y=0,y(0)=-2,y^'(0)=1

16 y^{^{''}} - 8 y^{^{'}} + 145 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

y = e^{\frac{t}{4}} (- 2 cos (3 t) + \frac{1}{2} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{\frac{t}{4}} (- 2 cos (3 t) + \frac{1}{2} sin (3 t))

y = e^{\frac{t}{4}} (- 2 cos (3 t) + \frac{1}{2} sin (3 t))

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