What is the solution for 2500y^{''}-81y=0,y(0)=9,y^'(0)=6 ?

The solution for 2500y^{''}-81y=0,y(0)=9,y^'(0)=6 is y=(9+73/6)e^{(9t)/(50)}-73/6 e^{-(9t)/(50)}

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2500y^{''}-81y=0,y(0)=9,y^'(0)=6

2500 y^{^{''}} - 81 y = 0, y (0) = 9, y^{^{'}} (0) = 6

y = (9 + \frac{73}{6}) e^{\frac{9 t}{50}} - \frac{73}{6} e^{- \frac{9 t}{50}}

Solution steps

2500 y^{''} (t) - 81 y = 0

Solve linear ODE:

y = (9 + \frac{73}{6}) e^{\frac{9 t}{50}} - \frac{73}{6} e^{- \frac{9 t}{50}}

y = (9 + \frac{73}{6}) e^{\frac{9 t}{50}} - \frac{73}{6} e^{- \frac{9 t}{50}}

y^{^{''}} + 36 y = 0, y (0) = 5, y^{^{'}} (0) = 12

y^{^{''}} + 3 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0, y (0) = 1

y^{^{''''}} - 3 y^{^{''}} - 4 y = 0

y^{^{''}} + 0.5 y^{^{'}} + y = 0

What is the solution for 2500y^{''}-81y=0,y(0)=9,y^'(0)=6 ?
The solution for 2500y^{''}-81y=0,y(0)=9,y^'(0)=6 is y=(9+73/6)e^{(9t)/(50)}-73/6 e^{-(9t)/(50)}