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

The solution for 16x^{''}+8x^'+145x=0,x(0)=-2,x^'(0)=1 is x=e^{-t/4}(-2cos(3t)+1/6 sin(3t))

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

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

x = e^{- \frac{t}{4}} (- 2 cos (3 t) + \frac{1}{6} sin (3 t))

Solution steps

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

Solve linear ODE:

x = e^{- \frac{t}{4}} (- 2 cos (3 t) + \frac{1}{6} sin (3 t))

x = e^{- \frac{t}{4}} (- 2 cos (3 t) + \frac{1}{6} sin (3 t))

y^{^{''}} + 14 y^{^{'}} + 50 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 5, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 13 \frac{d y}{d x} + 12 y = 0

\frac{3}{4} x^{^{''}} + 72 x = 0, x (0) = \frac{1}{3}, x^{^{'}} (0) = 0

y^{^{''}} - 2 y + y^{^{'}} = 0

What is the solution for 16x^{''}+8x^'+145x=0,x(0)=-2,x^'(0)=1 ?
The solution for 16x^{''}+8x^'+145x=0,x(0)=-2,x^'(0)=1 is x=e^{-t/4}(-2cos(3t)+1/6 sin(3t))